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Lakeshkumarpadhy

The document contains solutions to several math problems involving arithmetic progressions (APs) and geometric progressions (GPs). It summarizes key information about various AP and GP sequences, including their first terms, common differences/ratios, number of terms, and calculated sums. It also shows sample calculations for finding terms, differences, and sums of numeric AP and GP sequences.

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3 90 ⇒ β4 − β2 − 39β2 + 39 = 0 ⇒ (β2 − 1) (β2 − 39) = 0 Since, numbers are integers. ∴ β2 ≠ 39 ∴ β2 − 1 = 0 ⇒ β = ± 1 Hence, the four integers are 3, 5, 7 and 9. = 3 n2 − 3(n − 1)2 = 6 n − 3 On putting n = 1, 2, 3, ..., the sequence is 3, 9, 15, 21, ... which is clearly an AP. 27. The first two digit number which when divided by 4 leaves remainder 1 is 4⋅ 3 + 1 = 13 and last is 4 24⋅ + 1 = 97. Thus, we have to find the sum of the series 13 + 17 + 21 + + 97 which is an AP. ∴ 97 = 13 + (n − 1 4) ⇒ n = 22 n and Sn = [a + l ] = 1113[+ 97] 2 = 11 × 110 = 1210 28. The least and the greatest numbers of three digits divisible by 7 are 105 and 994, respectively. So, it is required to find the sum of the series 105 + 112 + 119 + + 994 Here, a = 105, d = 7, an = 994 ⇒ n − 1 = ∴ n − 1 = 127 ⇒ n = 127 + 1 = 128 ∴ Sum =[2a + (n − 1)d ] =[2 × 105 + (128 − 1) × 7] = 64(210 + 889) = 64 × 1099 = 70336 29. The odd numbers of four digits which are divisible by 9 are 1017, 1035, …, 9999. These are in AP with common difference 18. a = 1017, d = 18and l = 9999 ∴ nth term, an = a + (n − 1)d ⇒ 9999 = 1017 + (n − 1) × 18 ⇒ 18 n = 9999 − 999 = 9000 ⇒ n = 500 ∴ Sn = (a1 + an ) =(1017 + 9999) = 250 × 11016 = 2754000 3 30. Here, a = 10 and d = − Then, tn = 10 + (n − 1) 7 Now, tn is positive, if 3 10 + (n − 1) 0 7 ⇒ 70 − 3(n − 1) ≥ 0 ⇒ 73 ≥ 3 n ⇒ n So, first 24 terms are positive. Now, sum of the positive terms, 24 3 S24 = 2 × 10 + 23 7 2 69 852 = 12 20 − 7 =7 31. Let nth term be the first negative term. Then, nth term, tn < 0 ∴ 40 + (n − 1) (−2) < 0 ⇒ 42 − 2n < 0 ⇒ 2n > 42 ⇒ n > 21 ∴ Least value of n = 22 Hence, first 21 terms of an AP are non-negative. Since, sum will be maximum, if no negative terms is taken. ∴ Maximum sum, S 26. Given, Sn = 3 n2 ∴ Tn = Sn − Sn−1 Now, an = a + (n − 1)d ⇒ 994 = 105 + (n − 1) × 7 ⇒ 994 − 105 = 7(n − 1) ⇒ 889 = 7 (n − 1)
3 91 32. Let the first instalment be a and common difference of an AP be d. Given, 3600 = Sum of 40 terms ⇒ 3600 = [2a + (40 − 1)d ] ⇒ 3600 = 20 [2a + 39d ] ⇒ 180 = 2a + 39d …(i) After 30 instalments, one-third of the debt is unpaid. Hence, = 1200 is unpaid and 2400 is paid. Now, 2400 = [2a + (30 − 1)d ] ⇒ 160 = 2a + 29d …(ii) On subtracting Eq. (ii) from Eq. (i), we get 20 = 10 d ∴ d = 2 From Eq. (i), 180 = 2a + 39 2⋅ ⇒ 2a = 180 − 78 = 102 ∴ a = 51 Now, the value of 8th instalment = a + (8 − 1)d = 51 + 7 ⋅2 = ` 65 33. Let a be the length of the smallest side and d be the common difference. Here, n = 25 and S25 = 2100 Now, Sn =[2a + (n − 1)d ] ⇒ 2100 =[2a + (25 − 1)d ] ⇒ a + 12d = 84 ...(i) The largest side = 25th side = a + (25 − 1)d = a + 24d ∴ a + 24d = 20a [given] ...(ii) On solving Eqs. (i) and (ii), we get a = 8 , d = 6 n 34. Sn = [2a + (n − 1)d ] 2 Here, a = 1st term = 7 yr, d = 3 months = yr Sn = 250 yr n 1 ⇒ 250 = 2 2 × 7 + (n − 1) × 4 n n + 55 ⇒ 250 = 2 4 ⇒ 2000 = n2 + 55 n ⇒ n2 + 55 n − 2000 = 0 ⇒ (n − 25) (n + 80) = 0 ⇒ n = 25 ∴ Number of members in the club = 25 n 35. Sn = [2a + (n − 1)d ] 2 n ⇒ 60100 = [4 + (n − 1)(3)] 2 n ⇒ [3 n + 1] = 60100 2 ⇒ n = 200 36. Tn = Sn − Sn−1 = [3 n2 + 5 n] − [3(n − 1)2 + 5(n − 1)] = [3 n2 + 5 n] − [3(n2 − 2n + 1) + 5(n − 1)] = 6 n + 2 Let Tn = 164 Then, 164 = 6n + 2 ⇒ 6 n = 162 ⇒ n = 27 Hence, 164 is 27th term. 37. Here, a = 1, l = 11and Sn = 36 n  Sn n ∴ n ⇒ n = 6 38. Given series is 2 + 2 2 + 3 2 + 4 2 + ... = 2 [1 + 2 + 3 + 4 + upto n terms] n n( + 1) n n( + 1) = 2 (∑ n) = 2 2 ∑ n = 2 n n(+ 1) ∴ Sn = 1 1
3 92 2n + 1 − 1 = 2 40. Two digit odd numbers are 11, 13, ..., 99. These clearly form an AP with a = 11, d = 2 and n = 45. ∴ Required sum n 45 = (a + l ) = (11 + 99) = 2475 2 2 41. Sp − Sq = 2a p(− q) + (p2 − q2 − p + q d) = 2(q − p) ⇒ 2a + (p + q − 1)d = − 2 ∴ Sp+q = [2a + (p + q − 1)d ] = = − (p + q) 42. Given, Sn = 3 n2 − n If tn is the nth term of an AP, then tn = Sn − Sn−1 = (3 n2 − n) − [3(n − 1)2 − (n − 1)] = 6 n − 4 ∴ t1 = 6 − 4 = 2 n 43. We have, Sn = [2 p + (n − 1)Q] 2 By definition, d = Q 44. Given, Sn = 5 n2 + 2n ⇒ S1 = 5 + 2 = 7 and S2 = 5 4() + 2 2() = 24 ∴ T2 = S2 − S1 = 24 − 7 = 17 n 45. Given, tn = + y x 1 2 ⇒ t1 = + y and t2 = + y x x 1 ⇒ d = t2 − t1 = x where, d is common difference of given AP. r ∴ Sr = [2a1 + (r − 1)d ] 2 r r(+ 1) 1 = 2 x + ry a1 = x + y 46. Sn+ 3 − 3Sn+ 2 + 3Sn+1 − Sn = (Sn+ 3 − Sn+ 2 ) − 2(Sn+ 2 − Sn+1) + (Sn+1 − Sn ) = Tn+ 3 − 2Tn+ 2 + Tn+1 = (Tn+ 3 − Tn+ 2 ) − (Tn+ 2 − Tn+1) = d − d = 0 47. a + be y = b + ceyy = c + deyy y a − be b − ce c − de Applying componendo and dividendo rule, we get b c d = = a b c ⇒ a b c, , and d are in GP. 48. Given, Tm+ n = p and Tm n− = q ⇒ ar m+ n−1 = p …(i) and ar m n− −1 = q …(ii) On multiplying Eqs. (i) and (ii), we get a r2 2m−2 = pq ⇒ (ar m−1 2) = pq ⇒ ar m−1 = pq ⇒ Tm = pq 49. Since,1/2 cosec2θ, 2 cot θ and sec θ are in GP. ∴ ⇒ ⇒ 50. Let a and r be the first term and the common ratio respectively of GP. 51. Let A and R be the first term and the common ratio respectively of given GP. Then, T4 = p ⇒ AR3 = p T7 = q ⇒ AR6 = q and T10 = r ⇒ AR9 = r Now, (AR3 ) (AR9 ) = A R2 12 = (AR6 2) ⇒ pr = q2 Then, T10 = 9 ⇒ ar 9 = 9 and T4 = 4 ⇒ ar 3 = 4 ∴ T 7 = ar 6 = (ar 9 ⋅ ar 3 1 2) /= (9 × 4)1 2/ = 6 + − +
3 93 52. Let a and r be the first term and the common ratio respectively of GP. x1 y1 1 1 1 1 =x r1 y r1r r 1= 0 x r1 2 y r1 2 1 r 2 r 2 1 So, points are collinear, i.e. lie on a straight line. 54. Let ax = by = c z = k ⇒ a = k1/ x, b = k1/ y and c = k1/ z Since, a b, and c are in GP. ∴ b2 = ac ⇒ (k1/ y )2 = k1/ xk1/ z ⇒ k 2/ y = k1/ x + 1/ z ⇒ 2 = 1 + 1 y x z 1 1 1 ⇒ , and are in AP. x y z ⇒ x y, and z are in HP. 55. Let d be the common difference of an AP and R (≠ 0)be the common ratio of a GP. Then, q = p + d, r = p + 2d and y = xR, z = xR2  q − r = −d, r − p = 2d and p − q = −d ∴ xq−r ⋅ yr −p ⋅ zp − q = x− d ⋅(xR)2d ⋅(xR2 )− d = (x− d ⋅ x2d ⋅ x−d ) (R2d ⋅ R−2d ) = (x− d + 2d −d ) (⋅ R2d −2d ) = x0 ⋅ R0 = 1 × 1 = 1 56. Let the three numbers in GP be a, ar and ar 2. ∴ a + ar + ar 2 = 56 [given]…(i) On subtracting 1, 7, 21 from the numbers, we get a − 1, ar − 7 , ar 2 − 21, which are given to be in AP. ∴ (ar − 7) − (a − 1) = (ar 2 − 21) − (ar − 7) ⇒ ar − a − 6 = ar 2 − ar − 14 ⇒ a − 2ar + ar 2 = 8 …(ii) On subtracting Eq. (ii) from Eq. (i), we get 16 3 ar = 48 ⇒ a = …(iii) r 16 On substituting a = in Eq. (i), we get r 16 + 16 + 16 r = 56 r ⇒ 16 r 2 − 40r + 16 = 0 ⇒ 2r 2 − 5 r + 2 = 0 ⇒ (r − 2) (2r − 1) = 0 ∴ r = 2, If r = 2, then from Eq. (iii), a = 8 and the numbers are 8, 16, 32. If r = , then from Eq. (iii), a = 32 and the numbers are 32, 16, 8. 57. Let the sides of the right angled triangle be a, ar and ar 2, out of which ar 2 is the hypotenuse, then r > 1. Now, a r24 = a2 + a r2 2 ⇒ r 4 − r 2 − 1 = 0 ⇒ r 2 = 1 ± 5 2  r > 1 ∴ r 2 > 1 ⇒ r 2 = 1 + 5 2 Let ∠C be the greater acute angle. a 1 2 Now, cos C = ar 2 = r 2 = 1 + 5 58. Let the three digits be a, ar and ar 2. According to the hypothesis, 100a + 10ar + ar 2 − 792 = 100ar 2 + 10ar + a ⇒ a(1 − r 2 ) = 8 ...(i) and a, ar + 2, ar 2 are in AP. ∴ 2(ar + 2) = a + ar 2 ⇒ a r( 2 − 2r + 1) = 4 ...(ii) On dividing Eq. (i) by Eq. (ii), we get a(1 − r 2 ) 8 = a r( 2 − 2r + 1) 4 Then, a6 = 32 ⇒ ar 5 = 32 …(i) and a8 = 128 ⇒ ar 7 = 128 On dividing Eq. (ii) by Eq. (i), we get r 2 = 4 ⇒ r = ± 2 53. Let r be the common ratio of a GP, then Area formed by three points …(ii) =
3 94 (1 + r )(1 − r ) r + 1 ⇒ 2 = 2 ⇒ 1 − r = 2 (r − 1) ∴ r = 1/ 3 From Eq. (i), a = 9 Thus, digits are 9, 3, 1 and so the required number is 931. 59. Given a, b and c are in AP. ⇒ 2b = a + c …(i) Also, given a + 1, b, c are in GP and a, b, c + 2 are also in GP. b2 = (a + 1)c…(ii) and b2 = a c( + 2) …(iii) On solving Eqs. (i), (ii) and (iii), we get a = 8, b = 12 and c = 16 Hence, b = 12 60. The given series is a GP with a = 0.4. r = 40 a ∴ S = S∞ = ∴ =− x < 1, for x > 0] S 62. We have, = 255 [r = 2] r − 1 2 − 1 ⇒ 256 − a = 255 ⇒ a = 1 (x + 2)n − (x + 1)n 63. Clearly, (x + 2) − (x + 1) = (x + 2)n − 1 + (x + 2)n − 2(x + 1) + (x + 2)n − 3(x + 1)2 + … + (x + 1)n − 1 ∴Required sum = (x + 2)n − (x + 1)n [(x + 2) − (x + 1) = 1] 64. Here, c = ar, e = ar 2, d = bs, f = bs2 a b 1 1 2 c d 1 ∴ Area of triangle = e f 1 = ab s(− r )(s − 1)(r − 1) 65. Let S denotes the sum of all terms and S1 denotes the sum of odd terms. a S1 = (i.e. sum of odd terms) 1 1 Given, S = 5 ⋅S1 ⇒ 1 − r 5 1 − r 2 ⇒ ⇒ r 66. The given series is a GP with a = 2, r = 3 > 1 a r( 10 − 1) 2[( 3)10 − 1] ∴ S10 = = r − 1 3 − 1 2 = (242)( 3 + 1) = 121( 6 + 2 ) 2 n = 1 = = 212 .16 68. Let a and r be the first term and the common ratio respectively of given infinite GP. Then, ar = 2 …(i) a Also,= 8 …(ii) 2 From Eq. (i), r = a On putting the value of r in Eq. (ii), we get a = 4 69. Let a and r be the first term and the common ratio respectively of given GP. Then, n n =1 a(1 − r n ) 1 − r 99 61. The given series is a GP with 1 − r = x = − x x x = − =− − − − x x x − = + = + − −  − − = ⇒ − − = ⋅ −
3 95 ⇒ Sn = 1 − r ⇒ β = 2 1 − r ⇒ r 70. Given, f x( + y) = f x f y( ) ( ), ∀ x, y ∈N For any x ∈N, f x() = [ ( )]f 1x = 3x n ∴ ∑ f x() = 120 x = 1 n ⇒ ∑ 3x = 120 x = 1 ⇒ 31 + 32 + 33 + + 3n = 120 ⇒ 3n − 1 = 80 [f(1) = 3] ⇒ 3n = 81 = 34 ⇒ n = 4 71. 0.423 = 0.4 + 0.023 + 0.00023 +  72. Given, x = 1 + y + y2 +  1 1 ⇒ x = ⇒ 1 − y = 1 − y x 1 x − 1 ⇒ y = 1 − = x x 73. Let the GP be a, ar, ar 2 and ar 3. Then, a (1 + r + r 2 + r 3 ) [given] ⇒ a = 2 74. Sum of areas of all the squares 2 a2 a2 75. Let a be the first term and r (| r | < 1)be the common ratio of infinite GP. Then, a a2 = 3 and 2 = 3 …(i) 1 − r Also, 2 = 9 …(ii) (1 − r ) × = − − = − = + − + = − + ⇒ − + = − + = + + +  = − = −
3 + 96 = ∀ ≥ On solving Eqs. (i) and (ii), we get 3 a = and r = 2 Hence, the sequence corresponding to the series will 3 3 3 be , , ,  2 4 8 1 1 1 76. Given product = x 2 4 8 = x 2 = x 77.Length of a side ofSn =Length of a diagonal ofSn + 1 ⇒ Length of a side ofSn = 2(Length of a side ofSn + 1) Length of a side ofSn + 1 1 ⇒ ⇒ Side of S1,S2, ,Sn form a GP with common ratio and first term 10. n − 1 1 ∴ Side of Sn = 10 2 10 = n − 1 2 2 ⇒ Area of Sn = (Sn )2 = 100n − 1 2 Area ofSn < 1 [given] ⇒ ⇒ 78. Let r be the common ratio of the given GP Then, terms of GP are α α α, r r 2 and αr 3. According to the question, α(1 + r ) = 1, α 2r = p, α r 2 (1 + r ) = 4, α 2r 5 = q On solving, we get (p q, ) = −( 2, − 32) 79. Consider, (1 + x) + (1 + x + x2 ) + (1 + x + x2 + x3 ) + upto n terms 1 − x2 1 − x3 1 − x4 = + + + upto n terms 1 − x 1 − x 1 − x 1 = [(1 + 1 + 1 + to n terms) 1 − x − (x2 + x3 + x4 + upto n terms)] 1 x2(1 − xn ) = n − 1 − x 1 − x 80. We have, 9 + 99 + 999 + upto n terms = (10 − 1) + (102 − 1) + (103 − 1) + upto n terms = (10 + 102 + 103 + upto n terms) − n =− n =− n = − 9n − 10) 81. Let Sn = 0.5 0.55 + 0.555 + upto n terms upto n terms] 82. Let the three digits be a, ar and ar 2. Given, a + ar 2 = 2ar + 1 ⇒ a r( 2 − 2r + 1) = 1 ⇒ a r(− 1)2 = 1 …(i) Also, given a + ar = (ar + ar 2 ) ⇒ 3 a(1 + r ) = 2ra (1 + r ) ⇒ (1 + r ) (3 − 2r ) = 0 [a ≠ 0] 3 ∴ r = , − 1 2 1 1 If r =, then from Eq. (i), a = 2 = 2 = 4 (r − 1) 3 − 1 2 ⇒ n − 1 ≥ 7 ⇒ n ≥ 8 + + + − − = − + − + +  + + + − + + = +   − = ⋅ − − = − × − − − = − − − +
+ 3 97 If r = − 1, then from Eq. (i), a = , which is not possible, as a is an integer. Hence, a and ar 83. Let the two numbers be a and b, then G = ab or G2 = ab Also, p and q are two AM’s between a and b. ∴ a, p, q and b are in AP. Now, p − a = q − p and q − p = b − q ∴ a = 2 p − q and b = 2q − p Hence, G2 = ab = (2 p − q) (2q − p) 84. Let the 3 n terms of a GP be a, ar, ar 2, …, ar n − 1, ar n, ar n + 1, ar n + 2, …, ar 2n − 1, ar 2n, ar 2n + 1, ar 2n + 2,…, ar 3n − 1. Then, S1 = a + ar + ar 2 + + ar n − 1 a(1 − r n ) = 1 − r S2 = ar n + ar n + 1 + ar n + 2 + + ar 2n − 1 ar n(1 − r n ) = 1 − r S3 = ar 2n + ar 2n + 1 + ar 2n + 2 + + ar 3n − 1 2n n ar (1 − r ) = 1 − r Now, (S2 )2 = a r2 2n a(1 − r n ) 2n (1 − r n ) = ⋅ ar = S S1 3 1 − r 1 − r Hence, S1, S2 and S3 are in GP. 1 85. (32)(32)1 6/ (32)1 36/ = 321 + 61 + 361 + = 32 1−1 6 = 326/ 5 = (25 6) / 5 = 26 = 64 4 On subtracting, we get S = 1 + ++ +  7 5 35 ⇒ 4 4 16 87. Let S = 1 + 2 x + 3x2 + 4x3 + x S = x + 2 x2 + 3x3 +  On subtracting, we get (1 − x S) = 1 + x + x2 + x3 +  1 1 ⇒ (1− x S)= 1 − x ⇒S = ( 1 − x)2 88. Let S = 1 + 2 ⋅2 + 3 2⋅ 2 + 4 2⋅ 3 + + 100 ⋅299 2S = 1 2⋅ + 2 ⋅22 + 3 2⋅ 3 + 4 2⋅ 4 + + 99 2⋅99 + 100 ⋅2100 On subtracting, we get − S = 1 + (1 2⋅+ 1 2⋅2 + 1 2⋅3 +to 99 terms − 100 ⋅2100 ) = 1 + 2 2(99 − 1) − 100 ⋅2100 ⇒ S = 99 2⋅ 100 + 1 n2(n + 1)2 89. Here, Tn n n ⇒ S k 4 k = 1 91. We can write S as S = (1 − 2)(1 + 2) + (3 − 4)(3 + 4) + + (2001 − 2002)(2001 + 2002) + 20032 = − [1 + 2 + 3 + 4 + + 2002] + 20032 = − (2002)(2003) + (2003)2 = 2007006 92. According to the given condition, 1 1984 2() (− 1)n (704)(2) 1 − 2n = 3 1 − 2n 86. + + + + =  ⇒ + + + =  = 90. = + ⋅ = + + Σ = + = − + ∴ − = + = + = × =
3 + 98 2112 (−1)n ⇒ 128 = n − 1984 n 2 2 If n is odd, then we get 2n = 32 ⇒ n = 5 128 If n is even, then we get 128 = n ⇒ n = 0 2 93. We have, 2n + 10 = 2 ⋅22 3 2⋅ 3 + 4 2⋅ 4 + + n⋅2n ⇒ 2 2( n + 10 ) = 2 ⋅23 + 3 2⋅ 4 + + (n − 1)⋅2n + n⋅2n + 1 On subtracting, we get ⇒ ⇒ n = 513 94. Let S = (1)(2003) + (2)(2002) + (3)(2001) + + (2003)(1) and K = 12 + 22 + 32 + + 20032 On adding, we get S + K = (2004)[1 + 2 + 3 + + 2003] ⇒ (2003)(334)(x) + (2003)(4007)(334) = (2004)(1002)(2003) ⇒ x = 2005 95. We can write the given equation as 1 1 1 1 1 + + + + +  log 2 x 2 4 8 16 = 4 − (a b1 1 + a b22 + + a bm m)2 = (a b1 2 − a b2 1)2 + (a b1 3 − a b3 1)2 ++ (am − 1bm − a bm m − 1)2 Thus, − (x x1 2 + x x2 3 + + xn−1xn )2 ≤ 0 ⇒ (x x1 3 − x x2 2 )2 + (x x1 4 − x x3 3 )2 + + (xn − 2 xn − xn − 1xn − 1)2 ≤ 0 As x1, x2, …, xn are real, this is possible if and only if x x1 3 − x22 = x x2 4 − x32 = = xn − 2 xn − xn2 − 1 = 0 x1 = x2 = x3 = = xn ⇒ x2 x3 x4 xn − 1 ⇒ x1, x2, …, xn are in GP. 97. We have, r = = 1 = 1 98. We have, 2 2 = 4 2 = 1 + (2r + 1)(2r − 1) tan−1 21r 2 = tan−1 1(2+r(2r1+) 1) ((22rr −11)) = tan−1 (2r + 1) − tan−1 (2r − 1) n n ⇒ r∑= 1tan−1 2 1 r 2 = r∑= 1{ tan−1 (2r + 1) − tan−1 (2r − 1)} n ⇒ lim→ ∞ r∑= 1tan−1 1 2 = nlim→ ∞ tan−1 (2n + 1) − π 4 n 2r 99. 13 = 1⋅(1 − 1) + 1 23 = (2 ⋅1 + 1) + 5 , 33 = (3 2⋅+ 1) + 9 + 11 , 43 = (4⋅ 3 + 1) + 15 + 17 + 19, etc ∴ n3 = {n⋅(n − 1) + 1} + , in which next term being 2 more than the previous ∴ n3 = (n2 − n + 1) + (n2 − n + 3) + + (n2 + n − 1) 100. ∑n r2 − r − 1 = ∑n r − 1 − r = −n r = 1 (r + 1 )! r = 1 r! (r + 1)! (n + 1)! n ⇒ log2(x2 ) = 4 ⇒ x2 = 24 ⇒ 96. We shall make use of the identity (a12 + a22 + + am2 )(b12 + b22 + + bm2 ) x = 4
+ 3 99 101. Applying AM ≥ GM Since, AM =GM ∴ ( 102. As, D < 0 2 )x = ( 2 − 2 )x i.e. for x = 0 [AM = GM iff a = b] ⇒ (a + b + c )2 − 4(ab + bc + ca) < 0 ⇒ (a − b + c )2 < 4 ac ⇒ − 2 ac < a − b + c ⇒ ( a + c ) > b 1 103. Given , a and b are in GP. 16 ⇒ a2 = b …(i) 16 Also, a, b and are in HP. ⇒ b = …(ii) On solving Eqs. (i) and (ii), we get a and b , 1 104. Since, a b( − c ) + b c( − a) + c a( − b) = 0 ∴ x = 1is a root of the given equation. Since, both the roots are equal, therefore the other root is also 1. ∴Their sum = 1 + 1 = 2 −b c( − a) ⇒ = 2 a b( − c ) ⇒ −bc + ab = 2ab − 2ac ⇒ 2ac = b a( + c ) ∴ 2ac b = a + c Hence, a, b and c are in HP. 105. Let two numbers be a and b. Then, a + b = 5 ⇒ a + b = 10 Also, ab = 4 ⇒ ab = 16 2ab 2 16( ) 16 ∴HM between a and b = = = a + b 10 5 106. Given, log(a c ), log (c − a) and log (a − 2b + c ) are in AP. ⇒ 2 log(c − a) = log (a + c ) + log (a − 2b + c ) ⇒ (c − a)2 = (a + c ) (a − 2b + c ) ⇒ (c − a)2 = (a + c )2 − 2b a( + c ) ⇒ 2b a( + c ) = (a + c )2 − (c − a)2 = 4 ac ⇒ 2ac b = a + c ⇒ a b, and c are in HP. 107. Given, cos (x − y), cos x and cos (x + y) are in HP. 2 cos(x − y) cos (x + y) ⇒ cos x = cos (x − y) + cos (x + y) 2 2 2 (cos x − sin y) ⇒ cos x = 2 cos x cos y ⇒ cos2 x = 1 + cos y = 2 cos2 y 2 ⇒ cos2 x sec2 = 2 ⇒ cos x sec 108. log2 6 = log2(3 × 2) = log2 3 + log2 2 = 1+ log2 3 and log2 12 = log2(22 × 3) = log2 3 + 2 log2 2 = 2 + log2 3 Since, log2 3, 1 + log2 3, 2 + log2 3 are in AP. ⇒ log2 3, log2 6, log2 12 are in AP. ⇒ log3 2, log6 2, log12 2 are in HP. 109. Given a, b and c are in HP ⇒ b is HM of a and c a + c ⇒ > b 2 ⇒ a2 + c 2 > 2b2 110.AM> GM> HM + × × + = + ⇒ >  > + > + + =±
3 + 100 = ∴ x > y > z ⇒ 111. Given a, b and c are in HP. 1 1 1 ⇒ , , are in AP. a b c z < y < x ⇒ = + ⇒ Hence, the straight line = 0 passes through a b c a fixed point (1, − 2). 112. Given, a10 = 3 ⇒ a1 + 9d = 3 ⇒ 2 + 9d = 3 [a1 = 2 ] ⇒ d = 7 ∴ a4 = a1 + 3d = 2 + = 3 1 h10 = 3 ⇒ h10 3 − + = x + +
3 101 ⇒ D = − ∴ ⇒ ∴ a h4 7 = 6 3 7 113. Since, x, y and z are in HP. ∴ y = ⇒ x − 2 y + = x + z ⇒ (x + z) (x − 2 y + z) = (z − x)2 ⇒ log(x + z) + log(x − 2 y + z) = 2 log(z − x) 1 1 114. Let− = K Hi + 1 Hi ∴ (− 1) ∑2=n1i HHii +− HHii ++ 11 i 2n (− 1)i 1 1 = i∑= 1 Hi + 1 + Hi = 2n K 115. Let the given AP be a1, a1 + d, a1 + 2d, … By substituting the value, we can find that only options (b) and (d) are correct. 116. Let the AP be a, a + d, a + 2d, a + 3d,  Given that, 3 a + 3d = 9 ⇒ a + d = 3 and a2 + (a + d )2 + (a + 2d )2 = 35 ⇒ a2 + (3)2 + (3 + d )2 = 35 ⇒ (3 − d )2 + 32 + (3 + d )2 = 35 ⇒ d = ± 2 and a = 1, 5 n ∴ Sn = [2a + (n − 1) d ] 2 where d = 2, a = 1 ⇒ Sn = n2 When d = − 2, a = 5 , then Sn = n(6 − n) 117. cot−1 1a+2 −a a1 2a1 + cot−1 1a+3 −a a2a23 + cot−1 1a+4 −a a3a34 + + cot−1 1 a+n −a anann−−11 = cot−1 a1 − cot−1 a2 + cot−1 a2 − cot−1 a3 + + cot−1 an − 1 − cot−1 an = cot−1 a1 − cot−1 an = tan−1 an − tan−1 a1 118. If a, bandc are pth,qth and rth terms of an AP, then b − c a − b is always a rational number. 119. Let the roots be a1, a2, a3, a4, a5, then 1 1 1 1 1 + + + + = 10 a1 a2 a3 a4 a5 Σ a a a a1 2 3 4 = 10 ⇒ a a a a a1 2 3 4 5 ⇒ 10S = Σa a a a1 2 3 4 Since, a1, a2, a3, a4, a5 are in GP, then consider a1, a2, a3, a4 and a5 as a, ar, ar 2, ar 3, ar 4 ⇒ 10S = 10a r5 10 = a r36 (a + ar + ar 2 + ar 3 + ar 4 ) ⇒ 10a r5 10 = a r36(40) [sum of roots = 40] ⇒ a r24 = 4 ⇒ ar 2 = ± 2 ⇒ S = a r36(4) = ±(2)3 (4) = ± 32 120. Since, a1, a2, a3, …, an, … are in AP and b1, b2, … bn, …are in GP. ⇒ 1, (1 + d ) , …, [1 + (n − 1)d ] … are in AP ⇒ 1, r, r 2, …, r n − 1, … are in GP. Also, a9 = b9 ⇒ 1 + 8d = r 8 = + − = = = × = x + x x − = + + − + + x x = − x
3 102 ∴ S9 = [2 + (9 − 1)d ] = 369 ⇒ 2 + 8d = 41 × 2 ⇒ d = 10 ⇒ r 8 = 81 ⇒ r 8 ⇒ ∴6 = 9 × 3 = 27 and 121. Given, sin β = sin α cos α 4 or 2 cos2 π + α 4 122. Since, ais AM between 1st and (2n + 1)th terms, bis GM between 1st and (2n + 1) th terms and c is HM between 1st and (2n + 1) th terms. ⇒ AM ≥ GM ≥ HM and AM × HM = (GM)2 123. If each of x, y and z is less than 1, then Statement I is obviously true. Also,1 − 2 x + 1 − 2 y + 1 − 2 z = 3 − 2(x + y + z ) = 1, The sum of the three given numbers is positive also at must one of x, y and z can be more than 1. If one of x, y and z is more than or equal to , then their product is less than equal to zero, hence still remains true. Statement II is always true but it does not explain Statement I. 124. Since, ax2 + bx + c = 0 and a x1 2 + b x1 + c1 = 0 have common root. (a c1 − ac1)2 = 4(ab1 − ba1) (bc1 − b c1) …(i) a c11 On putting these values in Eq. (i), we get c a1 1 = b12 125. Statement I is false, since each term of the series +++ + is smaller than 10−5 but its 6 6 6 6 sum upto infinity is infinity. n Statement II is true, since lim is not finite as n→ ∞ 105 n → ∞ 126. Statement I can be proved by taking the intersection of the inequalities. a > 0, ar > 0, ar 2 > 0 at a + ar > ar 2, ar + ar 2 > a, ar 2 + a > ar The inequalities follow from reason. 127. If we could show that Statement II is true, then Statement I will be false. Indeed if Statement II is false, then 2 − 3 = (p − q d) ...(i) and 3 − 5 = (q − r d) ...(ii) On dividing Eq. (i) by Eq. (ii) we get 2 − 3 p − q = 3 − 5 q − r ⇒ Rational =Irrational Hence, Statement I is false and Statement II is true. 128. Statement I is true, since for any x > 0, we can choose xn sufficiently larger n such that is small. Statement II is n ! (n !)2 false, since contains n ! in the denominator but n ! ⇒ sin2 β = sin α cos α Now, cos 2β = 1 − 2 sin2 β = 1 − 2 sin α cos α = (sin α − cos α )2 = 2 sin2 − = − = − = = = = =
3 103 diverges to ∞. 129. We can show that Statement I is true and follows from Statement II. Indeed a1 + a2 ++ an + an + 1 a1 + a2 + + an Sn + 1 − S = − n n(+ 1) [an + 1 > a1, an + 1 > a2, , etc. ] Solutions (Q. Nos. 130-132) Given, a b(− c x)2 + b c( − a x)+ c a(− b) = 0 ...(i) Since, a b( − c ) + b c( − a) + c a( − b) = 0, therefore x = 1 is a root of Eq. (i). Let other root be α, then c a( − b) 1 × α = a b(− c ) 2ac c a − a + c == 1 2ac a − c a + c ∴ α = 1 Hence, both roots of Eq. (i) are 1, 1. Then, roots of x2 − Px + Q = 0 are also 1, 1. Now, 1 + 1 = P, 1 1⋅= Q ∴ P = 2, Q = 1 ∴ [P] = [2] = 2 Now, [2P − Q] = [4 − 1] = 3 Roots of Eq. (i) are 1, 1, then a X′ ,0) = 4 an ⋅ 2 x 32 an/2 an 3 0 an/2 = 8 an anan − 3 2 2 2 2 ⋅ 8 an = 4 2 an2 sq units = 3 2⋅ 2 3 134. We have, y2 = 4x and y2 = 4 − 4x ⇒ 8x = 4 ⇒ x and y = ± 135. For an = 1, An =sq units 3 136. a b ca b c = (aa a times) (bb b times) (cc c times) Applying AM-GM inequality, (a b ca b c )1/ n a + a + + a times) + (b + b + + b times) ≤ + (c + c + + c times) n a2 + b2 + c 2 = n Similarly, (a b cb c a )1/ n ≤ ab + bc + ca ≤ a2 + b2 + c 2 + = − + + + + +  = − + − + + − > + + +  =− + − − ⇒ − − =− 133. = + − ∫ ∫ X ′ − ⋅ − x −
3 104 Also, (a b cc a b)1/ n ≤ n n So, (a b ca b c )1/ n + (a b cb c a )1/ n + (a b cc a b)1/ n (a + b + c )2  and G = H + …(ii)  G2 = AH ⇒ G2 = G + 3 G − 6 [from Eqs. (i) and (ii)] 2 5 ∴ G = 6, A = [from Eq. (i)] Since, a and b are the roots of x2 − 15x + 36 = 0. On solving, we get a = 12, b = 3 or a = 3, b = 12 ∴ α = 15, β = 9 ⇒ α + β2 = 15 + 81 = 96 and α 2 + β = 225 + 9 = 234 B.  A = G + 2 …(i) and H b [if b > a] 2 bA  G = AH = 5 ⇒ 5 ab = bA ⇒ A = 5 a a + b ⇒ ⇒ ab + 2 From Eq. (i), ⇒ 5 a = 3 a + 2 ∴ a = 1and b = 9 ⇒ α = a + b = 10 and β =|a − b| = 8 ∴ α + β2 = 10 + 64 = 74 C.  H = 4 ⇒ 2 A + G2 = 27 ⇒ 2 A + AH = 27 [G2 = AH] ⇒ 6A = 27 ⇒ A and G Since, a and b are the roots of x2 − 9x + 18 = 0. ∴ a = 6, b = 3 or a = 3, b = 6 Now, α = a + b = 9, β =|a − b| = 3 ∴ α 2 + β = 81 + 3 = 84 2 3 4 5 and 1 = 1 + 3 + 32 + 33 + 34 + 35 = 364 138. Let us add one more number an+1 to the given sequence. The number an+1 is such that | an+1| =| an + 1|. On squaring all the numbers, we have a12 = 0 a22 = a12 + 2a1 + 1 a32 = a22 + 2a2 + 1 a42 = a32 + 2a3 + 1     an2 = an2−1 + 2an−1 + 1 an2+1 = an2 + 2an + 1 On adding the above equalities, we get a a a a = a12 + a22 + + an2 + 2(a1 + a2 + + an ) + n ⇒ a a a n a n a1 + a2 + + an ≥ − 1 = − λ ⇒ µ ≤ = n n Further, a < n ⇒ aa < an ∴ (a b ca b c ) < (abc )n⇒(a b ca b c )1/ n < (abc ) 3 137. A. A = G + …(i) + + ≤ + + = = =
3 105 139. Let abe the first term andd be the common difference of an AP. Again, let x, y, z be the (m + 1) th, (n + 1) th and (r + 1) th terms of an AP. Then, x = a + md, y = a + nd 140. Let f x( ) = x4 + ax3 + bx2 + cx + 1 As a b, and c are non-negative, no root of the equation f x( ) = 0 can be positive. Further as f(0) ≠ 0, all the roots of the equation, say x1, x2, x3 and x4 are negative. We have, ∑ x1 = − a, ∑ x x12 = b, ∑ x x x1 23 = −c and x x x x1 2 3 4 = 1 Using AM ≥ GM for positive numbers −x1, −x2, −x3 and −x4, we get a Using AM ≥ GM for positive numbers x x1 2, x x1 3, x x1 4, x x2 3, x x2 4 and x x3 4, we get b Finally, using AM ≥ GM for positive numbers −x x x1 23, −x x x124, −x x x1 34 and −x x x2 3 4, we get ≥⇒ c ≥ 4 ∴ HCF of {a b c, , } = HCF of {4, 6, 4} = 2 Entrances Gallery b c 1. = = (integer) a b a a 2b b2 6 ⇒ 1 − a + a2 = a b 2 6 Let the three consecutive terms be a − d, a, a + d, where d > 0 Then, a2 − 2ad + d 2 = 36 + K a2 = 300 + K and a2 + 2ad + d 2 = 596 + K On subtracting Eq. (i) from Eq. (ii) we get d(2a − d ) = 264 On subtracting Eq. (ii) from Eq. (iii), we get d (2a + d ) = 296 …(i) …(ii) …(iii) …(iv) …(v) Again, subtracting Eq. (iv) from Eq. (v), we get 2d 2 = 32 ⇒ d 2 = 16 ⇒ d = 4 [d = − 4, rejected] From Eq. (iv), 4(2a − 4) = 264 ⇒ 2a − 4 = 66 ⇒ 2a = 70 ⇒ a = 35 ∴ K = 352 − 300 = 1225 − 300 = 925 ⇒ K − 920 = 5 ⇒ b2 = ac ⇒ c = b2 a ...(i) Also, a + b + c = b + 2 3 ⇒ a + b + c = 3 b + 6 ⇒ a − 2b + c = 6 ⇒ 2 2 b b a − 2b + = 6 from Eq. (i), c = + = ∴ = + + + = = − + − ⇒ = + ⇒ = + = + − − = − − = − =− λ ⇒ λ=
3 106 − 1 = = cot ∑ cot−1 (1 + 2 + 4 + 6 + 8 + ... + 2n) n = 1 23 = cot ∑ cot−1 {1 + n n(+ 1)} n = 1 = cot n∑23= tan−1 1 + n n(1 + 1) 1 = cot ∑23 tan 1 1 n+ +n n(1 −+n1) − n = 1 = cot ∑23 [tan−1(n + 1) − tan−1 n] n = 1 = cot [(tan−1 2 − tan−1 1) + (tan−1 3 − tan−1 2) + (tan−1 4 − tan−1 3) + + (tan−1 24 − tan−1 23)] = cot{tan−1 24 − tan−1 1} −1 24 − 1 = cot tan−1 23 = cot tan + ⋅ 25 1 24 ( )1 = cot cot−1 25 23 25 = 23 k k( + 1) 4n 2 3. Sn = ∑ (− 1) ⋅ k 2 k = 1 = − (1)2 − 22 + 32 + 42 − 52 − 62 + 72 + 82 +  = (32 − 1 )2+ (42 − 22 ) + (72 − 52 ) + (82 − 62 ) +  = 2{(4 + 12 + 20 + ... ) + (6 + 14 + 22 + ... )}   n terms n terms n n = 2 2 {2 × 4 + (n − 1 8) } + 2 {2 × 6 + (n − 1 8) } = 2[n(4 + 4 n − 4) + n(6 + 4 n − 4)] = 2 4[ n2 + 4 n2 + 2n] = 4 n(4n + 1) Here, 1056 = 32 × 33, 1088 = 32 × 34, 1120 = 32 × 35, 1332 = 36 × 37 Hence, 1056 and 1332 are possible answers. 4. Here, a1 = 5, a20 = 25 for HP ∴ = 5 and ⇒ ⇒ ∴ a + (n − 1)d ⇒+ (n − 1)d < 0 1 ⇒ (n − 1) < 0 ⇒ (n − 1) > a a ⇒ a = 6 only 2 a + a − 14 ⇒ = 4 a + 1 23 n 2. cot n∑= 1cot−1 1 + k∑= 12k 23 + = + = = − =− = − ×  < ⇒ < −
3 107 5 19 × 25 ⇒ n > 1 + ⇒ n > 24.75 Hence, the least positive value of n is 25. 5. Given, a1 = 3, m = 5 n and a1, a2, ..., a100 are in AP. Sm = S5n is independent of n. Also, Sn Sn 5 n [2 × 3 + (5 n − 1)d ] Sm = 2 Consider Sn n [2 × 3 + (n − 1)d ] 2 5{(6 − d ) + 5 nd} = (6 − d ) + nd For independent of n, put 6 − d = 0 ⇒ d = 6 ∴ a2 = a1 + d = 3 + 6 = 9 Sm is independent of n. Also, if d = 0, then Sn ∴ a2 = 3 6. Using AM ≥ GM, a−5 + a−4 + a−3 + a−3 + a−3 + 1 + a8 + a10 ⇒ a−5 + a−4 + 3 a−3 + 1 + a8 + a10 ≥ 8 1⋅ Hence, the minimum value is 8. k − 1 7. We have, Sk = 1 − k ! 1 = (k − 1 1)! k Now, (k 2 − 3k + 1)Sk = {(k − 2)(k − 1) − 1} × 1 (k − 1)! 1 1 = − (k − 3)! (k − 1)! ⇒ |(kk )Sk| k = 1 1 1 = 1 + 1 + 2 − 99! 98! ⇒ 100 ! k = 1 8. Since, ak = 2ak − 1 − ak − 2 So, a1, a2, a11 are in AP. a d a d ⇒ 225 + 35d 2 + 150 d = 90 ⇒ 35d 2 + 150 d + 135 = 0 ⇒ d Given, a ∴ d = − 3 and d a1 + a2 + + a11 = 1 [30 − 10 × 3] = 0 ⇒ 11 2 9. Given, m is the AM of l and n. ∴ l + n = 2m …(i) and G1, G2, G3 are geometric means between l and n. ∴ l, G1, G2, G3, nare in GP. Let r be the common ratio of this GP. ∴ G1 = lr G2 = lr 2 G3 = lr 3 n = lr 4 n ⇒ r l Now, G1 4 + 2G2 4 + G3 4 = (lr )4 + 2 (lr 2 4)+ (lr 3 4) = l 4 × r 4(1 + 2r 4 + r 8 ) = l 4 × r 4(r 4 + 1) = l 4 × n n + l 2 l l = ln × 4 lm2 = 4 lm n2
3 108 − = = 10. Given, series is Let Tn be the nth term of the given series. 13 + 23 + 33 + ... n3 ∴ Tn = 1 + 3 + 5 + ... + to n terms n n(+ 1) 2 2 (n + 1)2 = n2 = 4 9 2 (n + 1) 1 2 2 2 2 2 9 ∑ ∴ S = = [(2 + 3 + … + 10) + 1 − 1 ] n = 1 4 4 1 10 10(+ 1)(20 + 1) 384 = − 1 = = 96 11. Given, α and β are roots of px2 + qx + r = 0, p ≠ 0. −q r ∴ α + β = ,αβ = ...(i) p p Since, p q, and r are in AP. ∴ 2q = p + r ...(ii) Also, + = 4 ⇒ ⇒ α + β = 4αβ [from Eq. (i)] p p ⇒ q = − 4r On putting the value of q in Eq. (ii), we get 2(− 4 )r= p + r ⇒ p = − 9r −q 4r 4 Now, α + β = = = = − p 9 r and αβ = p − 9r − 9 ∴ 2 ⇒⇒ |α − β| = 13 9 12. Given, k ⋅109 = 109 + 2 11( ) (1 10)8 + 3 11( ) (2 10)7 + + 10 11()9 11 11 2 11 9 ⇒ k = 1 + 2 3 + + 10 ...(i) 10 10 10 11 11 11 2 11 9 11 10 k = 1 + 2 + 9 10 10 10 10 10 10 ...(ii) On subtracting Eq. (ii) from Eq. (i), we get 11 11 11 2 11 9 11 10 k 1 − 1  10 10 10 10 10 10 11 10 1 10 − 1 ⇒ k 10 − 11 = − 10 11 10 10 11 10 1 10 n a r( − 1) [in GP, sum of n terms =, when r > 1] r − 1 11 10 11 10 ⇒ − k = 10 10 10 − 10 − 10 10 ∴ k = 100 13. Let a ar, and ar 2 be in GP (r > 1). On multiplying middle term by 2, then the numbers a, 2ar and ar 2 are in AP. 4 6 4 α β α β αβ + = ⇒ − =
3 109 = [(1 + 1 + upto 20 terms) 1 1 1 − 10 + 102 + 103 + upto 20 terms 1 − = 7 1 101 20 20 − 9 10 1 − 1 10 7 110 1 − 101 20 = 20 − × 9 10 9 = 7 20 − 109(1 − 10−20 ) = 81 7 (180 − 1 + 10−20 ) 9 = (179 + 10−20 ) 15. Here, T100 = a + (100 − 1)d = a + 99d T50 = a + (50 − 1)d = a + 49d T150 = a + (150 − 1)d = a + 149d Now, according to the given condition, 100 × T100 = 50 × T50 ⇒ 100(a + 99d ) = 50(a + 49d ) ⇒ 2(a + 99d ) = (a + 49d ) ⇒ a + 149d = 0 ⇒ T150 = 0 16. Statement I Let S = (1) + (1 + 2 + 4) + (4 + 6 + 9) +... + (361 + 380 + 400) S = (0 + 0 + 1) + (1 + 2 + 4) + (4 + 6 + 9) + + (361 + 380 + 400) Now, we can clearly observe the first elements in each bracket. In second bracket, the first element is 1 = 12 In third bracket, the first element is 4 = 22 In fourth bracket, the first element is 9 = 32 In last bracket, the first element is 361 = 192 Hence, we can conclude that there are 20 brackets in all. Also, in each of the bracket, there are 3 terms out of which the first and last terms are perfect squares of consecutive integers and the middle term is their product. Now, the general term of the series is Tr = (r − 1)2 + (r − 1)r + (r )2 So, the sum of n terms of the series is n Sn = ∑ {(r − 1)2 + (r − 1)r + (r ) }2 r = 1 n 3 3 r − (r − 1) r = 1 n Now, let Sn = ∑ {k 3 − (k − 1) }3 k = 1 On substituting the value of k and rearranging the terms, we get Sn = − 03 + (13 − 13 ) + + [(n − 1)3 − (n − 1)3 ] + n3 ⇒ Sn = n3 Since, the number of terms is 20, hence substituting n = 20, we get S20 = 8000 Statement II We have already proved in the ⇒ Sn = ∑ r = 1 r − (r − 1) ⇒ n Sn = ∑ {r 3 − (r − 1) }3 ⇒ + = ⇒ = + − ⇒ = ± − = ± ⇒ + =  14. + = + +  + + + =  − = − + − + + 
3 110 Statement I that n Sn = ∑ {k 3 − (k − 1) }3 = n3 k = 1 17. Let the time taken to save ` 11040 be (n + 3) months. For first 3 months, he save `s 200 each month. In (n + 3) months, his total savings is n 3 × 200 + [ (2 240) + (n − 1) × 40] = 11040 2 ⇒ 600 + 20n n(+ 11) = 11040 ⇒ 20n n(+ 11) = 10440 ⇒ n n(+ 11) = 522 ⇒ n2 + 11n − 522 = 0 ⇒ n2 + 29n − 18n − 522 = 0 ⇒ n n(+ 29) − 18(n + 29) = 0 ⇒ n = 18 or n = − 29 ∴ n = 18 [neglecting n = − 29 ] Hence, total time = n + 3 = 18 + 3 = 21months 18. Number of notes that the person counts in 10 min = 10 × 150 = 1500 We have, a10, a11, a12, ... are in AP with common difference − 2. Let n be the time taken to count remaining 3000 notes, then n [2 × 148 + (n − 1) × −(2)] = 3000 2 ⇒ n2 − 149 n + 3000 = 0 ⇒ (n − 24)(n − 125) = 0 ⇒ n = 24,125 Hence, total time taken by the person to count all notes 19. Let S ⇒ S ...(i) S − 1 2 6 10 14 ⇒ = 2 + 3 + 4 + 5 +  3 3 3 3 3 On subtracting Eq. (ii) from Eq. (i), we get ...(ii) S = 2 ⇒ + 20. Since, a + ar = a (1 + r ) = 12 …(i) and ar 2 + ar 3 = ar 2 (1 + r ) = 48 …(ii) From Eqs. (i) and (ii), we get r 2 = 4 ⇒ r = − 2 On putting the value of r in Eq. (i), we get a = − 12 21. Since, each term is equal to the sum of the next two terms. ∴ ar n − 1 = ar n + ar n + 1 ⇒ 1 = r + r 2 ⇒ r 2 + r − 1 = 0 5 − 1 − 5 − 1 r = 2 r ≠ 2 a1 + a2 + + ap p2 22. Since, a1 + a2 + + aq = q2 p ∴ 2q [[22aa11 ++ ((qp−−11))dd]] = qp22 2 (2a1 − d ) + pd = p ⇒ (2a1 − d ) + qd q ⇒ (2a1 − d ) (p − q) = 0 + 5d 2 23. Since, a1, a2, a3, ..., an are in HP. 1 1 1 1 + + + = − + ⇒ − + + = + + − = + = ⇒ =  ≠ = + + = + = =
3 111 ∴ , , , , are in AP. a1 a2 a3 an Let d be the common difference of AP. 1 1 ∴ − = d a2 a1 ⇒ a1 − a2 = a a d1 2 Similarly, a2 − a3 = a a d2 3    an − 1 − an = an − 1a dn On adding all the equations, we get a1 − an = d {a a1 2 + a a2 3 + ...+ an − 1an} …(i) 1 1 Also, = + (n − 1)d an a1 a1 − an ⇒ d = a a1 n (n − 1) On putting the value of d in Eq. (i), we get a1 − an {a a1 2 + a a23 + + an − 1 an} a1 − an = a a1 n(n − 1) ⇒ 24. Given that, x = ∑ an, y = ∑ bn, z = ∑ c n n = 0 n = 0 n = 0 1 ...(i) Similarly, y ⇒ x = 1 + a + a2 + = =…(ii) and z =...(iii) 1 − c Now, a b, and c are in AP. ⇒ − a, − band − c are in AP. ⇒ 1 − a, 1 − band 1 − c are also in AP. 1 1 1 ⇒ , and are in HP. 1 − a 1 − b 1 − c ⇒ x y z, , are in HP. Aliter From Eqs. (i), (ii) and (iii), 1 1 1 x = , y = and z = 1 − a 1 − b 1 − c x − 1 y − 1 z − 1 ⇒ a = , b = and c = x y z Since, a b, and c are in AP. ∴ 2b = a + c ⇒ 1 1 ⇒ Hence, x y, and z are in HP. 25. We know that, e x + e − x x2 x4 x6  putting x On in both sides, we get 1 26. Given, Tm = n 1 ⇒ a + (m − 1)d =...(i) n 1 and Tn = m 1 ⇒ a + (n − 1)d =...(ii) m On solving Eqs. (i) and (ii), we get 1 a = d = mn ∴ a − d = 0 27. Given that the sum of n terms of given series is n n( + 1)2 , if n is even. 2 Let n be odd, i.e. n = 2m + 1 Then, S2m + 1 = S2m + (2m + 1)th term (n − 1)n2 = + nth term 2 + + = + + −  ⇒ + + = ⋅ + ⋅ + ⋅ +  − − ⇒ x x − = − + − − = − + − x x = +
3 112 (n − 1) n2 2 = + n 2 = n2 n − 1 + 2 2 (n + 1) n2 = 2 28. We know that, 1 1 1 1 e = 1 + + + + + ... ...(i) 1! 2 ! 3! 4! and e − 1 = 1 − 1 + 1 − 1 + 1 − ... ...(ii) 1! 2 ! 3! 4! On adding Eqs. (i) and (ii), we get e e e 2 + 1 2 2 ⇒ 11 ⇒ + ... e 2 ! 4! ⇒ = + + ... 2 2 ! 4! 29. Consider, 1 1 1 1 1 1 − − − ... 2 2 3 3 4 1 1 1 = 2 1 − + − + ... − 1 2 3 4 e 30. Since, 1, log3 31 − x + 2 and log3 (4⋅ 3x − 1) are in AP. ∴3x = t] t ⇒ 12 t 2 − 5t − 3 = 0 ⇒ (3t + 1) (4t − 3) = 0 ⇒ t ⇒ x cannot be negative] 3 ⇒ log3 4 = x ⇒ x = 1 − log3 4 31. Let S = 21/ 4 ⋅ 41/ 8 ⋅ 81/16... = 21/ 4 ⋅ 22/ 8 ⋅ 23/16... 1 2 3 1 + + + ... = 2 4 2 2 2 = 2 where, S It is an infinite arithmetico-geometric progression. a d ⋅ r ∴ S1 = 1 − r + (1 − r )2 = 1 1 2 = + = 1 − 1 − 2 2 ∴ S 32. Since, 5th term of a GP = 2 ∴ ar 4 = 2 …(i) where, a and r are the first term and common ratio respectively of a GP. Now, required product = a × ar × ar 2 × ar 3 × ar 4 × ar 5 × ar 6 × ar 7 × ar 8 = a r936 = (ar 4 9) = 29 = 512 [from Eq. (i)] (loge x )n loge x (loge x)2 n = 0 n ! =eloge x = x 34. e e x 35. Vr = Sum of first r terms of an AP whose first term is r and common difference(2r − 1) r − = + + + − + =
3 113 r 2 = [2r + 2r − 3r + 1] 2 r 2 = [2r − r + 1] n n n n 1 3 2 1 36. Since, a b, and c are in GP. ∴ b2 = ac Given equation is (loge a x) 2 − (2 loge b x) + (loge c ) = 0 ...(i) On putting x = 1in Eq. (i), we get loge a − 2 loge b + loge c = 0 ⇒ 2 loge b = loge a + loge c ⇒ loge b2 = loge ac ⇒ b2 = ac, which is true. Hence, one of the roots of given equation is 1. Let another root be α. 2 loge b = loge b2 ∴ Sum of roots, 1 + α = loge a loge a ⇒ α = loge ac − 1 = (loge a + loge c ) − 1 loge a loge a = loge c = loga c loge a Hence, roots are 1 and loga c. = x + = 37. Given, f x( ) ∴ f(2 x) = 2 x + f(2 x) = and f(4x) = 4x + ⇒ f(4x) = Since, f x( ), f(2 x) and f(4x) are in HP. 1 1 1 ⇒ (2 x + 1) (8x + 1) = (5x + 1) (4x + 1) ⇒ 16x2 + 10 x + 1 = 20 x2 + 9x + 1 ⇒ 4x2 − x = 0 ⇒ x (4x − 1) = 0 ⇒ x = [x ≠ 0] Hence, one real value of x for which the three unequal terms are in HP. 38.AM ≥ GM ⇒ M ≤ 1 ⇒ M ≤ 1 Also, (p + q) (r + s) > 0 [p q r s, , , > 0] ∴ M > 0 Hence, 0 < M ≤ 1 39. Case I Let α = ω and β = ω2 n ∴ S = − n = 0 ω n = 0 = 1 − ω2 + ω4 − ω6 + ω8 −ω10 + ω12 + ... + ω600 − ω602 + ω604 = 1 − ω2 + ω − 1 + ω2 − ω + 1 + ... + 1 − ω2 + ω = + − + + + = + − + + − = + + + ∴ x x ⇒ x x = + ⇒ x x x + = + + + ⇒ x x x x + = + + + ∴ + + + ≥ + + ⇒ ≥ = = = = ∑ ∑∑∑ + − = = ⋅ + − + + + + = + − + + + + x + ⇒ x + x +
3 114 = 0 + ... + 1 − ω2 + ω = − ω2 − ω2 = − 2ω2 [1 + ω + ω2 = 0] Case II Let α = ω2 and β = ω 302 n 2 n 302 n ω4 n ω ∴ S = ∑ (− 1) ∑ (− 1) 3 n = 0 n = 0 n n = 0 = 1 − ω + ω2 − ω3 + ω4 − ω5 + ω6 − ... + ω300 − ω301 + ω302 = 1 − ω + ω2 − 1 + ω − ω2 + 1 − ... + 1 − ω + ω2 = 0 + ... + 1 + ω2 −ω = − ω − ω = − 2ω α 2 + β2 2 α 3 + β3 3 x − Now, α + β = p and αβ = q α 2 + β2 2 α 3 + β3 3 Hence, (α + β) x − x x +  2 3 = log (1 + px + qx2 ) 41. Given, 1 1 1 1 = 1 − 1 − 2 + 3 − 4 + 5 − ... = 1 − loge 2 42. We have, e x = (1 − x) (B0 + B x1 + B x2 2 + ... + B x + B xn n + .... ) By the expansion of e x, we get x x2 xn 1 + + + ... + + ... 1! 2 ! n ! = (1 − x) (B0 + B x1+ B x2 2 + ... + Bn − 1xn − 1 + B xn n + ... ) On equating the coefficient of xn both sides, we get 1 Bn − Bn − 1 = n ! ∞ 3n ∞ 3n − 2 x x 43. We have, a = ∑ , b = ∑ n = 0(3 n)! n = 1(3 n − 2)! and c 1 ∞ 3n ∞ 3n − 2 ∞ 3n − 1 x x x Now, a + b + c = ∑ + ∑ + ∑ (3 n)! (3 n − 2)! (3 n − 1)! n = 0 n = 1 n = 1 x2 x3 x e a b c x x and a + bω2 + cω = e ω2x, ω is an imaginary cube root of unity. Now, a3 + b3 + c 3 − 3 abc = (a + b + c ) (a + bω + cω ) (2a + bω2 + cω) =e x ⋅ e ωx ⋅ e ω2x = e x (1 + ω + ω2) = e 0⋅x = 1 x 3 2 7 3 15 4 44. Given, f x( ) = + x + x + x + ... 1! 2 ! 3! 4! + ... − + + + + ... 1! 2 ! 3! 4! 40. β α − + + + x x x  − + α αx x + β β β x x x − + − = +β + +α x x + + + = β α αβ x x
3 115 x x2 x3 x4 + ... − 1 + + + + + ... 1! 2 ! 3! 4! ⇒ f x() = e 2 x − e x Put f x() = 0, we get e 2 x − e x = 0 ⇒ e x(e x − 1) = 0 ⇒ e x = 0 or e x = 1 ⇒ x = 0 Hence, exactly one real solution exists. 8 21 40 65 45. We have, S = 1 + + + + + ... 2 ! 3! 4! 5! Let S1 = 1 + 8 + 21 + 40 + 65 + .... + Tn ...(i) and S1 = 1 + 8 + 21 + 40 + ...+ Tn − 1 + Tn On subtracting Eq. (ii) from Eq. (i), we get ...(ii) 0 = 1 + 7 + 13 + 19 + 25 + ... − Tn Tn = 1 + 7 + 13 + 19 + 25 + ... upto nterms n = [ ( )2 1 + (n − 1 6) ] 2 = n [1+ 3 (n − 1)] = n (3 n − 2) n(3 n − 2) ∴ S = Σ = Σ = Σ ⇒ S = Σ = 3e + e = 4e (n − 2)! (n − 1)! 1 1 e = 1 + 1! + 2 ! + ... We know that, 2 < e < 3 ∴ 8 < 4e < 12 ⇒ 8 < S < 12 46. Since, a1, a2, ..., an are in AP, therefore a2 − a1 = a3 − a2 = ... = a2k − a2k − 1 = d [say] Now, a1 2 − a2 2 = (a1 − a2 ) (a1 + a2 ) = − d (a1 + a2 ) a32 − a42 = − d (a3 + a4 ) a22k −1 − a22k = − d (a2k−1 + a2k ) On adding, we get S = − d (a1 + a2 + ... + a2k ) 2k = − d 2(a1 + a2k ) = − a( + a dk ) = (a2 − a2 2 k ) = [(a1 − a2 ) + (a2 − a3 ) + (a3 − a4 ) + ...+ (a2k−2 − a2k−1) + (a2k−1 − a2k )] k 2 2 == (a1 − a2k ) (− d ) (2k − 1) 2k − 1 47. Given, log x (ax), log x (bx)and log x (cx) are in AP. ⇒ 1 + log x a, 1 + log x b, 1 + log x c are in AP. ⇒ log x a, log x b, log x care in AP. − − )! − + − )! + Σ − − − −
3 116 log a log c log b ∴ + = 2 log x log x log x ⇒ log a + log c = 2 log b ⇒ ac = b2 48. Let the six numbers in AP be a − 5d a, − 3d a, − d a, + d a, + 3d a, + 5d ∴ a − 5d + a − 3d + a − d + a + d + a + 3d + a + 5 d = 3[sum = 3] ⇒ 6 a = 3 ⇒ a = Also, T1 = 4T3, where T T1, 3 are respectively first and third terms of an AP. ⇒ a − 5d = 4 (a − d ) ⇒ d = − 3 a = − So, the fifth term = a + 3d 1 3 1 9 = + 3 − = − 4 2 2 2 2 49. Since,log10 2, log10 (2 x − 1) and log10 (2 x + 3) are in AP. ∴ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ Now, let 2 x = t ⇒ t 2 − 4t − 5 = 0 ⇒ (t − 5) (t + 1) = 0 ⇒ t or t = − 1 50. Sn = 0.2 + 0.22 + 0.222 + ... upto n terms = 2 [0.1 + 0.11 + 0.111 + ... upto n terms] 2 =[0 9. + 0 99.+0 999. + ... upto n terms] = [(1− 0 1. ) + {1 − (0 1. ) }2+ {1 − (0 1. ) }3 +...upto n terms] upto n terms}] = 2 n − (0 1. ) {11−−(0 10 1. ) }. n = 29 n − 19 (1 − 10−n ) 9 2 2 2 3 n 1 1 1 = 1 2 − 2 − ... + 2 − 2 3 n [1 + {2 + 2 +... + 2}] 1 1 1 = (n− 1) times − 2 + 3 + ... + n ⇒ 2 = 5 [neglecting 2 x = − 1as 2 x is always positive] ⇒ x log2 2 = log2 5 ⇒ x = log2 5 +… + + = ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + + = + + + + + + − = = = − 52. = + + + = + + + + −
3 117 1 1 1 = [1 + 2 (n − 1)] − 1 + 2 + 3 + ... + n + 1 1 1 1 = 2n − 1 + 2 + 3 + ... + n = 2n − Hn 53. Let the numbers be a and b. a + b 2ab ∴ = 27 and = 12 2 a + b ⇒ a + b = 54 and 2ab = 12 (a + b) ⇒ 2ab = 12 (54) ⇒ ab = 6 (54) = 324 ⇒ ab = 18 Thus, GM = 18. 54. Since, x y, and z are in GP. ∴ y2 = x z Now, taking log10 on both sides, we get 2 log10 y = log10 x + log10 z 1 1 1 ⇒ 2 + ⇒ log x 10 logy 10 logz 10 ⇒ log x 10, logy 10 and logz 10 are in HP. (2 log 2)2 + 2 1 + 2 log 2 + +... 2 ! = 2 (elog 2 ) + 2 (e 2 log 2 ) = 2 × 2 + 2elog 4 = 4 + 2 × 4 = 12 56. Let Tn be the nth term of the given series. n n(+ 1) n + 1 (n − 1) + 2 ∴ Tn = = = n ! (n − 1)! (n − 1)! ∴ S T n = 1 n = 2(n − 2)! n = 1(n − 1)! = e + 2e = 3 e 57. Given, sum of n terms of an AP = 240 n ⇒ n(2 + n − 1) = 240 ⇒ n n(+ 1) = 15 × 16 ⇒ n = 15 58. Given, S and a Let r be the common ratio. a xn + yn ⇒ xn + 1 + yn + 1 = (xy)1 2/ (xn + yn ) ⇒ xn ⋅ x + yn ⋅ y = xn ⋅ x1/ 2 y1/ 2 + yn ⋅ x1/ 2 y1/ 2 ⇒ = 1 y ⇒ n + = 0 ⇒ 60. Let the numbers be a and b. ∴ ab = 10 ⇒ ab = 100 2ab and= 8 a + b 200 ⇒= 8 a + b 55. + + + + + + + + = + + ∴ − = ⇒ = − ⇒ − = ⇒ = ⇒ = 59. x + + + = x =−
3 118 + + ⇒ a + b = 25 On solving Eqs. (i) and (ii), we get a = 5 and b = 20 or a = 20 and b = 5 61. Tn = (2n − 1)3 = 8 n3 − 13 − 3 2⋅n⋅1(2n − 1) = 8 n3 − 1 − 12 n2 + 6 n = 8 n3 − 12 n2 + 6 n − 1 ∴ Sn = ΣTn = 8 Σn3 − 12 Σ n2 + 6 Σ n − Σ1 n n(+ 1) 2 n n(+ 1)(2n + 1) = 8 2 − 12 ...(i) 2 − n = 2n2(n + 1)2 − 2n n( + 1)(2n + 1) + 3 n n( + 1) − n = n n(+ 1)[2n n(+ 1) − 2 (2n + 1) + 3] − n = n n(+ 1)[2n2 + 2n − 4 n − 2 + 3] − n ...(ii) 2 = n n(+ 1)[2n − 2n + 1] − n = n n(+ 1) 2⋅n n(− 1) + n n(+ 1) − n = 2n2(n2 − 1) + n2 = n2(2n2 − 1) 2n 1 1 62. Here, Tn = = − (2n + 1)! 2n ! (2n + 1)! 
4 Complex Numbers The Real Number System Natural Numbers (N ) The numbers which are used for counting are known as natural numbers (also known as set of positive integers), i.e. N = {1, 2, 3, …}. Whole Numbers (W) If ‘0’ is included in the set of natural numbers, then we get the set of whole numbers, i.e. W = {0, 1, 2, … } = {N} + { }0 . Integers (Z or I) If negative natural numbers are included in the set of whole numbers, then we get set of integers, i.e. Z or I = {…, −3, −2, −1, 0, 1, 2, 3, …}. Rational Numbers (Q) p The numbers which are in the form of , (where p, q ∈I, q q ≠ 0), are called as rational numbers. e.g. etc. Irrational Numbers (Q′ ) The numbers which are not rational, i.e. which cannot be p expressed in form or whose decimal part is q non-terminating, non-repeating but which may represent magnitude of physical quantities are called irrational numbers. e.g. 2, 51/3 , π e, etc. Real Numbers (R) The set of rational and irrational numbers is called a set of real numbers, i.e. N ⊂ W ⊂ Z ⊂ Q ⊂ R Ø ● The real number system is totally ordered for any two numbers a, b ∈R. We must say, either a< b or b < a or b = a. Chapter Snapshot ● The Real Number System ● Modulus of a Real Number ● Imaginary Number ● Complex Number ● Algebra of Complex Numbers ● Conjugateof a Complex Number ● Modulus of a Complex Number ● Argument (orAmplitude) of a Complex Number ● Various Forms of a Complex Number ● De-Moivre’sTheorem ● Roots of Unity ● GeometricalApplications of Complex Numbers ● Loci in Complex Plane ● Logarithmof Complex Numbers
4 120 ● All real numbers can be represented by points on a straight line. This line is called as number line. ● Division by zero is meaningless. ● Number zero is neither positive nor negative but it is an even number. ● Square of a real number is always non-negative. ● An integer is said to be even, if it is divisible by 2, otherwise it is an odd number. ● Number ‘0’ is an additive identity. ● The magnitude of a physical quantity may be expressed as a real number times, a standard unit. ● Number ‘1’ is multiplicative identity. ● A positive integer p is called prime, if its only divisors are ± 1 and ± p. ● Between two real numbers, there lie infinite real numbers. ● Infinity (∞) is the concept of the number greater than greatest you can imagine. It is not a number, it is just a concept, so we do not associate equality with it. X Example 1. The value of is (a)1 (b)2 (c)1050 (d)0 Sol. (d) As we know, ∞ is the number greater than greatest we imagine. Also, the value of 1 upon ∞ is tending to zero. Hence, . Intervals Let a, x, b are real numbers, so that x ∈[a b,] ⇒ a ≤ x ≤ b, [a, b] is known as the closed interval a, b. x ∈(a b, )⇒ a < x < b, (a, b) is known as the open interval a, b. x ∈(a b, ]⇒ a < x ≤ b, (a, b] is known as open, closed interval a b, . x ∈[a b,) ⇒ a ≤ x < b, [a, b) is known as closed, open interval a b, . X Example 2. If a = 3 +1, b = 2 2, then the value a of lies in the interval b (a) (−1 0, ) (b) (0,1] (c)[0.5,1.5] (d) (0,1) Sol. (b, c, d) Consider a = 3 + 1 = 0.966 b 2 2 Clearly, 0.966 ∈(0, 1), (0, 1] and [0.5, 1.5]. Modulus of a Real Number The modulus of a real number x is defined as follows: x, when x > 0 | |x 0, when x = 0 x, when x < 0 x − a, when x ≥ a Ø |x − a| (x − a), when x < a X Example 3. If f x( ) = |x − 2| + | |x + |x + 3|, then the value of f x( ) for x ≤ − 3 is (a)3x −1 (b) −3x +1 (c) x − 3 (d) −(3x +1) Sol. (d) |x − 2| = − (x − 2) = 2 − x for x ≤ − 3 | |x = − x for x ≤ − 3 and |x + 3| = − (x + 3) for x ≤ −3 ∴ f x( ) = 2 − x − x − x − 3 = − 3x − 1 = − (3x + 1) Imaginary Number Square root of a negative real number is imaginary number. While solving equation x2 +1 = 0, we get x = ± −1 which is imaginary. So, the quantity −1 is denoted by ‘i’ called ‘iota’. Thus, i = −1. e.g. −2, −3, −4, … may expressed as i 2, i 3, 2i, … . Integral powers of iota As we have seen i = −1, so i2 = −1, i3 = − i and i4 =1. Hence, n ∈ N, in = i, −1, −i, 1 attains four values according to the value of n, so i4n +1 = i i4n + 2 = −1 i4n + 3 = − i i4n or i4n + 4 =1 In other words, n (−1)n/2 , if n is an even integer i n −1 (−1) 2 i, if n is an odd integer
4 121 ( ) i in n + + ∑ 1 13 Ø ● i2 = −1× −1≠ 1 ● −a × −b ≠ ab, so for two real numbers a and b but a ⋅ b = a ⋅ b possible, if both a, b are non- negative. ● i is neither positive, zero nor negative. Due to this reason, order relations are not defined for imaginary numbers. ● The sum of any four consecutive powers of i is zero, i.e. i4n+1 + i4n+ 2 + i4n+3 + i4n+4= 0 X Example 4. The value of i2014 is (a) i (b) − i (c) 1 (d) −1 Sol. (d) Consider i 2014 = (i 2 1007) = −( 1)1007 = − 1 X Example 5. If a < 0, b > 0, then a ⋅ b is equal to (a) − |a b|⋅ (b) |a b i|⋅ ⋅ (c) |a b| (d) None of these Sol. (b) As we can only multiply the positive values in square root. ∴ a b = −| |a b, as a < 0 and b > 0 i.e. −1⋅ | |a b = i | |a b = | |a b⋅⋅ i X Example 6. The value of the sum , where i = −1 is n =1 (a) i (b) i −1 (c) − i (d)0 Sol. (b) Since, the sum of any four consecutive powers of i, is zero. 13 13 13 ∴ ∑(i n + i n + 1) = ∑ i n + ∑ i n + 1 …(i) n = 1 n = 1 n = 1 = (i + i 2 + i 3 + + i13 ) + (i 2 + i 3 + i 4 + + i14 ) = i − 1 [from Eq. (i)] X Example 7. The value of 2n (1 + i)2n + , n ∈I, is equal to 2n n (1 + i) 2 (a)0 (b)2 (c){1 + −( 1) }n ⋅ in (d) None of these Sol. (c) Here, 2n + (1 + i)2n X Example 8. If i = −1, then the number of values of in + i− n for different n ∈I is (a) 3 (b) 2 (c) 4 (d) 1 Sol. (a) As i n + i − n can be written as n 1 i 2n + 1 x = i + = n If n = 4, x = If n = 5, x = If n = 6, x = If n = 7, x = − 1 and so on. Which shows there exist three different solutions for n ∈ I. Complex Number The complex number z = a + ib = (a b, ) can be represented by a point P, whose coordinates are referred to rectangular axes XOX ′ and YOY ′, which are called real and imaginary axis respectively. The plane formed by rectangular axes is called argand plane or argand diagram or complex plane or Gaussian plane. (i) A number of the form z = x + iy = Re z + iIm z is called a complex number. (ii) Two complex numbers are said to be equal, ifand only if their real parts and imaginary parts are separately equal, i.e. a + ib = c + id ⇔ a = c and b = d. (iii) The complex numbers do not possess theproperty of order, i.e. + = + + + + + + = + = + = = − + = − + = ⋅ − + X′ X O Y′ Y Pab ( , ) b M a Realaxis + = + = + − =− + =
4 122 x + iy < (or) > c + id is not defined. (iv) A complex number z is purely real, if Im( )z = 0 and said to be purely imaginary, if Re( )z = 0. The complex number 0 = 0 + i. 0 is both purely real and purely imaginary. 1 X Example 9. If z = −( 5 )i i , then Im ( )z is 8 equal to (a)1 (b)0 (c) −1 (d) None of these Sol. (b) Consider z = −( 5 )i 1 i 8 Let us first express z in the format a + ib, then z = i 2 = − (−1) = + i0 ⇒ Im( )z = 0 X Example 10. If 4x + i(3x − y) = 3 + i(−6), where x and y are real numbers, then the value of x and y respectively are (a)3, 33 (b) (c) (d) None of these Sol. (c) We have, 4x + i(3x − y) = 3 + i(−6) …(i) Equating the real and imaginary parts of Eq. (i), we get 4x = 3 and 3x − y = − 6, which on solving simultaneously, we get x = 3 and y = Algebra of Complex Numbers Addition of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 + z2 = x1 + iy1 + x2 + iy2 = (x1 + x2 ) + i y( 1 + y2 ) ⇒ Re (z1 + z2 ) = Re (z1 ) + Re (z2 ) and Im (z1 + z2 ) = Im (z1 ) + Im (z2 ) Properties of Addition of Complex Numbers (a) z1 + z2 = z2 + z1 [commutative law] (b) z1 + (z2 + z3 ) = (z1 + z2 ) + z3 [associative law] (c) z + 0 = 0 + z (where, 0 = 0 + i0) [additive identity law] X Example 11. The value of 3(7 + i7) + i(7 + i7) is (a)15 + 27 i (b)14 + 28 i (c)14 − 28 i (d)14 + 23 i Sol. (b) We have, 3(7 + i7) + i(7 + i7) = 21 + 21i + 7 i + 7 i 2 = 21 + 28i − 7 [i 2 = − 1] = 14 + 28i Subtraction of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) = (x1 − x2 ) + i y( 1 − y2 ) ⇒ Re(z1 − z2 ) = Re(z1 ) − Re(z2 ) and Im (z1 − z2 ) = Im (z1 ) − Im (z2 ) Multiplication of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 ⋅ z2 = (x1 + iy1 )(x2 + iy2 ) = (x x1 2 − y y12 ) + i x y( 12 + x y2 1 ) ⇒ z1 ⋅ z2 = [Re (z1 )Re (z2 ) − Im (z1 )Im (z2 )] + i[Re (z1 )Im (z2 ) + Re (z2 )Im (z1 )] Properties of Multiplication of Complex Numbers (a) z1 ⋅ z2 = z2 ⋅ z1 [commutative law] (b) (z1 ⋅ z2 )z3 = z1 (z2 ⋅ z3 ) [associative law] (c) If z1 ⋅ z2 =1 = z2 ⋅ z1, then z1 and z2 are multiplicative inverse of each other. Thevalueof 1 3 7 3 4 1 3 4 3 + + + + − − i i i is (a) 5 3 17 3 −i (b) 17 3 5 3 −i c) ( 17 3 5 3 +i (d) 17 5 4 3 −i + + + + − = + + + − + + + = − + + = + + + + − + =
4 123 (d) (i) z1 (z2 + z3 ) = z1 ⋅ z2 + z1 ⋅ z3 [left distribution law] (ii) (z2 + z3 ) z1 = z2 ⋅ z1 + z3 ⋅ z1 [right distribution law] X Example 13. The real values of x and y, if (1 + i x)− 2i (2 − 3i y)+ i + = i, are respectively (3 + i) (3 − i) (a) 3, −1 (b) 3, 1 (c) −3, 1 (d) −3, −1 Sol. (a) (1 + i x) − 2i + (2 − 3i y) + i = i (3 + i) (3 − i) ⇒ {(1 + i x) − 2 i}(3 − i ) + {(2 − 3 i y) + i}(3 + i ) = i(3 + i)(3 − i) ⇒ (1 + i )(3 − i x)− 2 (3i− i ) + (2 − 3 )(3i+ i y) + i(3 + i) = 10 i ⇒ (4 + 2 i ) x − 6 i − 2 + (9 − 7 )i y + 3 i − 1 = 10 i ⇒ (4x − 2 + 9y − 1) + i(2 x − 6 − 7 y + 3) = 10 i ⇒ (4x + 9y − 3) + i(2 x − 7 y − 3) = 10 i On equating real and imaginary parts on both sides, we get 4x + 9y = 3 …(i) and x − 7 y = 13 …(ii) On solving Eqs. (i) and (ii), we get x = 3, y = − 1 X Example 14. The multiplicative inverse of 4 − 3i is 4 3i 4 3i (a) − (b) + 25 25 25 25 4 3i (c) + (d) None of these 16 25 Sol. (b) Let z = 4 − 3i Then, its multiplicative inverse is 1 1 1 4 + 3i 4 + 3i = = × = z 4 − 3i 4 − 3i 4 + 3i 16 − 9i 2 [(a − b)(a + b) = a2 − b2] =[i 2 = − 1] = 4 3i 25 25 Division of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 (≠ 0) be two complex numbers, then z1 x1 + iy1 = z2 x2 + iy2 1 = 2 2 [(x x1 2 + y y1 2 ) + i x y( 2 1 − x y1 2 )] x2 + y2 z1 X Example 15. The value of , where z1 = 2 + 3i z2 and z2 =1 + 2i, is 8 1 i (b) − i 5 5 i (d) None of these Sol. (b)z1 = 2 + 3i and z2 = 1 + 2i ∴ z2−1 = 1 = 1 − 2 i 1 + 2 i (1 + 2 i)(1 − 2 i) i z1 = z1 ⋅ z2 −1 = (2 + 3 i) 1 − 2 i Then, z2 5 5 2 + 6 + i 4 + 3 = 8 − 1 i 5 5 5 5 5 5 Aliter Here, x1 = 2, y1 = 3, x2 = 1 and y2 = 2. ∴ z1 = 1 {(x x12 + y y12 )+ i x y( 21 − x y12 )} z2 ={(2 × 1 + 3 × 2) + i(3 × 1 − 2 × 2)} = {(2 + 6) + i(3 − 4)} = 8 − 1 i 5 5 Conjugate of a Complex Number Geometrically, the conjugate of z is the reflection of point image of z in the real axis. ∴ z x iy = + and z x iy = − . X Y Realaxis z z O + + + = + x + +
4 124 e.g. If z = 3 + 4i, then z = 3 − 4 i. Properties of Conjugate z is the mirror image of z along real axis. (i) ( )z = z (ii) z = z ⇔ z is purely real (iii) z = − z ⇔ z is purely imaginary. (xii) If z = f z( 1 ), then z = f z( 1 ) X Example 16. If z1 = 9y2 − 4 −10 ix and z2 = 8y2 + 20i, where z1 = z2, then z = x + iy is equal to (a) −2 + 2 i (b) −2 ± 2 i (c) −2 ± i (d) None of the above Sol. (b) Given, z1 = z2 ⇒ 9y2 − 4 − 10 i x = 8y2 + 20 i ⇒ (y2 − 4) − 10i x( + 2) = 0 Since, complex number is zero. ⇒ y2 − 4 = 0 and x + 2 = 0 ∴ y = ± 2 and x = − 2 Thus, z = x + iy = − 2 ± 2 i X Example 17. If (1 + i z) = (1 − i z) , then z is (a) x(1 − i), x ∈R (b) x(1 + i), x ∈R x + (c) , x ∈R 1 + i (d) None of the above Sol. (a) (1 + i)(x + iy) = (1 − i)(x − iy) ⇒ (x − y) + i x( + y) = (x − y) − i x( + y) ⇒ x + y = 0 ∴ z = x − ix = x(1 − i) Modulus of a Complex Number Let z = x + iybe any complex number. Then, | |z = (x2 + y2 ) is called the modulus of the complex number z, where modulus | |z represents distance of z from origin. e.g. If z = 3 + 2i is a complex number, then | |z = 32 + 22 = 9 + 4 = 13 X Example 18. If z is a complex number satisfying the relation | z +1| = z + 2 1( + i), then z is Sol. (c) Let z = x + iy ∴ |x + iy + 1| = x + iy + 2 1( + i) ⇒ (x + 1)2 + y2 = (x + 2) + i y( + 2) ⇒ (x + 1)2 + y2 = x + 2 and y + 2 = 0 ⇒ (x + 1)2 + 4 = (x + 2)2 and y = − 2 ⇒ 2 x + 5 = 4x + 4 and y = − 2 ⇒ x =and y = − 2 ⇒(1 − 4 i) X Example 19. If z =1 + itan α, where π < α < , then | |z is equal to (a)secα (b) −secα (c)cosec α (d) None of the above Sol. (b) As z = 1 + i tan α ∴ | |z = 1 + tan2 α =|secα| ⇒ | |z = − sec α, as π < α < ( ) iv Re() Re() z z z z = = + 2 (v) Im()z z z i = − 2 ) vi ( z z z z 2 1 2 1 + + = (vii) z z z z 2 1 2 1 − − = ) viii ( zz zz 12 12 = ) ix ( z z z z 1 2 1 2 = ) ,( z2 0 ≠ (x) zz zz zz zz 12 12 12 12 2 2 + = = ) ) Re( Re( ) (xi z z z n n n = = () ( ) =
4 125 Argument (or Amplitude) of a Complex Number Argument of z = θ Argument of z is not unique. General value of argument of z is 2nπ + θ. Principal Value of Argument The value of θ of the argument, which satisfies the inequality −π < θ ≤ π is called the principal value of the argument. Principal values of the argument are θ, π − θ, −π + θ, − θ according as the complex number lies on the Ist, IInd, IIIrd or IVth quadrant. Here, θ = tan −1 | y| , where z = x + iy | |x X' X e.g. arg (1 + i) = tan −1 1 = π 1 4 arg (1 − i) = tan −1 −1 = −π 1 4 arg (−1 − i) = tan −1 − 1 = − π + π −1 4 arg (−1 + i) = tan −1 − 1 = π − 1 4 Ø ● Argument of 0 is not defined. ● If z1 = z2, then |z1| = |z2| and arg(z1) = arg(z2) ● Argument of purely imaginary number is or − ● Argument of purely real number is 0 or π . Example 20. The modulus and argument X of the 1 + 2i complex number is 1 − 3i (a) (b) (c) (d) None of the above [(a + b)(a − b) = a2 − b2] =[i 2 = − 1] = 1 + 9 10 2 ⇒ z = − + i ∴ | |z =a2 + b2 ] ∴ Now, tan θ = 2 1 2 θ = tan−1 Im( )z Re ( )z Y X θ O Realaxis z x iy =( + ) θ θ θ θ arg()= z θ arg()= – z θ π arg()=– z θ arg()=– + z π θ Y Y' = + − ∴ = + − × + + = + + + − 1 2 , 3 4 π 1 2 , − 3 4 π 1 2 , 3 4 π + + − − − + = + − = −+ − +  = + + = = =
4 126 π π = 1 ⇒ tan θ = 1 = tan ⇒ θ = Since, the real part of z is negative and imaginary part of z is positive, so the point lies in IInd quadrant. ∴ arg ( )z = π = θ = π − = 3π 4 Hence, modulus = and arg( )z = Work Book Exercise 4.1 1 If x − 3 + y − 3 = i, where x, y ∈R, then 3 + i 3 − i a x = 2 and y = − 8 b x = − 2 and y = 8 cx = − 2 and y = − 6 d x = 2 and y = 8 2 What is the real part of (1 + i )50? a 0b 225 c − 225 d − 250 3 The complex number z satisfies z + | z| = 2 + 8 i. Then, the value of| z| is a 10 b 13 c 17 d 23 4 If z + z3 = 0, then which of the following must be true on the complex plane? a Re ( )z < 0 b Re ( )z = 0 c Im ( )z = 0 d z4 = 1 5 The sequences S = i + 2i 2 + 3i 3 + upto 100 terms simplifies to, where i = −1 c (1 + i n)2 d None of these 7 Let i = −1. The product of the real part of the roots of z2 − z = 5 − 5 i is a − 25 b − 6 c − 5 d 25 8 Number of complex numbers z satisfying z3 = z is a 1 b 2 c 4 d 5 9Number of real solution of the equation, z3 + iz − 1 = 0 is a zero b one c two d three 10 The diagram shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of F, which is a A b B c C d D 11 Identify the incorrect statement. a No non-zero complex number z satisfies the equation z = − 4z b z = z implies that z is purely real c z = − z implies that z is purely imaginary d If z1, z2 are the roots of the quadratic equation az2 + bz + c = 0 such that Im(z z1 2) ≠ 0, then a, b, c must be real numbers 12 If z = (3 + 7 )(i p + iq), where p, q ∈l − {0}, is purely imaginary, then minimum value of| z|2 is a 0 b 58 c d 3364 13 Consider two complex numbers α and β as a + bi 2 a − bi 2 , where a, b ∈R and a − bi a + bi z − 1 β = , where| z| = 1, then z + 1 a bothα andβ are purely real b bothα andβ are purely imaginary c α is purely real andβ is purely imaginary d β is purely real andα is purely imaginary 14 If z is a complex number having the argument θ, 0 < θ < and satisfying the equality| z − 3 i| = 3. 6 − + − 6 − = + + − + − + + − − + + + − + − + + − − + + + − + + − + − − + + + + − + − + + − − + + O D F A C Imaginaryaxis Real axis B
4 127 Then, cot θ − is equal to z a 1 b −1 ci d − i Various Forms of a Complex Number Polar Form Let z = a + ib be any complex number, then by taking a = rcosθ and b = rsin θ We have, z = a + ib = r(cosθ + isin )θ (known as polar form) of z. X Example 21. The polar form of the complex −16 number is 1 + i 3 π 2π 2 (a) 4 cos + isin (b) cos + isin 3 3 3 3 2π 2 (c)8 cos + isin (d) None of these 3 3 By squaring and adding, we get 16 + 48 = r2(cos2 θ + sin2 θ) which gives, r2 = 64, i.e. r = 8 − 1 Hence, cosθ = 2 s inθ = π ⇒ θ = π − 3 3 Thus, the required polar form is 8 cos2 π + i sin 2 3 3 X Example 22. Let z and w be two non-zero complex numbers, such that | |z = |w| and arg( )z + arg( )w = π. Then, z is equal to (a) w (b) − w (c) − w (d) w Sol. (c) Given| |z =| |w = r and arg(w) = θ Also, arg ( )z + arg (w) = π ⇒ arg( )z = π − θ Now, z = r[cos(π − θ) + i sin(π − θ)] = r[−cosθ + i sin θ] = − r(cosθ + i sin )θ = − w where, r z = and θ= Principalvalueofargument Realaxis a θ r ( ) a,b b π =
4 128 − π Eulerian Form of a Complex Number We have, ei θ = cosθ + isin θ and e− i θ = cosθ − isin θ. These two are called Euler’s notations. Let z be any complex number, such that | |z = r and arg ( )z = θ. Then, in polar form, z can be written as z = r(cosθ + isin )θ Using Euler’s notations, we have z = rei θ This form of z is known as the Eulerian form. X Example 23. Express the following complex numbers in Eulerian form. (i)1 + i (ii) −2 + 2i Sol. (i) Given, z = 1 + i. Then, r =| |z = Let θ be the argument of z. Then, tan θ= =1⇒θ = i π So, Eulerian form of z is 2e 4 . (ii) Given, z = − 2 + 2 i Then, r =| |z = ( 2)−2 + 22 = 2 2 Let θ be the argument of z. Then, tan θ = = − 1 ⇒ θ = 3π i So, Eulerian form of z is 2 2e 4 . (iii)Given, z = − −1 i 3 Then, r =| |z = ( 1)− 2 + −( 3)2 = 2 Let θ be the argument of z. Then, tan θ = − 3 = 3 ⇒ θ = − 2 π −1 − i So, Eulerian form of z is 2e 3 . X Example 24. Complex numbers z1, z2, z3 are the vertices A B C, , respectively of an isosceles right angled triangle with right angle at C. Show that (z1 − z2 )2 = 2(z1 − z3 ) (z3 − z2 ). Sol. In an isosceles ∆ ABC, Bz( 2) AC = BC and BC perpendicular to AC. It means that AC is rotated through angleto occupy the position BC. xii. xiii. xiv. Ø ● If z is unimodular, then |z| = 1. Now, iff( )z is a unimodular, Cz( 3) Az( 1) then it is always be expressed as f( )z = cos θ + i sin θ, θ∈R.. We have, ● Square root of z = a + ib is given by z2 − z3 = e iπ /2 = i z | |z + a + i | |z − a , a =Re( )z z1 − z3 2 2 ⇒ z2 − z3 = + i z( 1 − z3 ) To find the square root of a − ib, replace i by − i in the above ⇒ z2 2 + z3 2 − 2 z z2 3 = − (z1 2 + z3 2 − 2 z z1 3 ) result. ⇒ z12 + z22 − 2 z z12 = 2 z z13 + 2 z z23 − 2 z z12 − 2 z32 ● If x, y ∈R = 2(z1 − z3 )(z3 − z2 ) ⇒ (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) | z1 + z2|2 = | z1|2 + | z2|2 z ⇔ 1 is purely imaginary. z2 |z1 + z2|2 + |z1 − z2|2 = 2{|z1|2 + |z2| }2 |az1 − bz2|2 + |bz1 + az2|2 = (a 2 + b2 )(| z1|2 + | z2| )2 , where a, b ∈R. π (iii) − − 3 1 i + = + + − + + = + − = − − + If x iy ib a c id = − − −
4 129 Properties of Modulus of Complex Number X Example 25. , then i.(x2 + y2 )2 is equal to 2 2 2 ii.(a) a− b 2 (b) a + b c2 − d 2 c2 + d 2 iii.a 2 + b2 (c) (d) None of these iv.c2 − d 2 Sol. (b) Given, x − iy = a − ib v.c − id 1/ 2 a − ib ⇒ x + i(− y) c − id vi.On taking modulus both sides, we get vii.|x + i(− y)| c − id a − ib 1 2/ 1 2/ ⇒ x2 + −(y)2 a − ib c − id iy| = x2 + viii. [|x + 1 2/ ⇒ x2 + y2 a − ib ix. c − id On squaring both sides, we get x.2 2 a − ib x + y c − id ⇒ x2 + y2 = |a − ib| |c − id| a2 + b2 y2 and|z n| =| |z n ]  z1 = z1 z2 z2 xi.⇒ (x2 + y2 ) =[|x − iy| = x2 + y2 ] c2 + d 2 On squaring both sides, we get 2 2 2 a2 + b2 (x + y ) = c2 + d 2 | |z ≥ 0 ⇒| |z = 0 iff z = 0 and | |z > 0 iff z ≠ 0 −| |z ≤ Re ( )z ≤ | |z and −| |z ≤ Im ( )z ≤ | |z | |z = | z| = −| z| = −| z| zz = | |z 2 | z z1 2| = | z1|| z2| In general, | z z z1 2 3 zn | = | z1|| z2|| z3|| zn | z1 = | z1 |, (z2 ≠ 0) z2 | z2| | z1 ± z2| ≤ | z1| + | z2| In general, | z1 ± z2 ± z3 ±± zn | ≤ | z1| + | z2| + | z3| ++ | zn | | z1 ± z2| ≥ || z1| − | z2|| | zn | = | zz |n || z1| − | z2|| ≤ | z1 + z2| ≤ | z1| + | z2| Thus, | z1| + | z2| is the greatest possible value of | z1 + z2| and || z1| − | z2|| is the least possible value of | z1 + z2|. | z z | (z z )(z z )
4 X 130 Example 26. For a complex number z, the minimum value of | |z + | z − 2| is (a) 1 (b) 2 (c) 3 (d) None of these Sol. (b) By using | |z1 + |z2|≥|z1 − z2| We have, | |z + |z − 2|≥|z − (z − 2)| ∴ | |z + |z − 2|≥ 2 X Example 27. If | z1 −1| <1, | z2 − 2| < 2 and | z3 − 3| < 3, then | z1 + z2 + z3| (a) is less than 6 (b) is more than 3 (c) is less than 12 (d) lies between 6 and 12 Sol. (c)|z1 + z2 + z3|=|(z1 − 1) + (z2 − 2) + (z3 − 3) + 6| ≤|z1 − 1| + |z2 − 2| + |z3 − 3| + 6 < 1 + 2 + 3 + 6 = 12 X Example 28. If α, β are two complex numbers, then |α |2 + | |β 2 is equal to (a) (|α+ β|2 − α− β|| )2 (b) (|α+ β|2 + α− β| | )2 (c)|α + β|2 + α − β| |2 (d) None of these Sol. (b)|α + β|2 = α + β α + β = α + β( )( ) ( )(α + β ) = αα + ββ + αβ + αβ =| |α 2 + | |β 2 + αβ + αβ …(i) |α − β|2 = α − β α − β = αα + ββ − αβ − αβ( )() =| |α 2 + | |β 2 − αβ − αβ …(ii) Adding Eqs. (i) and (ii), we get | |α 2 + | |β 2 = {|α + β|2 + |α − β| }2 X Example 29. If z1 and z2 are two complex numbers, such that | z1| <1 < | z2|, then prove that 1 − z z1 2 <1. z1 − z2 Sol. Given,| |z1 < 1and|z2|> 1 …(i) Then, to prove  z1 = | |z1 < 1 z1 − z2 z2 |z2| ⇒ |1 − z z12|<|z1 − z2| …(ii) On squaring both sides, we get (1 − z z12 )(1 − z z12 )< (z1 − z2 )(z1 − z2 ) (| |z 2 = zz) ⇒ 1 − z z1 2 − z z1 2 + z z z z1 1 2 2 < z z1 1 − z z1 2 − z z2 1 + z z2 2 ⇒ 1 + | | |z1 2 z2|2 <| |z1 2 + |z2|2 ⇒ 1 −| |z1 2 −|z2|2 + | | |z1 2 z2|2 < 0 ⇒ (1 −| | )(1z1 2 −|z2| )2 < 0 …(iii) which is true by Eq. (i) as| |z1 < 1 and|z2|> 1. ∴ (1 −| | )z1 2 > 0 and (1 −|z2| )2 < 0 ∴ Eq. (iii) is true, whenever Eq. (i) is true. 1 − z z12 < 1 Hence proved. ⇒ z1 − z2 X Example 30. The value of −8 − 6i is equal to (a)1 ± 3i (b) ± (1 − 3 )i (c) ± (1 + 3 )i (d) ± (3 − i) Then, ⇒ ⇒ X (4 + 3 (a) ±6 (d) ± 3 Sol. (a) We may write, (4 + 3 − 20) = (4 + 6i 1 − z z1 2
4 131 Let (4 + 3 − 20) Then, (4 + 6i ⇒ ⇒ ∴ On solving the equations x2 = y2 = 14 and x2 − y2 = 4, we get x2 = 9 and y2 = 5 ∴ x = ± 3 and y = ± 5 Since, xy > 0, it follows that x and y are of the same sign. = Properties of Arguments i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii. Ø Proper value of k must be chosen, so that RHS of (i), (ii), (iii) and (iv) lies in (−π, π). 3 + i X Example 32. If z =, then the 3 − i fundamental amplitude of z is π (a) − (b) (c)(d) None of these Sol. (b) amp( )z = amp 3 + i 3 − i = amp( 3 + i) − amp( 3 − i) = tan−1 1 − tan−1 −1 = π + π = π 3 3 6 6 3 X Example 33. The value of amp(iω) + amp(iω2 ), where i = −1 and ω = 3 1 = non-real, is arg (z z1 2 ) = arg (z1 ) + arg (z2 ) + 2kπ (k = 0 or 1 or −1) In general, arg(z z z1 2 3 zn ) = arg(z1 ) + arg(z2 ) + arg(z3 ) ++ arg(zn ) + 2kπ (k = 0 or 1 or −1) z1 arg arg(z1 )− arg(z2 )+ 2kπ z 2 (k = 0 or 1 or −1) z arg = 2arg( )z + 2kπ (k = 0 or 1 or −1) z arg(zn ) = narg( )z + 2kπ (k = 0 or 1 or −1) z2 z1 If arg , then arg 2kπ − θ, where z1 z2 k ∈I. 1 arg( )z = − arg( )z = arg 2 If arg( )z = 0 ⇒ z is real. arg(z z1 2 ) = arg(z1 ) − arg(z2 ) | z1 + z2| = | z1 − z2| ⇒ arg(z1 ) − arg(z2 ) = | z1 + z2| = | z1| + | z2| ⇒ arg(z1 ) = arg(z2 ) If | z1| ≤1, | z2| ≤1, then (a) | z1 − z2|2 ≤ (| z1| − | z2|)2 + [arg(z1 ) − arg(z2 )]2 (b) | z1 + z2|2 ≥ (| z1| + | z2|)2 − [arg(z1 ) − arg(z2 )]2 | z1 ± z2|2 = | z1|2 + | z2|2 ± 2| z1|| z2| cos (θ1 − θ2 ) z z1 2 + z z1 2 = 2| z1|| z2|cos (θ1 − θ2 ) where, θ1 = arg( z 1 ) and θ2 = arg( z 2 ) 3 π 6
4 X 132 (a) 0 ( b) (c) π (d) None of these Sol. (c) amp(iω +) amp(iω2 ) = amp(i 2 ⋅ω3 )= amp( 1)− = π X Example 34. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then z1 z2 arg arg equals to z4 z3 (a) 0 (b) (c) (d) π Sol. (a) We have, z2 = z1 and z4 = z3 ∴ z z12 =| |z1 2 and z z3 4 =|z3|2 z1 + arg z2 = arg z z1 2 Now, arg z4 z3 z z43 = arg || |zz13|22 = arg zz31 2 = 0 [argument of positive real number is zero] De-Moivre’s Theorem (a) If n ∈I (the set of integers), then (cosθ + isin )θ n = cos nθ + isin nθ. (b) If n ∈Q (the set of rational numbers), then cos nθ + isin nθ is one of the values of (cosθ + isin θ)n . Remark (i) The theorem is also true for (cos θ − isin θ), i.e. (cos θ − isin θ)n = cos nθ − isin nθ, because (cosθ − isin θ)n = [cos(−θ) + isin(−θ)]n = cos( (n −θ) + isin ( (n −θ)) = cos(−nθ) + isin(−nθ) = cos nθ − isin nθ (ii) = (cosθ + isin )θ −1 = cosθ − isin θ (iii) If z = (cos θ1 + isin θ1 ) (cos θ2 + isin θ2 ) (cos θn + isin θn ) Then, z = cos (θ1 + θ2 ++ θn ) + isin(θ1 + θ2 ++ θn ) (iv) If z = r(cosθ + isin )θ and n is a positive integer, then 1/n 1/n 2k 2k z = r cos n + isin n , where k = 0, 1, 2, …, (n −1) Ø ● (sin θ ± i cos θ)n ≠ sin nθ ± i cos nθ n ● (sinθ + i cosθ)n = cos π 2 − + i sin π 2 − nπ nπ cos n i sin − n 2 2 ● (cosθ + i sin φ)n ≠ cos nθ + i sin nφ X Example 35. If z . Then, arg ( )z is (a)2θ(b)2θ − π (c) π + 2θ (d) None of these Sol. (a) z = (cos θ + i sin θ)2 = cos2θ + i sin 2θ where, < 2θ < π Clearly, arg ( )z = 2θ  π < arg( )z < π 2 Example 36.If xr = cos 2r + isin 2r , then the value of x x x1 2 3 ∞ is (a) −1 (b) 1 (c) 0 (d) None of these Sol. (a)xr = cos πr + i sin 2 πr 2 ∴ x i
4 133 x i   (cosθ + isin )θ 4 X Example 37. is equal to (sin θ + icos )θ 5 (a)cosθ − isin θ (b)cos9θ − isin9θ (c)sin θ − icosθ (d)sin9θ − icos9θ (cos θ + i sin θ)4 (cos θ + i sin θ)4 Sol. (d) = 5 5 (sin θ + i cos θ) 5 1 i sin θ + cos i (cos θ + i sin θ)4 (cos θ + i sin θ)4 = = i(cos θ − i sin θ)5 i(cos θ + i sin θ)− 5 = 1 (cosθ + i sin )θ 9 = sin9θ − i cos9θ i 3 i 5 3 i 5 X Example 38. If z + − , 2 2 2 2 then (a) Re ( )z = 0 (b)Im ( )z = 0 (c) Re ( )z > 0, Im ( )z > 0(d) Re ( )z > 0, Im ( )z < 0 Sol. (b) Given, 5 5 3 i 3 i z − 2 2 2 2 1 The maximum and minimum values of| z + 1|, when| z + 3| ≤ 3 are a (5, 0) b (6, 0) c (7, 1) d (5, 1) 5 5 = cos 6 + i sin 6 + cos 6 − i sin 6 π π π π = cos+ i sin+ cos− i sin= 2cos Hence, Im( )z = 0. X Example 39. The product of all the values of π 3 4/ cos + isin is 3 3 (a) −1 (b) 1 (c) 3/2 (d) −1/2 3 4/ Sol. (b) Given, cos π + i sin = [cos π + i sin π]1 4/ 3 3 Since, the expression has only 4 different roots, therefore on putting n = 0, 1, 2, 3 in 2n 2 The region represented by inequalities arg z, Im (z) ≥ 1in the argand diagram is given by ⇒ x x x …= + + π π π π  + + = + + + π π π π   = − + − π π =− + = π π
4 X 134 Work Book Exercise 4.2 3 If 1 + 2i = r(cos θ + i sin θ), then 2 + i a r = 1,θ = tan−1 3 b r = 5,θ = tan−1 c r = 1,θ = tan−1 d None of these 4 If z1 and z2 are two non-zero complex numbers, such that| z1 + z2| =| z1| + | z2|, then arg (z1) − arg (z2 ) is equal to a − π b − c 0 d 5 If z = 1 − sin α + i cos α, where 0, , then 2 the modulus and the principal value of the argument of z are respectively a 2(1 − sin α), π + b 2(1 − sin α), π − 4 2 4 2 c 2(1 + sin α), π + d 2(1 + sin α), π − 4 2 4 2 π 8 6 The expression 1 + sin π 8 + i cos π 8 is equal to 1 + sin − i cos 8 8 a 1 b − 1 c i d − i 7 If zr = cos π + i sin π, r = 0, 1, 2, 3, 4, …, then z z z z z1 2 3 4 5 is equal to a − 1 b 0 c 1 d None of these 8 If zn , then lim (z1 ⋅ z2 ⋅ z3 zn ) is equal to n → ∞ ab cos π + i sin π 6 6 c d None of these 9 If z1, z2 are two complex numbers and a, b are two real numbers, then|az1 − bz2|2 + |bz1 + az2|2 is equal to a (a + b) [2 | z1|2 + | z2| ]2 b (a + b)[| z1|2 + | z2| ]2 c (a2 − b2)[| z1|2 + | z2| ]2 d (a2 + b2)[| z1|2+ | z2| ]2 10 All real numbers x, which satisfy the inequality 1 + 4i − 2− x ≤ 5, where i = −1,x ∈R are a [−2, ∞) b (−∞, 2] c [0, ∞) d[−2 0, ] 11 Square root of x2 + x 1 2 − 4 i x − 1 x − 6, where x ∈R is equal to a x − 1 + 2i b x − 1 − 2i x x c ± x + 1 + 2i d ± x + 1 − 2i x x 12 The minimum value of|1 + z| + |1 − z|, where z is a complex number, is 3 a 2 b 2 c 1 d 0 13 If arg(z + a) = and arg(z a) , then a z is independent of a b | |a =| z + a| c z = a, c is d z = a, c is b a ° 60 2 1 1 2 60° 2 1 1 2 Y X X Y c d 60° 1 2 1 2 60° 1 2 1 2 Y X Y X
4 135 14 Let z be a complex number satisfying the equation (z3 + 3)2 = − 16, then| z| has the value equal to a 51/2 b 51/3 c 52/3 d 5 d | z1|4 + | z2|4 =| z3|8 Roots of Unity Cube Roots of Unity Let x = 3 1 ⇒ x3 −1 = 0 ⇒ (x −1) (x2 + x +1) = 0 −1 + i 3 −1 − i 3 Therefore, x =1, , 2 2 If second root is represented by ω, then third root will be ω2. Therefore, cube roots of unity are 1, ω, ω2 and ω, ω2 are called the imaginary cube roots of unity. Properties i. ii. iii. iv. Important Identities (i) x2 + x +1 = (x − ω)(x − ω2 ) (ii) x2 − x +1 = (x + ω)(x + ω2 ) (iii) x2 + xy + y2 = (x − yω)(x − yω2 ) (iv) x2 − xy + y2 = (x + yω) + (x + yω2 ) (v) x2 + y2 = (x + iy x)( − iy) (vi) x3 + y3 = (x + y x)( + yω)(x + yω2 ) (vii) x3 − y3 = (x − y) = (x − yω)(x − yω2 ) (viii) x2 + y2 + z2 − xy − yz − zx = (x + yω + zω2 )(x + yω2 + zω) or (xω + yω2 + z)(xω2 + yω + z) or (xω + y + zω2 ) (xω2 + y + zω) (ix) x3 + y3 + z3 − 3xyz = (x + y + z)(x + ωy + ω2 y x)( + ωy2 + ωz) (x) Two points P z( 1 ) and Q z( 2 ) lie on the same side or opposite side of the line az + az + b accordingly as az1 + az1 + b and az2 + az2 + b have same sign or opposite sign. Example 40. If x2 − x +1 = 0, then the value of 5 2 n 1 ∑ x + n is n =1 x (a) 8 (b) 10 (c) 12 (d) None of the above Sol. (a) x2 − x + 1 = 0 ⇒ x = 1 ± 3i = −ω, − ω2 2 5 ∴ n∑= 1 x2n + 12n + 2 = x2 + x2 + x4 + 1 + 2 x6 + 1 + 2 0, if r is not a multiple of 3 1 + ωr + ω2 r 3, if r is not a multiple of 3 ω3 =1or ω3 r =1 ω3r +1 = ω, ω3r + 2 = ω2 It always forms an equilateral triangle. 15 = − + = − + = + − Σ = = + − Σ + = − x + x + x +
4 X 136 x8x8 + 2 x10 x110 + 2 = (ω2 + ω4 + ω6 + ω8 + ω10 ) + 1 2 + 1 4 + 1 6 + 1 8 + 1 10 + 10 ω ω ω ω ω = − 1 − 1 + 10 = 8 [x = − ω or − ω2 ] X 49 + 3 + 3 100 ⇒ 2 2 ∴ x + iy = −i ⇒ x = − ∴ k = − X Example 42. If z2 − z +1 = 0, then zn − z− n , where n is a multiple of 3, is (a) 2( 1)− n (b) 0 (c)(−1)n +1 (d) None of the above Sol. (b) As z2 − z + 1 = 0 ⇒ z = − ω, − ω2 ∴ zn − z− n = −( 1)nωn − −(1)−n ω− n = −( 1)n (ωn − ω−n ) , if n is a multiple of 3. = −(1)n (1 − 1) = 0 nth Roots of Unity Let z =11/n . Then, z = (cos0° + isin0° )1/n z = (cos2rπ + isin2rπ)1/n , r ∈Z 2rπ 2rπ ⇒ z = cos + isin , r = 0, 1, 2, …, (n −1) n n [using De-Moivre’s theorem] i r2 π ⇒ z = e n , r = 0, 1, 2, …, (n −1) ⇒ z = {ei2π/n }r , r = 0, 1, 2, …, (n −1) i2π z = α r , α = e n , r = 0, 1, 2, …, (n −1) Thus, nth roots of unity are 1, α, α 2 ,…, α n −1 , where i2π n = cos 2π + isin 2π α = e n n Properties of nth Roots of Unity i. ii. iii. iv. v. vi. Example 41. x = ky, then k is If 3 (x iy) i 2 2 and nth roots of unity form a GP with common i2π ratio e n . Sum of nth roots of unity is always zero. Sum of pth powers of nth roots of unity is zero, if p is not a multiple of n. Sum of pth powers of nth roots of unity is n, if p is a multiple of n. Product of nth roots of unity is (−1)n −1 . nth roots of unity lie on the unit circle | |z =1 and divide its circumference into n equal parts. ) (a − 1 3 (b) 3 (c) − 3 ) (d − 1 3 x + + = = − +
4 137 X Example 43. If z1, z2, z3, …, zn are nth, roots of unity, then for k =1, 2, …, n (a)| zk | = k z| k +1| (b)| zk +1| = k z| k | (c)| zk +1| = | zk | + | zk +1| (d)| zk | = | zk +1| Sol. (d) The nth roots of unity are given by i2π ( k − 1) zk = e, k = 1, 2, …, n ∴ |zk| e n = 1, ∀ k = 1, 2, …, n ⇒ |zk| =|zk + 1|, ∀ k = 1, 2, …, n X Example 44. If n is a positive integer greater than unity and z is a complex number satisfying the equation zn = (z +1)n , then (a) Re ( )z < 0 (b) Re ( )z > 0 (c) Re ( )z = 0 (d) None of these n Sol. (a) We have, zn = (1 + z)n ⇒ z = 1 X Example 45. If α is non- real and α = 5 1, then z + 1 1+ α + α 2 + α − 2 − α −1 z Sol. (a)α5 = 1 z z ⇒ x + = 0 [taking z = x + iy] ⇒ x =Re( )z < 0 ∴ Work Book Exercise 4.3 1 If ω is a non-real cube root of unity, then the 10 expression (1 − ω)(1 − ω2 )(1 + ω4 )(1 + ω8 ) is equal to a 0 b 3 c 1 d 2 2 x3m + x3n − 1 + x3r − 2, where m, n, r ∈N, is divisible by a x2 − x + 1 b x2 + x + 1 c x2 + x − 1 d None of these 11 3 If ω is a non-real cube root of unity, then is equal to 12 a − 1 b 2ω c 0 d − 2ω 4 If ( 3 + i )n = ( 3 − i )n, n ∈N, then least value of n is a 3 b 4 c 6 d None of these 13 5 If x3 − 1 = 0 has the non-real complex roots α β, , then the value of (1 + 2α + β)3 − (3 + 3α + 5β)3 is a − 7 b 6 c − 5 d 0 ⇒ = 11/ n z + 1 z ⇒ is nth root of unity. z + 1 the value of 2 (a) 4 (c) 1 is equal to (b) 2 (d) None of these ⇒ 1 ⇒ | | = 1 z + 1 |z + 1| ⇒ | |z =|z + 1| − π ∴ + + = + + + − − + − − α α α α α α α α + + + + − = α α α α α = − − − = = = × = 4 α α α α α = − ⇒
4 X 138 6 If ( 3 − i )n = 2n, n ∈I, the set of integers, then n is 14 a multiple of a 6b 10 c 9 d 12 7 If z is a complex number satisfying z4 + z3 + 2 z2 + z + 1 = 0, then| z| is equal to 15 a b c 1 None of these 8 If z is a non-real root of − 1, then z86 + z175 + z289 is equal to 16 a 0 b − 1 c 3 d 1 9 Non-real complex number z satisfying the equation z3 + 2 z2 + 3z + 2 = 0 are a c 2 2 If α is the non-real nth root of unity, then 1 + 3α + 5α 2 + + (2n − 1) α n − 1 is equal to n 1 − α d None of these −1∞ is equal to, where ω is the imaginary cube of root of unity and i = − 1. a ω orω2 b − ω or − ω2 c 1 + i or 1 − i d −1 + i or −1 − i If α = e i 2π / n, then (11 − α )(11 − α 2 ) (11 − α n − 1) is equal to 1 bc 11n − 1 − 1 d 11n − 1 − 1 a 11n − 10 11 The complex number w satisfying the equation ω3 = 8 i and lying in the IInd quadrant on the complex plane is 3 1 a − 3 + i b − + i 2 2 c −2 3 + i d − 3 + 2 i Let z be a complex number satisfying the equation z6 + z3 + 1 = 0. If this equation has a root re i θ with 90° < θ < 180°, then the value of θ is a 100° b 110° c 160° d 170° If ω is an imaginary cube root of unity, then the value of (p + q)3 + (pω + qω2 3) + (pω2 + qω)3 is a p3 + q 3 b 3(p3 + q 3) c 3(p3 + q 3) − pq p( + q ) d 3(p3 + q 3) + pq p( + q ) If z2 − z + 1 = 0, then the value of 1 2 2 1 2 3 1 2 z + z z + 2 z+ z3 z +  z24 + 1 24 is equal to z a 24 b 32 c 48 d None of these − ± − + − − − + − − −α −α − − − −
4 139 27 If p = a + bω + cω2 q = b + cω + aω2 and r = c + aω + bω2, where a, b, c ≠ 0 and ω is the complex cube root of unity, then a p + q + r = a + b + c b p2 + q 2 + r2 = a2 + b2 + c2 c p2 + q 2 + r2 = − 2(pq + qr + rp) d None of the above 18 If a and b are imaginary cube of unity, then α n + βn is equal to roots a 2cosb cos c 2i sind i sin 19 If the six solutions of x6 = − 64 are written in the form a + bi, where a and b are real, then the product of those solutions with a > 0 is a 4 b 8 c 16 d 64 20 If cos θ + i sin θ is a root of the equation xn + a x1 n − 1 + a x2 n − 2 + + an − 1x + an = 0, then n the value of ∑ ar cos r θ is r = 1 a 0 b 1 c − 1 d None of the above 21 If ω is a complex nth root of unity, then n ∑ (ar + b)ωr − 1 is equal to r = 1 a n n( + 1)a b nb 22 If α, β respectively are the fifth and fourth non-real roots of unity, then the value of (1 + α )(1 + β)(1 + α 2 )(1 + β2 )(1 + β3 )(1 + α 3 ) is a 0 b (1 + α + α2)(1 − β2) c(1 + α)(1 + β + β2) d 1 2 − n na 23 When the polynomial 5x3 + Mx + N is divided by x2 + x + 1, the remainder is 0. The value of (M + N) is equal to a −3 b 05 c −5 d 15 24 If z and w are two complex numbers simultaneously satisfying the equations, z3 + w5 = 0 and z2 ⋅ w 4 = 1, then a z and w both are purely real b z is purely real and w is purely imaginary c w is purely real and z is purely imaginary dz and w both are imaginary 25 Number of ordered pairs (z, ω) of the complex numbers z and ω satisfying the system of equations, z3 + ω7 = 0 and z5 ⋅ω11 = 1is a 7 b 5 c 3 d 2 26 If 1, z 1, z 2, z 3, …, z n − 1 is the nth roots of unity and w is a non-real complex cube root of unity, n − 1 then the product of ∏ (ω − zr ) is cannot be r = 1 equal to a 0 b 1 c −1 d 1 + ω 27. If Zr , r = 1, 2, 3, …, 50 are the roots of the equation, then the value of cd None of these n − 1 r = 0 is r = 1 r a − 85 b− 25 c 25 d 75 π π π π
4 140 Z − 1 X Example 47. If A, B, C are three points in the argand plane representing the complex numbers z1, z2, z3 such that z1 = , where λ ∈R, then the distance of A from the line BC is (a) λ (b) (c) 1 (d) 0 Sol. (d) As z1 = which shows z1 divides z2, z3 in the ratio of 1: λ. Thus, the points are collinear. ∴Distance of A from line BC is zero. X Example 48. Find the relation, if z1, z2, z3, z4 are the affixes of the vertices of a parallelogram taken in order. Sol. As the diagonals of a parallelogram bisect each other, therefore affix of the mid-point of AC is same as the affix of the mid-point of BD. z1 + z3 = z2 + z4 i.e. 2 2 ⇒ z1 + z3 = z2 + z4 Equation of the Straight Line Equation of the Line Passing through the Points z1 and z2 Let z be any point on the line joining z1 and z2, then z − z1 arg = π or 0 z2 − z1 z − z1 ⇒ must be real. z2 − z1 ⇒ z z( 1 − z2) − z z( 1 − z2 ) + z z1 2 − z z2 1 = 0 General Equation of a Line az + az + b = 0, represents a straight line, where b is a real number and a is a complex number. Geometrical Applications of Complex Numbers Basic Concepts in Geometry Distance formula The distance between two points P z( 1 ) and Q z( 2 ) is given by PQ = | z2 − z1| = | affixof Q − affix of P | Qz( 2) Pz( 1) For any complex number z, | z | = | z − 0 | = | z − (0 + i0)| Thus, modulus of a complex number z represented by a point in the argand plane is its distance from origin. X Example 46. Length of the line segment joining the points −1 − i and 2 + 3i is (a) −5 (b) 15 (c) 5 (d) 25 Sol. (c) Let z1 = − 1 − i and z2 = 2 + 3 i Then, required distance = | z2 − z1 | = | 2 + 3i + 1 + i | = 5 Section formulae If R z( ) divides the line segment joining P z( 1 ) andQ z( 2 ) in the ratio m1 :m2 (m1, m2 > 0), then m z + m z (i) For internal division, z = 1 2 2 1 m1 + m2 m z − m z (ii) For external division, z = 1 2 2 1 m1 − m2 If R z( ) is the mid-point of PQ, then affix of R is z1 + z2 2 ∴ Requiredequationis z z z z z z z z − − = − − 1 2 1 1 2 1 ⇒ z z z z z z 1 1 1 0 1 1 2 2 =
4 141 1 2 2 1 Parametric Equation of a Line z = z1 + t z( 2 − z1 ), where t is real parameter = (1 − t z) 1 + t z2, represents the complete line through z1and z2. X Example 49. Find the general equation of line joining the points z1 = (1 + i) and z2 = (1 − i). Sol. Clearly, the equation of a line is given by z z( 1 − z2 ) − z z( 1 − z2 ) + z z1 2 − z z2 1 = 0 where, z1 = 1 + i and z2 = i − i On substituting the values of z1 and z2, we get z(1 − i − 1 − i) − z(1 + i − 1 + i) + (1 + i)(1 + i) − (1 − i)(1 − i) = 0 ⇒ z( − 2i) − z(2i) + (1 − 1 + 2i) − (1 − 1 − 2i) = 0 ⇒ − 2iz − 2iz + 4i = 0 ⇒ z + z − 2 = 0, which is the required equation. Condition of Collinearity Three points z1, z2 and z3 are collinear, if 1 = 0 X Example 50. If z1, z2, z3 are three complex numbers such that 5z1 −13z2 + 8z3 = 0, then prove that ` 1 = 0. Sol. 5z1 − 13z2 + 8z3 = 0 ⇒ = z2 ⇒ z1, z2 and z3 are collinear. z1 z1 1 ⇒ z2z2 1 = 0 [condition of collinear points] z3 z3 1 Hence proved. Length of Perpendicular The length of perpendicular from a point z1 to az + az + b = 0 is given by az1 + az1 + b 2 a X Example 51. The length of perpendicular from P(2 − 3 )i to the line (3 + 4 )i z + (3 − 4 )i z + 9 = 0 is equal to (a)9 (b) (c) (d) None of these Sol. (c) Let PM be the required length, then PM = |(2 − 3i)(3 + 4i) + (3 − 4i)(2 + 3i) + 9| = 45 9 10 2 Slope of a Line Slope of the Line Segment Joining Two Points If A, B represent complex numbers z1, z2 in the argand plane, then the complex slope of AB is defined by z1 − z2 z − z Re (z − z ) and real slope is defined by . Im (z2 − z1 ) Slope of Line az + az + b = 0 The complex slope of the line −a −Coefficient of z az + az + b = 0 is = a Coefficient of z and real slope of the line az + az + b = 0 is Re (a) −i a( + a) – = Im ( )a (a − a) Ø ● If w1 and w2 are the complex slope of two lines on the argand plane, then the lines are (a)perpendicular, if w1 + w2 = 0 (b)parallel, if w1 = w2 ● The equation of a line parallel to the line az + az + b = 0 is az + az + λ = 0, where λ ∈R . ● The equation of a line perpendicular to the line az + az + b = 0 is az − az + λ =i 0, where λ ∈R . z z z z z z 1 1 2 2 3 3 1 1 z z z z z z 1 1 2 2 3 3 1 1 − =
4 142 X Example 52. If a point z1 is the reflection of a point z2 through the line b z + bz = c b, ≠ 0 in the argand plane, then bz2 + b z1 is equal to (a) 4c (b)2c (c) c (d) None of these Sol. (c) If P (z1)is the reflection of Q z( 2 )through the line bz + bz = c in the argand plane. Then, R z1 + z2 lies on the line. 2 b z1 + z2 + b z1 + z2 = c 2 2 ⇒ b z1 + b z1 + b z2 + b z2 = 2c …(i) Since, PQ is perpendicular to the line bz + bz = c. Therefore, Slope of PQ + Slope of the line = 0 z2 − z1 b = 0 ⇒ z2 − z1 b ⇒ b (z2 − z1) − b ( z2 − z1) = 0 ⇒ b z2 − bz1 − bz2 + bz1 = 0 …(ii) Adding Eqs. (i) and (ii), we get 2(bz1 + bz2 ) = 2c ⇒ bz1 + bz2 = c Concept of Rotation In this section, we shall learn about the effect of multiplication of a complex number by eiα which will also be interpreted geometrically. Complex Number as a Rotating Arrow in the Argand Plane 1. Let z = r (cosθ + isin θ) = reiθ be a complex number, represented by a point P in the argand plane. Then, OP = r and ∠XOP = θ Now, zeiα = reiθ ⋅ eiα = rei (θ + α) This shows that zeiα is the complex number whose modulus is r and argument θ + α. Clearly, zeiα is represented by a point Q in the argand plane such that OQ = r and ∠XOQ = θ + α. In other words, to obtain the point representing zeiα , we rotate OP through angle α in anti-clockwise sense. Thus, multiplication by eiα to z rotates the vector OP in anti-clockwise sense through an angle α and vice-versa. Similarly, multiplication of z with e− αi will rotate the vector OP in clockwise sense. Remark Let z1 and z2 be two complex numbers represented by points P and Q in the argand plane, such that X¢X ∠POQ = θ. Then, z e1 iθ is vector of magnitude | z1| = OP along OQ and z e1 iθ is a unit vector | z1| z eiθ along OQ. Consequently, | z |⋅ 1 2 is a vector of | z1| magnitude | z2| = OQ along OQ. i.e. z= | z2| z ei θ ⇒ z = z2 ⋅ z ei θ 2 1 21 Pz ( ) 1 z+z 1 2 2 R bz+bz=c Qz ( ) 2 ( ) Y O Qze i ( ) α Pz () Y′ X′ X x α θ Y O Q (z ) 2 P(z ) 1 Y¢ q
4 143 | z1| z1 X Example 53. The point represented by the Then, z3 − z1 = OQ (cosα + isin α) complex number 2 − i is origin z2 − z1 OP rotated about CA iα through an anglein the clockwise direction, the = e BA new position of point is = | z 3 − z 1 | iα Sol. (b) Here, z = 2 − i Let z1 be the required complex number. ∴z1 = (2 − i) cos 2 + i sin 2 = (2 − i) cos π – i sin 2 2 = (2 − i)(0 − i) = − (2i − i 2 ) = – 2i − 1 ∆OPQ and ∆ABC are congruent, X Example 54. A particle P starts from the point OQ CA z0 =1 +2i, where i = −1. It moves first horizontally ∴ OP = BA away from origin by 5 units and then vertically z3 − z1 away from origin by 3 units to reach a point z1. or amp From z1, the particle moves 2 units in the z2 − z1 direction of the vector  i +  j and then it moves X Example 55. A man walks a distance of through an angle π/2 in anti-clockwise direction on 3 units from the origin towards the North-East a circle with centre at origin to reach a point z2. (N 45° E) direction. From there, he walks a The point z 2 is given by distance of 4 units towards the North- West (a)6 + 7i (b) −7 +6i (N 45°W) direction to reach a point P. Then, the (c) 7 + 6i (d) −6 + 7i position of P in the argand plane is (a)1 + 2i (c)2 + i (b) −1 − 2i (d) −1 + 2i Y e | z − z | 2 1  = −π z z Q ) – ( 1 3 z P z ( – ) 1 2 Az ) ( 1 Bz ( ) 2 Cz ) ( 3 X O α
4 144 Sol. (d) Imaginary axis (a)3eiπ/4 + 4i (b) (3 − 4i e) iπ/4 (c) (4 + 3i e) iπ/4 (d) (3 + 4i e) iπ/4 associated with A is 3 eiπ / 4. If z is the complex number associated with P, then 2 2 sin45°) = (7, 6) = 7 + 6i By rotation about (0, 0), z2 = eiπ / 2 ⇒ z2 = z2 eiπ2 z2′ z2 = (7 + 6i) cos π + i sin 2 2 = (7 + 6 )( )ii = − 6 + 7i 2. Let z1, z2 and z3 be the vertices of a ∆ABC described in anti- clockwise sense. Draw OP and OQ parallel and equal to AB and AC, respectively. Then, point P is z2 − z1 and Q is z3 − z1. If OP is rotated through ∠α in anti-clockwise sense, it ⇒ coincides with OQ. ⇒ z = (3 + 4i e) iπ / 4 X Example 56. The complex numbers z1, z2 and z1 − z3 1 − i 3 z3 satisfying =are the vertices of a z2 − z3 2 triangle which is (a) of area zero (b) right angled isosceles
4 145 1 2 = z1 − z3 z1 − z3 3 Hence, the triangle is an equilateral. Area of Triangle (i) Area of the triangle with vertices z1, z2 and z3 is (z2 − z3 )| z1 |2 sq unit. 4iz1 (ii) The area of triangle whose vertices are z, izand z + iz is | |z 2 . (iii) The area of triangle whose vertices are z, ωz and 3 2 z + ωz is| z | . 4 X Example 57. If the area of the triangle on the complex plane formed by the points z, iz and z + iz is 50 sq units, then | z | is (a) 5 (b) 10 (c) 15 (d) None of these Sol. (b) We know that, the area of the triangle formed by + iz is 2. the points z, iz and z ∴ | z| = 50 ⇒ | z| = 10 X Example 58. If the area of the triangle on the complex plane formed by complex numbers z, ωz and z + ωz is 4 3 sq units, then | z | is (a) 4 (b) 2 (c) 6 (d) 3 Sol. (a) The area of the triangle formed by z, ω z and z + ωz 3 2 is| |z . 4 ∴ | |z 2 = 4 3 4 ⇒ | |z = 4 Applications of Triangle 1. Centroid The centroid of the triangle (in the argand plane) formed by z1, z2 and z3 is given by 1 (z1 + z2 + z3 ) 3 2. Incentre The incentre of the triangle (in the argand plane), formed by z1, z2 and z3 is az1 + bz2 + cz3 , where a + b + c a = | z2 − z3|, b = | z3 − z1|, c = | z1 − z2| 3. Excentres The excentres of the triangle (in the argand plane), for meet by z1, z2 and z3 are given by −az1 + bz2 + cz3 (i) I1 = −a + b + c az1 − bz2 + cz3 (ii) I2 = a − b + c az1 + bz2 − cz3 (iii) I3 = a + b − c where, a = | z2 − z3|, b = | z3 − z1| and c = | z1 − z2| 4. Circumcentre The circumcentre of the triangle (in the argand plane), formed by z1, z2, z3 is given by 5. Orthocentre The orthocentre of the triangle (in the argand plane), formed by z1, z2, z3 is given by X Example 59. If z1, z2 and z3are affixes of the vertices A, B and C, respectively of a ∆ABC having centroid at G such that z = 0 is the mid-point of AG, then (a) z1 + z2 + z3 = 0 (b) z1 + 4z2 + z3 = 0 (c) z1 + z2 + 4z3 = 0 (d) 4z1 + z2 + z3 = 0 Sol. (d) The affix of G is z1 + z2 + z3 . Since, z = 0 is the 3
4 146 mid-point of AG. Therefore, affix of the mid-point of AG is 0. z1 + z2 + z3 + z1 ⇒ 3 = 0 ⇒ 4z1 + z2 + z3 = 0 1 + 1 X Example 60. The centre of a square ABCD is at the origin and point A is represented by z1. The centroid of ∆BCD is represented by z1 z1 (a) (b) − 3 3 iz1 iz1 (c) (d) − 3 3 Sol. (b) The affixes of the vertices B, C and D are iz1 − z1 and − iz1, respectively. Therefore, the affix of the centroid of ∆BCD is iz1 – z1 − iz1 = − z1 3 3 X Example 61. Let P e( i θ1 ),Q e( i θ2 ) and R e( i θ3 ) be the vertices of ∆PQR in the argand plane. Then, the orthocentre of the ∆PQR is (a) ei (θ1 + θ2 + θ3) (b) ei (θ + θ + θ12 3) (c) ei θ1 + ei θ2 + ei θ3 (d) None of these Sol. (c) We have, |ei θ1| =|ei θ2| =|ei θ3| = 1 ⇒ OP = OQ = OR = 1, where O is the origin. ⇒ Origin O is the circumcentre of ∆PQR. The affix of the centroid is (ei θ1 + ei θ2 + ei θ3 ) Let z be the affix of the orthocentre. Since, centroid divides the segment joining circumcentre and orthocentre in the ratio 1: 2. ⇒ z = ei θ1 + ei θ2 + ei θ3 6. Equilateral Triangle (i) The triangle whose vertices are the pointsz1, z2 and z3 on the argand plane, is an equilateral triangle, if z12 + z22 + z32 = z z1 2 + z z2 3 + z z3 1 1 1 1 or + + = 0 z1 − z2 z2 − z3 z3 − z1 (ii) If the complex numbers z1, z2 and z3 are the vertices of an equilateral triangle and z0 is the circumcentre of the triangle, then z12 + z22 + z32 = 3z02. 7. Isosceles Triangle (i) If z1, z2 and z3 are the vertices of a right angled isosceles triangle, then (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) (ii) If z1, z2 and z3 are the vertices of an isosceles triangle, right angled at z2, then z12 + z22 + z32 = 2z2 (z1 + z3 ) X Example 62. Prove that the complex numbers z1, z2and the origin form an equilateral triangle only, if z12 + z22 − z z1 2 = 0. Sol. If z1, z2 and z3 form an equilateral triangle, then ⇒ z z z z z z z z z ⇒ z z z z z z ⇒ z + z = z z⇒z z z z Hence proved. X Example 63. Let z1and z2be two complex z1 z2 numbers such that + =1, then z2 z1 (a) z1, z2are collinear (b) z1, z2 and the origin form a right angled triangle (c) z1, z2 and the origin form an equilateral triangle (d) None of the above Sol. (c) We have, z1 + z2 = 1⇒ z12 + z22 = z z1 2 z2 z1 ⇒ z12 + z22 + z32 = z z12 + z z1 3 + z z2 3, where z3 = 0 ⇒ z1, z2 and the origin form an equilateral triangle. O D iz (– ) 1 C z (– ) 1 Biz ( ) 1 Az ( ) 1
4 147 Circle Equation of a Circle (i) The equation of a circle whose centre is at pointhaving affix z0 and radius r, is | z − z0 | = r. (ii) If the centre of the circle is at origin and radius r, then its equation is | z | = r. (iii) | z − z0 | < r represents interior of a circle | z − z0| = r and | z − z0 | > r represents exterior of the circle | z − z0| = r. (iv) General equation of a circle The general equation of the circle is zz + az + az + b = 0, where a is complex number and b ∈R. ∴ Centre and radius are −a and | a |2 − b, respectively. (v) Equation of circle in diametric form If end points of diameter represented by A z( 1 )and B z( 2 ) and P z( ) is any point on the circle, then (z − z1 )(z − z2 ) + (z − z2 )(z − z1 ) = 0 which is required equation of circle in diametric form. X Example 64. A circle whose radius is r and centre z0, then the equation of the circle is (a) zz − zz0 − zz0 + z z0 0 = r2 (b) zz + zz0 − zz0 + z z0 0 = r2 (c) zz − zz0 + zz0 − z z0 0 = r2 (d) None of the above Sol. (a) Equation of circle|z − z0|2 = r2 ⇒ (z − z0 )(z − z0 ) = r2 ⇒(z − z0 )(z − z0 ) = r2 zz − zz0 − zz0 + z z0 0 = r2 . X Example 65. The set of values of k for which the equation zz + ( 3− + 4 )i z − (3 + 4 )i z + k = 0 represents a circle is (a) (−∞ , 25) (b) (25, ∞) (c) (5, ∞) (d) (−∞, 5) Sol. (a) We have, zz + −( 3 + 4 )i z − (3 + 4 )i z + k = 0 This equation represents a circle with centre a + (3 − 4 )i and Radius = ( 3−− 4 )i 2 − k = 25 − k For circle to exist, we must have 25 − k > 0 ⇒ k < 25 Hence, the given equation will represent a circle if k < 25. Loci in Complex Plane If z is a variable point and z1, z2 are two fixed points in the argand plane, then (i) | z − z1| = | z − z2| ⇒ Locus of z is the perpendicular bisector of the line segment joining z1and z2. (ii) | z − z1| + | z − z2| = k, if | k | > | z1 − z2| ⇒ Locus of z is an ellipse. (iii) | z − z1| + | z − z2| = | z1 − z2| ⇒Locus of z is the line segment joining z1 and z2 (iv) | z − z1| − | z − z2| = | z1 − z2| ⇒ Locus of z is a straight line joining z1and z2 but z does not lie between z1 and z2. (v) | z − z1| – | z − z2| = k, where k <| z1 − z2| ⇒ Locus of z is a hyperbola. (vi) | z − z1|2 + | z − z2|2 = | z1 − z2|2 ⇒ Locus of z is a circle with z1 and z2 as the extremities of diameter. (vii) | z − z1| = k z| − z2|, (k ≠1) ⇒ Locus of z is a circle. z − z1 (viii) arg = α (fixed) z − z2 ⇒ Locus of z is a segment of circle. z − z1 (ix) arg = ± π / 2 z − z2 ⇒ Locus of z is a circle with z1and z2 as the vertices of diameter. z − z1 (x) arg = 0 or π z − z2 ⇒ Locus of z is a straight line passing through z1 and z2. X Example 66. The complex numbers z = x + iy z − 5i which satisfy the equation=1, lie on z + 5i (a) the X-axis (b) the straight line y = 5 r Cz ( ) 0 Pz ()
4 148 (c) a circle passing through the origin (d) None of the above Sol. (a) Given, z − 5i = 1 ⇒ |z − 5i| =|z + 5i| z + 5i [if|z − z1| =|z – z2|, then it is a perpendicular bisector of z1 and z2] ∴ Perpendicular bisector of (0, 5) and (0, − 5) is X-axis. X Example 67. The equation | z − i| + | z + i| = k k,> 0 can represent an ellipse, if k 2 is (a) <1 (b) < 2 (c) > 4 (d) None of these Sol. (c)| z − z1| + | z – z2 | = k represents ellipse, if | k|>|z1 − z2| Thus,| z − i | + | z − i | = k represents ellipse, if | k|>|i + i| or | k|>|2i| ∴ | k|> 2 or k2 > 4 X Example 68. The equation | z + i| − | z − i| = k represents a hyperbola if (a) −2 < k < 2 (b) k >2 (c)0 < k < 2 (d) None of these Sol. (a)|z − z1| −|z − z2| = k, represents hyperbola, if| k|<|z1 − z2| Thus, |z + i| −|z − i| = k, represents hyperbola, if | k|<| i + i | or| k|< 2 ⇒ −2 < k < 2 X Example 69. If z =1 − t + i t 2 + t + 2, where t is a real parameter. The locus of z in the argand plane is (a) a hyperbola (b) an ellipse (c) a straight line (d) None of these Sol. (a) x + iy = 1 − t + i ⇒ x = 1 − t , y = Eliminating t, y2 = t 2 + t + 2 = (1 − x)2 + 1 − x + 2 2 = x − 3 + or y2 − x − 7 X 2 4 which is a hyperbola. X Example 70. Identify the locus of z, if ⇒ | z − a|2 = r2 ⇒ | z − a| = r X Hence, locus of z is circle having centre a and radius r. X Example 71. If the equation | z − z1|2 +| z − z2|2 = k represents the equation of a circle, where z1 = 2 + 3i, z2 = 4 +3i are the extremities of a diameter, then the value of k is (a) (b) 4 (c) 2 (d) None of these Sol. (b) As z1 and z2 are the extremities of diameter. ⇒ |z − z1|2 + |z − z2|2 =|z1 − z2|2 ⇒ k =|z1 − z2|2 =|2 + 3i − 4 − 3i|2 =| − 2|2 = 4 X X Example 72. If | z +1| = 2 | z – |1, then the locus described by the point z in the argand diagram is a (a) straight line (b) circle (c) parabola (d) None of these Sol. (b)| z + 1| = 2 |z − 1| Putting z = x + iy ⇒ x + iy + 1 = 2|x + iy − 1| ⇒ |(x + 1) + iy| = 2 |(x − 1) + iy| ⇒ (x + 1)2 + y2 = 2[(x − 1)2 + y2 ] ⇒ x2 + y2 − 6x + 1 = 0 which is the equation of a circle. Y (0, 5) (0,– 5) Y¢ X¢ X O z a r a z = + − > 2 0 , . = + − ⇒ − = − ⇒ − − = + + + + =
4 149 π + − π ≤ ≤ z −1 X Example 73. The locus of z given by=1 z − i is (a) a circle (b) an ellipse (c) a straight line (d) a parabola ⇒ 2 x = 2 y or x − y = 0 which is the equation of a straight line. Example 74. If z = x + iy and | z − 2 + i| = | z − 3 − i|, then locus of z is (a)2x + 4y − 5 = 0 (b)2x − 4y − 5 = 0 (c) x + 2y = 0 (d) x − 2y + 5 = 0 Sol. (a) |z − 2 + i| =|z − 3 − i| ⇒ |(x − 2) + i (y + 1)| =|(x − 3) + i (y − 1)| ⇒ (x − 2)2 + (y + 1)2 = (x − 3)2 + (y − 1)2 ⇒ x2 + 4 − 4x + y2 + 1 + 2 y = x2 + 9 − 6x + y2 + 1 − 2 y ⇒ 2 x + 4y − 5 = 0 Example 75. If z = z0 + A z( − z0 ), where A is a constant, then prove that locus of z is a straight line. Sol. z = z0 + A (z − z0 ) ⇒ Az − z − Az0 + z0 = 0 …(i) ⇒ A z− z − Az0 + z0 = 0 …(ii) Adding Eqs. (i) and (ii), we get (A − 1) z + (A − 1) z − (Az0 + Az0 ) + z0 + z0 = 0 This is of the form az + az + b = 0, where a = A − 1 and b = − (Az0 + Az0 ) + z0 + z0 ∈R Hence, locus of z is a straight line. Example 76. Plot the region represented by π z +1 2π ≤ arg in the argand plane. 3 z −1 3 Sol. Let us take z + 1 2 π z arg , clearly z − 1 3 lies on the minor arc of the circle passing through (1, 0) and (−1 0, ). z + 1 π Similarly, arg z − 1 3 means that z is lying on the major arc of the circle passing through (1,0) and (−1 0, ). Now, if we take any point in the region included between the two arcs, say P z1( 1), we get ≤ arg 2 π π z + 1 Thus, represents the shaded region 3 z − 1 3 excluding the points (1, 0) and (−1 0, ). Some Important Results i. Ø Points z1, z2, z3 and z4 (not necessarily in order) will be concyclic, z2 − z4 z1 − z3 if is positive or negative. z1 − z4 z2 − z3 X Example 77. Show that the points 3 + 4i, 3 − 4i, −4 + 3i, −4 − 3i are concyclic. Sol. Let z1 = 3 + 4i, z2 = 3 − 4i, z3 = − 4 + 3i and z4 = − 4 − 3i Then, z2 − z4 z1 − z3 =(7 − i)(7 + i) z1 − z4 z2 − z3 (7 + 7i)(7 − 7i) = 50 = 25 49 + 49 49 = a real number Hence, the given points are concyclic. X Example 78. If z1, z2 and z3 are complex Four points z1, z2, z3 and z4 in anti-clockwise order will be concyclic, if and only if z − z z − z θ = arg 2 4 = arg 2 3 z1 − z4 z1 − z3 z − z z − z ⇒ arg 2 4 −arg 2 3 = 2nπ , n ∈I z1 − z4 z1 − z3 z − z z − z ⇒ arg 2 4 1 3 = 2nπ z1 − z4 z2 − z3 z − z z − z 2 4 1 3 is real and positive. z1 − z4 z2 − z3 − − = ⇒ − − = ⇒ |( x x + = − + − ⇒ x x − + = + − (1, 0) (–1, 0) 2 /3 π π/3
4 150 2 1 1 numbers, such that = + , then show that z1 z2 z3 the points represented by z1, z2 and z3 lie on a circle passing through the origin. ⇒ 2 1 = 1 3 ⇒ 2 1 = − 2 z z1 2 z z3 1 z3 − z1 z3 z2 − z1 = arg z2 ⇒ arg z3 − z1 z3 z2 − z1 = π + arg z2 ⇒ arg z3 − z1 z3 z2 − z1 = π − arg z3 ⇒ arg z3 − z1 z2 ⇒ α = π − β ⇒ α + β = π Hence, the points are concyclic. ii. X Example 79. greatest and least values of | |z . ∴ X Example 80. Let z1 and z2 be two non-real complex cube roots of unity and z − z1 2 + z − z2 2 = λ be the equation of a circle with z1, z2 as ends of a diameter, then the value of λ is = = Logarithm of Complex Numbers Let z = α + iβ = rei(θ + 2nπ) log z = log(rei(θ + 2nπ) ) = log r + i(θ + 2nπ) = log| |z + iarg z + 2n iπ If we put n = 0, we get principal value of log z. ∴ Principal value of log z = log| |z + i arg z. If z + = a, then the greatest and least If 4 z + z = a, then find the z 1 valuesof z arerespectively a a + + 2 4 2 and − + + a a 2 4 2 + = ⇒ − = − − − − Rz ( ) 2 S(0) Qz ) ( 1 Pz ) ( 3 β α = + = + = = + + = + + = =− + +
4 151 π π π Work Book Exercise 4.4 1 z − 1 = 1, represents z + 1 a a circle b an ellipse c a straight line d None of these 2 | z − 4| <| z − 2|, represents the region given by a Re ( )z > 0 b Re ( )z < 0 c Re ( )z > 2 d None of these 3 If 2 z1 − 3z2 + z3 = 0, then z1, z2, z3 are represented by a three vertices of a triangle b three collinear points c three vertices of a rhombus d None of the above 4 If z = x + iy, such that| z + 1| =| z − 1| and z − 1 π amp , then z + 1 4 a x = 2 + 1, y = 0 b x = 0, y = 2 + 1 c x = 0, y = 2 − 1 d x = 2 − 1, y = 0 5 If z8 = (z − 1)8, then the roots of this equation are a collinear b concyclic c the vertices of irregular polygon d None of the above 6 The equation zz + az + az + b = 0, b ∈R represents a circle, if a | |a 2 = b b | |a 2 ≥ b c | |a 2 < b d None of these 7 If z = (λ + 3) + i locus of z is 3 − λ2 , where|λ| < 3, then a circle b parabola c line d None of these 8 If a point P denoting the complex number z moves on the complex plane such that |Re (z)| + |Im (z)| = 1, then the locus of z is a a square b a circle c two intersecting lines d a line 9 The figure formed by four points 1 + 0i, −1 + 0i, 25 3 + 4i and on the argand plane is −3 − 4i a a parallelogram but not a rectangle b a trapezium which is not equilateral c a cyclic quadrilateral d None of the above 10 If i = −1 , then define a sequence of complex number by z1 = 0, zn + 1 = zn 2 + i for n ≥ 1. In the complex plane, how far from the origin is z111 ? a 1 b 2 c 3 d 110 11 The complex numbers whose real and imaginary parts are integers and satisfy the relation zz3 + z z3 = 350, forms a rectangle on the argand plane, the length of whose diagonal is a 5 units b 10 units c 15 units d 25 units 12 The locus of z, for arg z = − is a same as the locus of z for arg z = b same as the locus of z for arg z = c the part of the straight line 3x + y = 0 with y < 0, x > 0 d thepartofthestraightline3x + y = 0 with y > 0, x < 0 13 If z1 and z2 are two complex numbers and z1 + z2 = π but| z1 + z2| ≠| z1 − z2|, then the arg z1 − z2 2 figure formed by the points represented by 0, z1, z2 and z1 + z2 is a a parallelogram but not a rectangle or a rhombus b a rectangle but not a square c a rhombus but not a square d a square 14 z1 and z2 are two distinct points in an argand plane. If a z| 1| = b z| 2|, where a, b ∈R, then the az1 + bz2 is a point on the point bz2 az1 a line segment [−2, 2] of the real axis b line segment [−2, 2] of the imaginary axis c unit circle| z| = 1 d the line with arg z = tan−1 2 15 The roots z1, z2, z3 of the equation z3 + 3αz2 + 3βz + γ = 0 correspond to the points ⇒ z = e−π / 2 Example 82. If z = ilog (2e − value of cos z is 3), then the (a)2 (b) −2 (c)2i (d) −2i Example 81. The value of ii is (a) e−π 2/ (b) eπ / 2 (c) eπ/4 (d) None of these Sol. (a) Let z = i i, loge z = loge i i = i loge i = i loge eiπ / 2 = i 2 loge e = −
4 152 α A, B and C on the complex plane. Then, the triangle is equilateral, if a α2 = β b α = β2 c α2 = 3β2 d 3α2 = β2 16 If z1 2 + z2 2 + 2 z z1 2 cos θ = 0, then the points represents by z1, z2 and the origin form a equilateral triangle b right angled triangle c isosceles triangle d None of these 17 Complex numbers z1, z2, z3 are the vertices A, B and C, respectively of an isosceles right angled triangle, right angled at C. Then, (z1 − z2 )2 is equal to a 2(z1 − z3)(z3 − z2) b (z1 + z3)(z3 − z1) c ( z 1 − z 3)( z 3 − z 2) d None of these 18 A, B and C are points represented by complex numbers z1, z2 and z3. If the circumcentre of the ∆ABC is at the origin and the altitude AD of the triangle meets the circumcircle again at P, then the complex number representing point P is z z1 2 b z = − z z3 2 a z = z3 z1 z z1 2 d z = z z3 2 c z = − z3 z1 19 If z1 and z2 are the roots of the equation z2 + pz + q = 0, where the coefficients p and q may be complex number. Let A and B represent z1 and z2 in the complex plane. If ∠AOB = α ≠ 0 and OA = OB, where O is the origin, then p2 is equal to a 4q cos2 α b 2q cos2α 2 c q cos d None of these 20 If Re z + 4 = 1, then z is represented by a 2 z − i 2 point lying on a a circle b an ellipse c a straight line d None of these 21 Suppose z1, z2 and z3 are the vertices of an equilateral triangle inscribed in the circle| z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the clockwise sense, then a z2 = 1 − 3i, z3 = 2 b z2 = 2, z3 = 1 − 3i c z2 = − 1 + 3i, z3 = − 2 d None of the above 22 Suppose z , z and z are the vertices of an 1 2 3 equilateral triangle circumscribing in the circle | z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the anti-clockwise sense, then z2 is a b c 3 i ) d None of the above 23 The straight line (1 + 2 )i z + (2i − 1)z = 10 i on the complex plane, has intercept on the imaginary axis equal to a 5 b c − d −5 24 If the complex number z satisfies the condition | z| ≥ 3, then the least value of z + is equal to 5 ab cd None of these 25. If z1 = 2 , z3 = a − bi, for a, b ∈R and 1 − i 2 + i z1 − z2 = 1, then the centroid of the triangle formed by the points z1, z2, z3 in the argand plane is given by a b (1 + 7i ) c (1 − 3i ) d (1 − 3i ) 26. Let λ ∈R. If the origin and the non-real roots of 2 z2 + 2 z + λ = 0 form three vertices of an equilateral triangle in the argand plane, then λ is a 1b 2 /3 c 2 d −1 27 If a, b, c and u, v, w are complex numbers representing the vertices of two triangles, such that c = (1 − r a) + rb and w = (1 − r u) + rv . When r is a complex number, then the two triangles a have the same area b are similar c are congruent d None of these 28. On the complex plane, ∆OAP and ∆OQR are = 1. If similar and I OA( ) the points P and Q denote the1) complex numbers z1 and z2, then the complex numbersX z denoted by the point R is given by z1 a z z1 2 b z2 c z2 d z1 + z2 z1 z2 29 Intercept made by the circle zz + αz + αz + r = 0 on the real axis on complex plane, is a (α + α −) r b (α + α)2 − 2r c (α + α)2 + r d (α + α)2 − 4r − − = Rz () Pz ( Qz ( ) 2 θ θ O A (1, 0) Y
4 153 30 The equation of the radical axis of the two circles represented by the equations,| z − 2| = 3 and | z − 2 − 3 i| = 4 on the complex plane is a 3y + 1 = 0 b 3y − 1 = 0 c 2 y − 1 = 0 d None of the above
154 α α 2 2 WorkedOut Examples Type 1. Only One Correct Option Ex 1. If a = cosα + isin α, b = cosβ + isin ,β a b c c = cos γ + isin γ and + + =1, then b c a cos (α − β +) cos (β − γ ) + cos (γ − α) is equal to (a) (b) − (c) 0 (d) 1 + + Sol. cis β cis γ = 1, cis γ cis α where, cis θ represents cos θ + i sin θ. ⇒ cis(α − β) + cis(β − γ) + cis(γ − α) = 1 Equation real parts of both sides cos (α + β) + cos (β − γ) + cos(γ − α) = 1 Hence, (d) is the correct answer. Ex 2. The locus of the centre of a circle which touches the circles | z − z1| = a and | z − z2 | = b externally (z z, 1 and z2 are complex numbers) will be (a) an ellipse (b) a hyperbola (c) a circle (d) None of these Sol. Let A z( 1), B z( 2)be the centres of given circles and Pbe the centre of the variable circle which touches given circles externally, then |AP| = a + r and |BP| = b + r, where r is the radius of the variable circle. On subtraction, we get |AP|− |BP| = a − b ⇒ ||AP | − | BP || = | a − b |, a constant. Hence, locus of P is (i) right bisector AB, if a = b. (ii) a hyperbola, if | a − b | < |AB| = |z2 − z1 |. (iii) an empty set, if |a − b| > |AB | = | z2 − z1 |. (iv) set of all points on line AB except those which lie between A and B, if |a − b| = |AB| ≠ 0. Hence, (d) is the correct answer. Ex 3. If arg (z1 ) = arg (z2 ), then (a) z2 = kz1 −1 , k > 0 (b) z2 = kz1 , k > 0 (c) | z2 | = | z1 | (d) None of these z z1 1 = |z1|2 z1−1 Sol. z1 = z1 ⇒ arg (z1 −1 ) = arg(z1) = arg(z2) ⇒ z2 = kz1 −1 (k > 0) Hence, (a) is the correct answer. tan α − i sin + cos Ex 4. Ifis purely α 1 + 2isin 2 i maginary, then α is given by π (a) nπ + (b) nπ − (c) (2n + 1) π (d) 2nπ + α tanα − i sin + cos 1 − 2isin 2 0 2 α ⇒ sinα(1 − cosα) = cosα(1 − cosα) 4 4 π ⇒ Re tan cos sin sin c sin sin tan α α α α α α α − + − − ⋅ + + 2 2 2 2 2 2 2 i os sin α α 2 1 4 2 0 2 + = ⇒ sin sin tanα α α + = 2 2 2 ⇒ sin cos sin cos α α α α − + = 1 ⇒ sin cos cos cos sin α⋅ α= α+ α− α 2 a b b c c a =1 ⇒ cis cis α β + + Re tan cos sin sin α α α α + − + = i i 2 2 2 1 2 0 ⇒ Re 2 2
155 ⇒ sinα = cosα, cosα = 1 ⇒ or α = 2nπ, n ∈I Hence, (a) is the correct answer. a − ib Ex 5. The expression tan ilog a + ib reduces to 2ab (a)(b) a2 − b2 ab 2ab (c)(d) a2 − b2 a2 + b2 a − ib Sol. Given, tan ilog a + ib Let a + ib = reiθ , r2 = a2 + b2 ⇒ a − ib = re− θi , tan θ = b a a − ib −2i θ ⇒= e a + ib a − ib −2iθ = 2θ ilog ilog(e ) a + ib a − ib ⇒ tan ilog a + ib tan2θ Hence, (b) is the correct answer. Ex 6. If z1 = a + ib and z2 = c + id are complex numbers such that | z1| = | z2| =1 and Re (z z1 2 ) = 0, then the pair of complex numbers a + ic = w1 and b + id = w2 satisfies (a) |w1 | ≠ 1 (b) |w2 | ≠ 1 (c) Re(w w12 ) = 0 (d) None of these Sol. z1 = a + ib z, 2 = c + id |z1| = |z2| = 1 ⇒ a2 + b2 = c2 + d2 = 1 w1 = a + ic w,2 = b + id As Re ( ) ⇒ ac + bd = 0 ⇒ ac = − bd w w12 = (a + ic) (b − id) = (ab + cd) + i bc(− ad) We have, a2 + b2 = c2 + d2 ⇒ a2 − c2 = d2 − b2 ⇒ a2 − c2 + 2i ac = d2 − b2 − 2ibd [as ac = − bd] ⇒ (a + ic)2 = (d − ib)2 ⇒ a + ic = (d − ib) or −d + ib ⇒ a = d and c = − b or a = − d,b = c ⇒ c2 + d2 = b2 + d2 a2 + c2 = a2 + b2 ⇒ a2 + c2 = 1, b2 + d2 = 1 ⇒ |w1| = |w2| = 1 Also, ab + cd = − cd + cd = 0 ⇒ Re (w w12) = 0 Hence, (c) is the correct answer. z1 Ex 7. If =1and arg (z z1 2 ) = 0, then z2 (a) z1 = z2 (b) | z2 |2 = z z1 2 (c) z z1 2 = 1 (d) None of these z1 = 1 Sol. Let z1 = r1 (cos θ1 + isinθ1), then z2 ⇒ |z1| = |z2| ⇒ |z1| = |z2| = r1 Now, arg (z1, z2) = 0 ⇒ arg(z1) + arg (z2) = 0 ⇒ arg (z2) = −θ1 Therefore, z2 = r1[cos(−θ1) + isin (−θ1)] = r1(cos θ1 − isinθ1) = z1 ⇒ z2 = (z1) = z1 ⇒ |z2|2 = z z1 2 Hence, (b) is the correct answer. Ex 8. If z is a complex number, then z2 + z 2 = 2 represents (a) a circle (b) a straight line (c) a hyperbola (d) an ellipse Sol. Let z = x + iy, then z2 + z2 = 2 ⇒ (x + iy)2 + (x − iy)2 = 2 ⇒ x2 − y2 = 1 which represents a hyperbola. Hence, (c) is the correct answer. ab b a 2 2 + = − 2 1 2 tan tan θ θ = − = − 2 1 2 2 2 2 2 ba b a ab a b /
4 156 Ex 9. If = A + iB, then A2 + B 2 equals (a) 1 (b) α2 (c) −1 (d) −α2 Sol. A + iB = ⇒ A − iB = ⇒ (A + iB) (A − iB) = = 1 ⇒ A2 + B2 = 1 Hence, (a) is the correct answer. Ex 10. If| z1| = | z2|and arg (z1 ) + arg (z2 ) = π/2, then (a) z z1 2 is purely real (b) z z1 2 is purely imaginary (c) (z1 + z2 )2 is purely real (d) arg(z1 −1 ) + arg(z2 −1 ) = Sol. Let |z1| = |z2| = r ⇒ z1 = r(cosθ + isin )θ and z2 = r cos 2 − isin 2 − ⇒ z z1 2 = r i2 , which is purely imaginary. z1 + z2 = r [(cosθ + sin )θ + i(cosθ + sin )]θ ⇒ (z1 + z2)2 = 2r2 ⋅ (cosθ + sin )θ 2 ⋅ i which is purely imaginary. Also, arg (z1 −1 ) + arg (z2 −1 ) = − Hence, (b) is the correct answer. Ex 11. The value of the expression 1 1 1 1 2 1 1 + ω2 + 3 2 2 + ω2 1 1 + 4 3 3 + 2 +… + (n +1) ω 1 1 n n + 2 , ω where ω is an imaginary cube root of unity, is n n( 2 + 2) n n( 2 − 2) (a) (b) 3 n n n (d) None of these Sol. tn = (n + 1) n + ω1 n + ω12 n3 + n2 ω 1 2 + ω 1 + 1 1 1 + n 1 + 2 + 1 ω = n3 + n2 (ω + ω2 + 1) + n(ω + ω2 + 1) + 1 = n3 + 1 n n 2 2 3 n n( + 1) ∴ Sn = ∑tr = ∑ (r + 1) = n r = 1 r = 1 Hence, (c) is the correct answer. Ex 12. If z1 and z2 are two complex numbers satisfying the equation z1 + iz2 =1, then z1 z1 − iz2 z2 is (a) purely real (b) of unit modulus (c) purely imaginary (d) None of the above Sol. (z1 + iz2)(z1 − iz2) = (z1 − iz2)(z1 + iz2) ⇒ z z1 2 = z z1 2 z1 = z1 ⇒ z2 z2 ⇒ z1 is purely real. z2 Hence, (a) is the correct answer. Ex 13. If z = − 2 + 2 3i, then z2n + 22n zn + 24n may be equal to (a) 22n (b) 0 (c) 3 4⋅ 2n , n is multiple of 3 (d) None of the above Sol. z = − 2 + 2 3i = 4ω ∴z2n + 22n nz + 24n = 42nω2n + 22n ⋅ 4n ⋅ωn + 24n = 42n[ω2n + ωn + 1]
157 x y y x m n + 1 y x n m = 0, if n is not a multiple of 3. = 3 4⋅ 2n , if nis a multiple of 3. Hence, (c) is the correct answer. Ex 14. The complex number z =1 + i is rotated through an angle 3π/2 in anti-clockwise direction about the origin and stretched by additional 2 units, then the new complex number is (a) − 2 − 2 i (b) 2 − 2 i (c) 2 − 2 i (d) None of the above Sol. If z1 is the new complex number, then |z1| = | |z + 2 = 2 2 z1 = |z1|⋅ ei3π/ 2 Also, z | |z 3π 3 ⇒ z1 = z⋅ 2 cos 2 + isin 2 = 2(1 + i)(0 − i) = − 2i + 2 = 2(1 − i) Hence, (d) is the correct answer. 1 1 Ex 15. If 2cosθ = x + and 2cosφ = y + , then x y (a) + = 2cos (θ + φ) (b) x y = 2cos (mθ + nφ) m n x y xm yn (c) + = 2cos (mθ + nφ) 1 (d) xy + = 2cos (θ − φ) xy 1 1 Sol. 2cosθ = x + , 2cosφ = y + x y ⇒ x2 − 2xcos θ + 1 = 0 2cos θ ± 4cos2 θ − 4 ⇒ = ⇒ + = 2cos(θ − φ) x ym n = 2cos(mθ + nφ) = 2cos(mθ − nφ) 1 xy + = 2cos(θ + φ) xy Hence, (b) is the correct answer. Ex 16. The complex numbers z1 and z2 are such that z1 ≠ z2 and | z1| = | z2|. If z1 has positive real part and z2 has negative imaginary part, then z1 + z2 may be z1 − z2 (a) zero (b) real and positive (c) real and negative (d) purely imaginary Sol. Given, |z1| = |z2| Re (z1) > 0, Im (z2) < 0 z1 + z2 = 1 z1 + z2 + z1 + z2 Re z1 − z2 2 z1 − z2 z1 − z2 1 (z1 + z2)(z1 − z2) + (z1 + z2)(z1 − z2) = 2 (z1 − z2)(z1 − z2) z z1 1 − z z1 2 + z z2 1 − z z2 2 + z z1 1 + z z1 2 1 = z z2 1 − z z2 2 2 |z1 − z2| z1 + z2 is purely imaginary. z1 − z2 Hence, (d) is the correct answer. Ex 17. If z z1 − z2 = k, (z1, z2 ≠ 0), then x 2 ⇒ x = cosθ ± isinθ = e±iθ Similarly, y = e± φi 2 |z1 − z2|2 1 2|z |2 − |z |2 = 0 x y y x xy mn + 1 x y y x m n n m +
4 158 z z1 + z2 (a) for k ≠ 1, locus z is a straight line (b) for k ∉{1, 0}, z lies on a circle (c) for k ≠ 0, z represents a point (d) for k ≠ 1, z lies on the perpendicular bisector z2 − z2 of the line segment joining and z1 z1 z − z2 Sol. Given, z z z z1 +− z z2 2 = k z z1 2 = k 1 z + z1 Clearly, if k ≠ 0, 1 then z would lie on a circle. If k = 1, zwould lie on a perpendicular bisector of the z2 and − z2 . line segment joining z1 z1 If k = 0, z represents a point. Hence, (b) is the correct answer. Ex 18. If A z( 1 ), B z( 2 ) and C z( 3 ) are the vertices of a π AB ∆ABC in which ∠ABC = and = 2, then 4 BC the value of z2 is equal to (a) z3 + i z( 1 + z3 ) (b) z3 − i z( 1 − z3 ) (c) z3 + i z( 1 − z3 ) (d) None of the above AB Sol. Given, = 2 BC Considering the rotation about B, we get z1 − z2 = |z1 − z2|⋅ eiπ/ 4 = AB ⋅ eiπ/ 4 z3 − z2 |z3 − z2| BC 1 i = 2 + = 1 + i 2 2 ⇒ z1 − z2 = (1 + i z)( 3 − z2) ⇒ z1 − (1 + i z) 3 = z2(1 − 1 − i) ⇒ iz2 = − z1 + (1 + i z) 3 ⇒ z2 = iz1 − i(1 + i z) 3 ∴ z2 = z3 + i z( 1 − z3) Hence, (c) is the correct answer. Ex 19. If z z1 2 ∈C, z z R, z1 (z1 2 − 3z2 2 ) = 2 and z z z , then the value of z1 2 + z2 2 is (a) 5 (b) 6 (c) 10 (d) 12 Sol. Here, z z1( 1 2 − 3z2 2 ) = 2 …(i) z2(3z1 2 − z2 2 ) = 11 …(ii) On multiplying Eq. (ii) by i and adding it to Eq. (i), we get z1 3 − 3z z2 2 1 + i(3z z1 2 2 − z2 3 ) = 2 + 11i ⇒ (z1 + iz2) = 2 + 11i …(iii) Again multiplying Eq. (ii) by i and subtracting it from Eq. (i), we get z1 3 − 3z z2 2 1 − i(3z z1 2 2 − z2 3 ) = 2 − 11i ⇒ (z1 − iz2)3 = 2 + 11i On multiplying Eqs. (iii) and (iv), we get (z12 + z22 3) = 4 + 121 ⇒ z12 + z22 = 5 Hence, (a) is the correct answer. …(iv) Ex 20. 1 − c2 = nc −1 and z = eiθ , then c n (1 + nz) 1 + is equal to 2n z (a) 1− ccosθ (b) 1+ 2ccosθ (c) 1+ ccosθ (d) 1− 2ccosθ Sol. Here, 1 − c2 = nc − 1 ⇒ 1 − c2 = n c2 2 − 2nc + 1 c 1 ∴ = …(i) or (1 + c 1 nz) 1 + + n z + z 2n + n⋅ (2cosθ )} [using Eq. (i)] = 1 + ccos θ Hence, (c) is the correct answer. Ex 21. Consider an ellipse having its foci at A z( 1 ) and B z( 2 ) in the argand plane. If the eccentricity of the ellipse is e and it is known that origin is an interior point of the ellipse, then | z1 + z2 | (a) e 0, n n 2 1 2 + n z n 1 1 2 = + 1 2 + n = + + 1 1 1 2 2 n n { = + + + (1 ) 2 cos 1 2 2 n n n θ = + + 1 2 1 2 n n cos θ
159 2 2 | z1 | + |z2 | | z1 − z2 | (b) e 0, | z1 | + | z2 | | z1 + z2 | (c) e 0, | z1 | − |z2 | (d) Can’t be determined Sol. If P z( )is any point on the ellipse. Then, equation of the ellipse is |z1 − z2 | …(i) |z − z1| + |z − z2| = e If we replace z by z1or z2, Eq. (i) becomes |z1 − z2|. For P z( )to lie on ellipse, we have |z1 − z2 | |z − z1| + |z − z2|< e It is given that origin is an interior point of the ellipse. |z1 − z2 | ⇒ |0 − z1| + |0 − z2|< e |z1 − z2 | ⇒ e 0, |z1| + |z2| Hence, (b) is the correct answer. Ex 22. If| z − 2 − i| = | |z sin π − arg ( )z , then 4 locus of z is (a) a pair of straight lines (b) a circle (c) a parabola (d) an ellipse Sol. We have, |(x − 2) + i y(− 1)| = | |z 1 cosθ − 1 sinθ where, θ = arg ( )z |x − y | which is a parabola. Hence, (c) is the correct answer. Ex 23. α1, α 2, α 3, …, α100 are all the 100th roots of unity. The numerical value of ∑∑ (α αi j )5 is 1 ≤ i < j ≤100 (a) 20 (b) 0 (c) (20)1 20/ (d) None of these Sol. − (α101+ α102+ α103+ α104 + α510 + …) = 0 − 0 = 0 r r r 100, if r = 100k Because (α1 + α2 + … + α100) 0 , if r ≠ 100k Hence, (b) is the correct answer. Ex 24. The maximum area of the triangle formed by the complex coordinates z, z1 and z2, which satisfy the relations | z − z1| = | z − z2|, z − z1 + z2 r, where r >| z1 − z2|, is 2 1 By the given conditions, the area of ∆ABC is |z1 − z2| r Hence, (b) is the correct answer. Ex 25. Locus of z, if 3π , when | z | ≤ −| z 2| 4 arg[z − + =(1i)] π , when | z | > −| z2|, is 4 (a) straight line passing through (2, 0) (b) straight line passing through (2, 0), (1, 1) (c) a line segment (d) a set of two rays Sol. The given equation is written as ( ) ( ) x y − + − = 2 1 1 2 2 2 ( ) a 1 2 2 1 2 | | z z − ) b ( 2 2 1 | | z z r − ( ) c 1 2 1 2 2 2 | | z z r − (d) 1 2 1 2 2 | | z z r − Az () r Cz ) ( 2 Bz ) ( 1 zz 12 + 2 Y O X
4 160 1 2 3π − (1 + i)] 4 , when x ≤ 2 arg[z −π , when x > 2 4 The locus is a set of two rays. Hence, (d) is the correct answer. 3 n A(∩ B) is equal to (a) 1 (b) 2 (c) 3 (d) 0 Sol. We can observe that, 3 + 3i ∈ A but ∉ B. ∴ n A(∩ B) = 0 Hence, (d) is the correct answer. Ex 27. Dividing f z( )by z − i, we obtain the remainder i and dividing it by z + i, we get the remainder 1 + i. The remainder upon the division of f z( ) by z2 +1, is 1 1 (a)(z + 1) + i (b)(iz + 1) + i 2 (c)(iz − 1) + i (d)(z + i) + 1 Sol. f z( ) = g z( ) (z − i z)( + i) + az + b; where a, b ∈C f i( ) = i ⇒ ai + b = i …(i) f (−i) = 1 + i ⇒ a(−i) + b = 1 + i …(ii) From Eqs. (i) and (ii), we get i 1 a = , b = + i 2 2 Hence, required remainder = az + b = iz + + i Ex 28. If z1 = a1 + ib1 and z2 = a2 + ib2 are complex numbers such that | z1| =1, | z2| = 2 and Re (z z1 2 ) = 0, then the pair of complex ia2 numbers ω1 = a 1 + and ω2 = 2 b 1 + ib 2, 2 satisfy (a) |ω1 | = 1 (b) |ω 2 | = 2 (c) Re (ω ω12 ) = 0 (d) Im (ω ω12 ) = 0 Sol. a12 + b12 = 1, a22 + b22 = 4 and a a1 2 = b b1 2 a22 + b22 = 4a12 + 4b12 (a2 + 2ia1)2 = (2b1 + ib2)2 ⇒ a2 = ± 2b1 2 |ω1|2 = a1 + a12 + a2 = a12 + b12 = 1 4 ⇒ |ω1| = 1 and |ω2|2 = 4b1 2 + b2 2 = 4 ⇒ |ω2| = 2 a b2 2 = 0 Re (ω ω12) = 2a b1 1 − 2 Im (ω ω1 2) = a b1 2 + a b2 1 = 2a1 2 + 2b1 2 = 2 Hence, (a), (b) and (c) are the correct answers. Ex 29. If from a point P representing the complex number z1 on the curve | |z = 2, pair of tangents are drawn to the curve | |z =1, meeting at point Q z( 2 ) and R z( 3 ), then z1 + z2 + z3 (a) complex number will lie on the Ex 26. If A z |arg ( )z = 4 and 2 B z |arg (z − 3 − 3 )i = , z ∈C. Then, Hence, (b) is the correct answer. Type 2. More than One Correct Option (0, 2) (1, 1) (2, 0) Re()=1 z O (3, 3) A B Y X 2 1 2
161 z3 3 (d) orthocentre and circumcentre of ∆PQR will coincide Sol. Options (a) and (d) are true as PQR is an equilateral triangle, so orthocentre, circumcentre and centroid will coincide. (b) z1 + z2 + z3 = 1 ⇒ |z1 + z2 + z3|2 = 9 3 ⇒ (z1 + z2 + z3) (z1 + z2 + z3) = 9 4 1 1 4 1 1 ⇒ + + + + 9 z1 z2 z3 z1 z2 z3 (c) ∠QOR = 120° Hence, (a), (b), (c) and (d) are the correct answers. Ex 30. One vertex of the triangle of maximum area that can be inscribed in the curve | z − 2i| = 2, is 2 + 2i, remaining vertices is/are (a) − +1i(2 + 3) (b) − −1i(2 + 3) (c) − +1i(2 − 3) (d) − −1i(2 − 3) Sol. Clearly, the inscribed triangle is equilateral. Hence, options (a) and (c) are the correct answers. Ex 31. Let z be a complex number and a be a real parameter, such that z2 + az + a 2 = 0, then (a) locus of z is a pair of straight lines (b) locus of z is a circle (c) arg( )z = ± (d) |z | = |a | Sol. z2 + az + a2 = 0 ⇒ z = aω, aω2 where, ω is non-real root of cube unity. ⇒ Locus of z is a pair of straight lines and arg( )z = arg ( )a + arg ( )ω or arg ( )a + arg (ω2 ) ⇒ arg ( )z = ± Also, | |z = | | |a ω | or | | |a ω2 | ⇒ | |z = | |a Hence, (a), (c) and (d) are the correct answers. Type 3. Assertion and Reason Directions (Ex. Nos. 32-36) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement IIis a correct explanation for Statement I (b) Statement I is true,Statement II is true; Statement II isnot a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true Ex 32. Statement I If A z( 1 ), B z( 2 ), C z( 3 ) are the vertices of an equilateral ∆ABC, then z2 + z3 − 2z1 π arg z3 − z2 4 Statement II If ∠B = α, then z2 + z3 z + − 2z − z1 Sol. arg 2 z3 1 = arg 2 z3 − z2 z3 − z2 3 curve | | z =1 ) b ( 1 1 1 1 4 4 9 1 3 3 2 2 1 z z z z z z + + + + = (c) arg z2 2 = π Y Qz ( ) 2 Pz ( ) 1 Rz ( ) 3 O Y¢ X¢ X ⇒ z z z z e i 2 0 0 1 2 3 − − = π , z z z z e i 0 3 0 1 2 3 − − = − π ⇒ z i 2 3 2 1 + =− + ) ( and z i 3 1 2 3 =− + − ( ) z2 z i 1(2+2) z3 zi 0(2) z z z z AB BC ei 1 2 3 2 − − = α or arg z z z z 2 1 2 3 − − =α α Bz () 2 Cz () 3 Az () 1
4 162 [as AD ⊥ BC ] 2 Hence, (d) is the correct answer. 1 Ex 33. Statement I If x + =1 and x p = x4000 + 1 and q is the digit at unit place 4000 x 2n in the number 2 +1, n ∈ N and n >1, then the value of p + q = 8. Statement II ω, ω2 are the roots of 1 3 1 x + = −1, x + = 2. x x3 1 2 Sol. x + = 1 ⇒ x − x + 1 = 0 x ∴ x = −ω − ω, 2 Now, for x p Similarly, for x = −ω2 , p = −1 For n > 1, 2n = 4k ∴ (2)2n = 24k = (16)k = anumber with last = 6 ⇒ q = 6 + 1 = 7 Hence, p + q = −1 + 7 = 6 Hence, (d) is the correct answer. Ex 34. Statement I If z1, z2, z3are complex numbers represent the points A B C, , such that 2 1 1 = + . Then, A B C, , passes through z1 z2 z3 origin. Statement II If 2z2 = z1 + z3, then z1, z2, z3 are concyclic. 2 1 1 1 1 1 1 Sol. = + ⇒ − = − z1 z2 z3 z1 z2 z3 z1 z2 − z1 = z1 − z3 ⇒ z z1 2 z z1 3 z2 − z1 = − z2 ⇒ z3 − z1 z3 z2 −z1 = arg z2 ⇒ arg z3 − z1 z3 z2 −z1 = π − arg z2 = π − arg z2 ⇒ arg z3 − z1 z3 z3 ⇒ ∠CAB = π − ∠COB ⇒ ∠CAB + ∠COB = π ⇒ Points O A B C, , , are concyclic. Hence, (b) is the correct answer. Ex 35. Statement I 3 + ix y2 and x2 + y + 4i are complex conjugate numbers, then x2 + y2 = 4. Statement II If sum and product of two complex numbers is real, then they are conjugate complex number. Sol. If 3 + ix y2 and x2 + y + 4i are conjugate, then x y2 = − 4 and x2 + y = 3 ⇒ x2 = 4, y = − 1 ⇒ x2 + y2 = 5 Hence, (d) is the correct answer. Ex 36. Statement I If | |z < 2 −1, then | z2 + 2z cos α | <1 Statement II | z1 + z2| ≤ | z1| + | z2|, also |cos α ≤| 1. Sol. |z2 + 2zcos α | < |z2 | + |2zcos α | < | |z 2 + 2| | |coszα| Hence, (a) is the correct answer. Type 4. Linked Comprehension Based Questions = π Az ( ) 1 Cz ( ) 3 Bz ( ) 2 D zz 23 + 2
163 Passage I (Ex. Nos. 37-39) In argand plane, | z | represents the distance of a point z from the origin. In general, | z1 − z2| represents the distance between two points z1and z2. Also, for a general moving point z in argand plane, if arg( )z = θ then z = | |z ei θ , where ei θ = cos θ + i sinθ. Ex 37. The equation | z − z1| + | z − z2| =10, if z1 = 3 + 4i and z2 = − 3 − 4i represents (a) point circle (b) ordered pair (0, 0) (c) ellipse (d) None of these Sol. As z1 − z2 = 6 + 8i =10 ∴ z − z1 + z − z2 = z1 − z2 ⇒ z lies between z1 and z2. i.e. a line segment. Hence, (d) is the correct answer. Ex 38. z − z1| − | z − z2 = t, t >10 where t is a real parameter, always represents (a) ellipse (b) hyperbola (c) circle (d) None of these Sol. Given equation can represent a hyperbola, since |z1 − z2| = 10 ⇒ z lies on hyperbola. Hence, (b) is the correct answer. π Ex 39. If | z − (3 + 2i)| = z cos − arg ( )z ,then 4 locus of z is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola Sol. Given, |z − (3 + 2i)|= |(x − 3) + i y( − 2)| which is a parabola. Hence, (b) is the correct answer. Passage II (Ex. Nos. 40-42) The complex slope of a line passing through two points represented by complex z2 − z1 numbers z1 and z2 is defined by and we shall z2 − z1 denote by ω. If z0 is complex number and c is a real number, then z z0 + z z0 = 0 represents a straight line. Its z0 . Now ,consider two lines complex slope is − z0 αz +αz + iβ = 0…(i) and az + az + b = 0 …(ii) where, α β, and a b, are complex constants and let their complex slopes be denoted byω1 andω2, respectively. Ex 40. If the lines are inclined at an angle of 120° to each other, then (a) ω ω21 = ω ω1 2 (b) ω ω212 = ω ω1 22 (c) ω12 = ω 22 (d) ω1 + 2ω 2 = 0 Sol. ⇒ω31 = ω32 ⇒ω13 ω ω1222 = ω23 ω ω12 22 ⇒ ω1 ω1 2 ω22 = ω2 ω2 2ω12 ⇒ω ω112= ω ω212,as ω1= ω2 ∴ ω2 ω12 = ω1 ω22 Hence, (b) is the correct answer. Ex 41. Which of the following must be true? (a) a must be pure imaginary (b) β must be pure imaginary (c) a must be real (d) β must be imaginary Sol. Since, i β is real. ∴ β is pure imaginary. Hence, (b) is the correct answer. Ex 42. If line (i) makes an angle of 45° with real axis, 2 then (1 + i) is (a) 2 2 (b) 2 2i (c) 2(1− i) (d) − 2(1+ i) Sol. e i 2 ∴ (1 + i) = ± 2 (− 1 + i) Hence, (c) is the correct answer. Passage III (Ex. Nos. 43-45) Consider ∆ABC in argand plane. Let A(0), B(1) and C(1+ i) be its vertices and M be the mid-point of CA. Let z be a variable complex number in the plane. Let us another variable complex number defined as, u = z2 +1. Ex 43. Locus of u, when z is on BM, is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola Ex 44. Axis of locus of u, when z is on BM, is = + || cos sin z 1 2 1 2 θ θ ,where θ= arg()z ⇒ ( ) ( ) | | y x x y = − + − + 2 3 1 2 2 2 − = =± ± α α π i 2
4 164 (a) real axis (b) imaginary axis (c) z + z = 2 (d) z − z = 2i Ex 45. Directrix of locus of z, when z is on BM, is (a) real axis (b) imaginary axis (c)z + z = 2 (d) z − z = 2i Sol. (Ex. Nos. 43-45) BM ≡ y − 0 = − 1(x − 1) x + y = 1 ∴ u − 1 = t + i (1 − t) u = 2t + 2i t (1 − t) x = 2t and y = 2 (1t − t), where u = x + iy (x − 1)2 = − 2 y − 1 , which is parabola. 2 Axis is x = 1, i.e. z + z = 2 Directrix is y = 1, i.e. z − z = 2i 43. (b) 44. (c) 45. (d) Type 5. Match the Columns Ex 46. Match the following : ⇒ 75 × 25 = z1 + 1 ∴ z1 − (− 1) = 1875 ⇒ z1 lies on circle. Locus of the point equationRe (z ) = Re(z + z),isa/an Ex. 47. If z1, z2, z3, z4 are the roots of the equation B. Locus of the point z satisfying the q. straight line z4 + z3 + z2 + z +1 = 0, then equation Sol. A. Put z = x + iy ∴ Re (x + iy)2 = Re (x + iy + x − iy) 2 x2 2 − y2 = 2x Sol. The given equation is z 5 − 1 = 0, which means that or x − y − 2x = 0 z − 1 Rectangular hyperbola, eccentricity = 2 z1, z2, z3, z4 are four out of five roots of unity except 1. B. For ellipse, λ > |z1 − z2| A. z14 + z24 + z34 + z44 + 14 = 0 and for straight line, 4 λ = |z1 − z2| ⇒ ∑zi4 = 1 4 B. ∑ zi 5 is equal to q. i = 1 4 4 r. C.(zi + 2)is equal to i = 1 1 D. Least value of [| z1 + z2|], where s. [ ] represents greatest integer function, is 11 − + − = λ λ∈ + − λ< − + = = − ∈ + = − + ∏ = ∑
165 1 3 1 3 C.  i = 1 B. z z z z ⇒ ⇒ i = 1 C. z4 + z3 + z2 + z + 1 = (z − z1) (z − z2) (z − z3) (z − z4) Putting z = − 2on both the sides, we get ⇒ 4 ∏ (zi + 2) = 11 i = 1 D. |z1 + z2| =2 + 2cos144° for minimum 5 − 1 D. Given, z = 25 = 2cos72° = 2 Let whose greatest integer is 0. A → r; B → q; C → s; D → p Type 6. Single Integer Answer Type Questions Ex 48. Consider an equilateral triangle having vertices Sol. (5) A z( 1) = 2i at points incircle, then AP 2 + BP 2 + CP 2 is equal to _______ . BC D Hence, any point on incircle i.e. P z( )is cos α, sinα i.e. (cosα + isinα) Solving for |AP|2 + |BP|2 + |CP| ,2 we get AP2 + BP2 + CP2 = 5 Ex 49. Let A1, A2,..... An be the vertices of a regular polygon of n sides in a circle of radius unity and a = | A A1 2|2 + | A A1 3|2 + .... + | A A1 n |2 , a b = | A A1 2|| A A1 3|....| A A1 n |, then = ____. b Sol. (2) Let us assume thatO is the centre of the polygon and z0, z1,…, zn − 1 represent the affixes of A1, A2,…, An such that i2π z0 = 1, z1 = α, z2 = α2 ,..., zn−1 = αn−1 , where α = e n Now, |A A1 r |2 = α| r − 1|2 = |1− αr |2 2rπ 2rπ 2 = 1 − cos − isin n n 2r 2 2r 2 2rπ i Bz i ) ( 2 2 3 3 2 1 2 1 3 − = − = Cz i i ) ( 3 2 3 3 2 2 1 3 − = − − =− Radiusofincircleof ∆ABC , i.e. r = 1 3 unit. 1 3 2 1 i z z m − + = z i z m − + = 2 1 2 For m = 2, z z − + = 1 2 1 1 ⇒ z i z − = + 2 1 | |,i.e.straightline z z 1 1 75 =− + ∴ 75 1 1 z z = + or 75 1 1 z z = + 3 O A r B e A e i i 2 3 2 3 6 2 π π − , and C e i 2 3 5 6 − π .If Pz () isanypointonits
4 166 1 − cos sin = 2 − 2 cos n n n n n ∴r∑= 1 |A A12|2 = ∑r = 1 2 − 2 cos rn 2π 4π 2(n − 1) = 2 (n − 1) cos+ cos + ... + cos n n n = 2 (n − 1) − 2⋅ Real part of (α + α2 + ... + αn−1 ) = 2 (n − 1) − 2 (1) [since, {1 + α + α2 + + αn−1 = 0}] ∴ |A A12|2 + |A A13|2 + ... |+ A A1n|2 = 2n Also, let E = |A A12| |A A13||A A1 n| = |1 − α| |1 − α2 | |1 − α3 |... |1 − αn−1 | = (|1 − α)(1 − α2 )(1 − α3 )(1 − αn−1 )| Since, 1 α α, 2 ,…, αn−1 are the roots of zn − 1 = 0. n ⇒ (z − 1) (z − α) (z − α2 )... (z − αn−1 ) = z − 1 z −1 ⇒ (z − α) (z − α2 ) (z − αn−1 ) = zn − 1 = 1 + z + z2 + … + zn−1 Substituting z = 1,we have (1 − α) (1 − α2 )... (1 − αn−1 ) = n |1 − α| |1 − α2 |... |1 − αn−1 | = n a 2n Hence, the value of = = 2 b n
167 1 5 (c) 1 10 Target Exercises Type 1. Only One Correct Option i592 + i590 + i588 + i586 + i584 1. The value of − 1 is i582 + i580 + i578 + i576 + i574 (a) −1 (b) −2 (c) −3 (d) −4 2. i57 + 1 , when simplified has the value 125 i (a) 0 (b) 2i (c) −2i (d) 2 3. in + in + 1 + in + 2 + in + 3 is equal to (a) 1 (b) − 1 (c) 0 (d) None of these 4. 1+ i2 + i4 + i6 + + i2n is (a) positive (b) negative (c) 0 (d) Can’t be determined 5. The value of i19 1 25 2 is i (a) 4 (b) − 4 (c) 2 (d) − 2 6. If n is any positive integer, then the value of i4n + 1 − i4n − 1 equals 2 (a) 1 (b) −1 (c) i (d) −i 7. For a positive integer n, the expression (1− i)n 1− 1 n equals i (a) 0 (b) 2in (c) 2n (d) 4n (1− i)n 8. If the number is real and positive, then nis (1+ i)n − 2 (a) any integer (b) 2λ (c) 4λ + 1 (d) None of these 9. The smallest positive 1+ i n = − 1is 1− i integer n for which (a) 1 (b) 2 (c) 3 (d) 4 10. The smallest positive (1+ i)2n = (1− i)2n is number n for which (a) 4 (b) 8 (c) 2 (d) 12 1+ 2i 11. The complex number lies in the 1− i (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant (1− i)3 12. The value of is equal to 1− i3 (a) i (b) − 1 (c) 1 (d) − 2 1+ i 2 1− i 2 13. is equal to 1− i 1+ i (a) 2i (b) −2i (c) −2 (d) 2 (1+ i)2 14. The value of Re is equal to 3 − i (a) − (b) (d) − 15. If a2 + b2 = 1, then (1 + b + ia) is equal to (1+ b − ia) (a) 1 (b) 2 (c) b + ia (d) a + ib 16. If b + ic = (1+ a z) and a2 + b2 + c2 = 1, then equals a − ib a − ib a + ib a + ib 1 1 + − iz iz
4 168 6 20 i 5 12 π (a)(b)(c) (d) 1 − c 1 + c 1 − c 1 + c −3i 1 17. If 4 3i −1 = x + iy, then 3 i (a) x = 3, y = 1 (b) x = 1, y = 3 (c) x = 0, y = 3 (d) x = 0, y = 0 18. 2i equals (a)1 + i (b)1 − i (c) − 2i (d) None of these 19. If z is a complex number such that | |z ≠ 0 and Re( )z = 0, then (a) Re (z2 ) = 0 (b) Im (z2 ) = 0 (c) Re (z2 ) = Im (z2 ) (d) None of these 20. If (x + iy)1 3/ = a + ib, then x + y is equal to a b (a) 2 (a2 − b2 ) (b) 4 (a2 − b2 ) (c) 8 (a2 − b2 ) (d) None of these 21. If 8iz3 + 12z2 − 18z + 27i = 0,then 32. If x + iy = u + iv , then x2 + y2 is equal to 2u − iv (a) | |z = (b) | |z = (c) | |z = 1(d) | |z = 3(a) 1 (b) −1 (c) 0 (d) None of these 22. If z = 1+ i, then the multiplicative inverse of z2 is 33. If (1+ i) (1+ 2 ) (1i + 3 )... (1i + ni) = α + iβ, then i i 2 (a)1 − i (b) (c) − (d) 2i 2 5 10⋅ ⋅... (1+ n )is equal to 2 2 2 2 (a) α − βi (b) α − β 1+ i 3 1− i 3 (c) α2 + β2 (d) None of these 23. If = x + iy, then (x y, )is equal to 1− i 1+ i 34. The principal argument of the complex number (a) (0, 2) (b) (− 2, 0) (c) (0, −2) (d) None of these 24. The multiplicative inverse of (6 + 5 )i 2 is 11 60 11 60 (a) (c) −(d) (a) − i (b) − i 60 61 61 61 9 60 35. The complex number i + 3 in polar form can be (c) − i (d) None of these 61 61 written as 3 + 4i (a) 1 sin π + icos (b) 2 cos π + isin 25. The multiplicative inverse ofis 2 6 6 6 6 4 − 5i 1 π π 8 31 8 31 (c) 2 sin 6 + icos 6 (d) 4 cos 6 + icos 6 (a) − + i (b) − i 25 25 25 25 (c) 8 31i (d) None of these 36. If z is a complex number, then − − 25 25 (a) |z2 | > | |z 2 (b) |z2 | = | |z 2 (c) |z2 | < | |z 2 (d) |z2 | ≥ | |z 2 ) ( ( ) ) ( 1 3 1 2 3 5 2 + + + − − i i i i is 19 12 π ( ) b − 7 12 π
169 z 26. If z = x + iy lies in III quadrant, thenalso lies in 37. If z = x + iy x, and y are real, then | |x + | y|≤ k z| |, 27. The conjugate complex number of is the least value and greatest value of |z + 1|are (a) 1, 6 (b) 0, 6 (c) 2, 8 (d) None of these 39. If z satisfies |z + 1|< | z − 2|, then w = 3z + 2 + i satisfies (a) |w + 1| < |w − 8| (b) |w + 1| < |w − 7| (c) |w + w| > 7 (d) |w + 5| < |w − 4| 28. 40. For any complex number z, the minimum value of | z | + |z − 1|is (a) 1 (b) 0 (c) (d) 29.41. The greatest positive argument of complex number satisfying | z − 4| = Re ( )z , is 1− ix 2 2 (a) 30. If = a − ib and a + b = 1, where a and b are 1+ ix (c) real, then x is equal to 2a 2b (a)(b)42. The square roots of − +2 2 3 i are (1 + a)2 + b2 (1 + a)2 + b2 (a) ± (1 + 3 )i (b) ± (1 − 3 )i 2a 2b ± ( 1− + (c)(d)(c) 3 )i (d) None of these 31. The modulus ofis (a) (b) (c) (d) (d) None of these III quadrant, if (a) x > y > 0 (c) y < x < 0 z (b) x < y < 0 (d) y > x > 0 where k is equal to (a) 1 (b) 2 (c) 3 (d) None of these 38. If z is any complex number such that | z + 4 |≤ 3, then (1 + b)2 + a2 (1 + b)2 + a2 (3 + 2i)2 43. If z = a + ib, where a > 0, b > 0,then ( ) b ( ) ||z a b ≥ + 1 2 ( ) 4 3 − i 11 5 π 3 π 2 (b) 2π 3 (d) π 4 ) a ( ) ( ||z a b ≥ − 1 2 ( ) c ( || ) z a b < + 1 2
4 170 π 2 π 3 π 6 π 6 2 18 44. If for the complex numbers z1 and z2, |1− z z1 2 |2 − |z1 − z2 |2 = k (1− | z1 |2 ) (1− | z2 |2 ), then k is equal to (a) 1 (b) −1 (c) 2 (d) 4 45. The range of real number α for which the equation z + α| z − 1| + 2i = 0 has a solution, is (a) (b) (c) 0, 2 (d) , − 2 2 , 46. If | |z = 2 and locus of 5z − 1 is the circle having radius aand z1 2 + z2 2 − 2z z1 2 cos θ = 0, then | z1 |:| z2 | is equal to (a) a:1 (b) 2a:1 (c) a:10 (d) None of these 47. The maximum value of | z | when z satisfies the (a) 48. If z1 ≠ z2 and | z1 + z2 | = (a) atleast one of z1, z2 is unimodular (b) z1 ⋅ z2 is unimodular (c) z1 ⋅ z2 is non-unimodular (d) None of the above 49. If 1,ω ω, 2 ,,ω n − 1 are the n, nth roots of unity and z1, z2 are any two complex numbers, then n−1 ∑ | z1 + ω k z2 |2 is equal to k =0 (a) n z[| 1|2 + |z2| ]2 (b) (n − 1)[|z1|2 + |z2| ]2 (c) (n + 1)[|z1|2 + |z2| ]2 (d) None of these 50. If z satisfies the equation | z | − z = 1+ 2i, then z equal to is (a) + 2i (b) − 2i 3 3 (c) 2 − i (d) 2 i 2 2 51. If 3 + i = (a + ib) (c + id ), then tan −1 b + tan −1 d a c is equal to (a) + 2nπ, for some integer n (b) − + nπ, for some integer n (c) + nπ, for some integer n (d) − + 2nπ, for some integer n 52. If z is any complex number satisfying | z − 1| = 1, then which of the following is correct? (a) arg (z − 1) = 2 arg ( )z (b) 2 arg ( )z = 2 3/ arg (z2 − z) (c) arg (z − 1) = arg (z + 1) (d) arg ( )z = 2arg (z + 1) 53. If z = −1, then the principal value of arg (z2 3/ ) is equal to (a) (b) or 2π or (c)(d) π z − z π 54. If z1 = 8 + 4i z, 2 = 6 + 4i and arg 1 = , then z − z2 4 z satisfies (a) |z − 7 − 4i | = 1 (b) |z − 7 − 5i | = (c) |z − 4i | = 8 (d) |z − 7i | = 55. If arg ( )z < 0, then arg (−z)− arg (z) is equal to (a) π (b) − π (c) − (d) 56. If z is a point on the argand plane such that | z − 1| = 1 z − 2 , then is equal to z (a) tan (arg z) (b) cot (arg z) (c) itan (arg z) (d) None of these 57. If C 2 + S 2 = 1, then 1 + C + iS is equal to 1+ C − iS (a) C + iS (b)C − iS (c) S + iC (d) S − iC 58. The imaginary part of (z − 1) (cosα − isin α) + (z − 1)−1 × (cosα + isin α)is zero, if (a) |z − 1| = 2 (b) arg (z − 1) = 2α (c) arg (z − 1) = α (d) | |z = 1 8 sin + icos 8 8 condition z z + = 2 2is 3 1 − (c) 3 3 2 3 2 , ∪ 5 5 5 2 5 2 , 5 ( ) b 3 1 + ) d ( 2 3 + z z 2 1 1 1 + ,then π 3 5 3 π
171 2 2 2 2 59. The value ofis 8 sin − icos 8 8 (a) −1 (b) 0 (c) 1 (d) 2i 4 π 4 π 1 + a 3n 60. If a = cos + isin , then the value of is 3 3 2 (a) (−1)n (b) ( − 1 3n )n n (d) (−1)n + 1 2 1+ i 8 1− i 8 61. The value of is equal to 2 2 (a) 4 (b) 6 (c) 8 (d) 2 62. The value of ∑10 sin 2πk − icos 2πk is k = 1 11 11 (b) −1 (c) i (a) 1 (d) −i 63. The value ofis 10 (a)(1 + i) (c) (1 + i) (d) None of these 64. The value of (cos 2θ − isin 2θ)7 (cos 3θ + isin 3θ)−5 is (cos 4θ + isin 4θ)12 (cos 5θ + isin 5 )θ −6 (a) cos 33θ + i sin 33θ (b) cos 33θ − i sin 33θ (c) cos 47θ + i sin 47θ (d) cos 47θ − i sin 47θ 65. (cos 2θ + isin 2 )θ −5 (cos 3θ − isin 3 )θ 6 (sin θ − icos θ)3 is equal to (a) cos 25θ + i sin 25θ (b) cos 25θ − i sin 25θ (c) sin 25θ + i cos25θ (d) sin 25 θ − i cos 25θ 66. The principal value of the arg ( )z and | z | of the (1+ cosθ + isin θ)5 complex number z =are (cosθ + isin θ)3 (a) − θ , 32 cos5 θ (b) θ , 32 cos5 θ 2 2 (c) − θ ,16 cos4 θ (d) None of these 67. If z = cos θ + isin θ then (a) zn + 1 = 2cos nθ (b) zn + 1 = 2n cos nθ nn zz n 1 n n 1 n (c) z − n = 2 i sin nθ (d) z − n = (2 )i sin nθ z z 2n z − 1 68. If z = cos θ + isin θ then , where n is an z2n + 1 integer, is equal to (a) icotnθ (b) itannθ (c) tannθ (d) cotnθ 69. If a = cos 2α + isin 2 ,α b = cos 2β + isin 2 ,β c = cos 2γ + isin 2γ and d = cos 2δ + isin 2δ, then 1 abcd + is equal to abcd (a) 2 cos (α + β + γ + δ) (b) 2cos (α + β + γ + δ) (c) cos (α + β + γ + δ) (d) None of these m 70. e2mi cot−1 p ⋅ pi + 1 is equal to pi − 1 (a) 0 (b) 1 (c) − 1 (d) None of these 71. If α = cos 8 isin 8 , then 11 11 Re (α + α2 + α3 + α4 + α5 )is equal to (a) (b) − (c) 0 (d) None of these 72. If ω (≠ 1) is cube root of unity satisfying 1 1 1 2 + + = 2ω and a + ω b + ω c + ω 1 1 1 + + = 2 ω, then the value of 2 2 2 a + ω b + ω c + ω 1 1 1 + + is equal to a + 1 b + 1 c + 1 2 5 2 75 4 75 30 30 04 ) (cos sin .( cos sin ) ° °+ °+ ° i i (b) 10 2 (1 ) − i
4 172 (a) 2 (b) −2 (c) − 1 + ω2 (d) None of these 73. Ifα and β are the roots of the equation x2 − 2x + 4 = 0, then the value ofα6 + β6 is (a) 64 (b) 128 (c) 256 (d) None of these 1 74. If z is any complex number such that z + = 1, then z the value of z99 + 1 is 99 z (a) 1 (b) −1 (c) 2 (d) −2 75. The value of (− 1+ −3)62 + −( 1− − 3)62 is (a) 262 (b) 264 (c) −262 (d) 0 76. If ω is a cube root of unity, then (3 + 5ω + 3ω 2 )2 (3 + 3ω + 5ω 2 )2 is equal to (a) 4 (b) 0 (c) − 4 (d) None of these 1+ i 3 6 1− i 3 6 77. The value of is 1− i 3 1+ i 3 (a) 2 (b) −2 (c) 1 (d) 0 78. If z1 = 3 + i 3 and z2 = 3 + i , then the complex z 1 50 number lies in the quadrant number z2 (a) I (b) II (c) III (d) IV 79. If x = a + b, y = aω + bω 2 , z = aω 2 + bω, then xyz is equal to (a) (a + b)3 (b) a3 + b3 (c) a3 − b3 (d) (a + b)3 + 3ab a( + b) 80. If 1, ω and ω 2 are the cube roots of unity, then the value of (1+ ω)3 − (1+ ω 2 )3 is (a) 2ω (b) 2 (c) −2 (d) 0 81. If 1, ω and ω 2 are the three cube roots of unity and α β, and γ are the roots of p p, < 0, then for any x y, xα + yβ + zγ and z, the expression equals xβ + yγ + zα (a) 1 (b) ω (c) ω2 (d) None of the above 82. The value of the expression 1 2( − ω) (2 − ω 2 ) + 2 3( − ω) (3 − ω 2 )+ .... + (n − 1) (n − ω) (2 − ω 2 ), where ω is an imaginary cube root of unity, is n n( + 1) 2 n n( + 1) 2 (a) (b) − n 2 2 n n( + 1) 2 (c) + n (d) None of these 2 83. If α and β are the complex cube roots of unity, then α3 + β3 + α−2 β−2 is equal to (a) 0 (b) 3 (c) − 3 (d) None of these 84. If ω is a complex cube root of unity, then (1− ω + ω 2 ) (1− ω 2 + ω 4 ) (1− ω 4 + ω 8 ) (1− ω 8 + ω16 )is equal to (a) 12 (b) 14 (c) 16 (d) None of these 85. If x = ω − ω 2 − 2, then the value of x4 + 3x3 + 2x2 − 11x − 6 is (a) 1 (b) − 1 (c) 2 (d) None of these 86. If 1, ω and ω 2 are the three cube roots of unity, then (1+ ω) (1+ ω 2 ) (1+ ω 4 ) (1+ ω 8 ) … to 2n factors is equal to (a) 1 (b) − 1 (c) 0 (d) None of these 87. If x = a + b + c y, = aα + bβ + c and z = aβ + bα + c where, α and β are complex cube roots of unity, then xyz is equal to (a) 2 (a3 + b3 + c3 ) (b) 2 (a3 − b3 − c3 ) (c) a3 + b3 + c3 − 3abc (d) a3 − b3 − c3
173 88. The common roots of the equations z3 + 2z2 + 2z + 1= 0and z1985 + z100 + 1= 0are (a) −1, ω (b) −1, ω2 (c) ω ω, 2 (d) None of the above 89. The values of (16)1 4/ are (a) ± 2, ± 2i (b) ± 4, ± 4i (c) ±1, ± i (d) None of the above 90. If p q, and r are positive integers and ω is an imaginary cube root of unity and f x( ) = x3p + x3q + 1 + x3r + 2 , then f (ω)is equal to (a) ω (b) −ω2 (c) 0 (d) None of the above 91. If1,α α, 2 ,,αn − 1 are the n, nth roots of unity, then n − 1 1 ∑ is equal to i = 1 2 − αi (b) (n − 2)⋅ 2n (n − 2)⋅ 2n − 1 (c) (d) None of these 2n − 1 1+ i 92. The triangle formed by the points 1,and i as vertices in the argand diagram is a/an (a) scalene (b) equilateral (c) isosceles (d) right angled 93. If the points represented by complex numbers z1 = a + ib z, 2 = a′ + ib′ and z1 − z2 are collinear, then (a) ab′ + a b′ = 0 (b) ab′ − a b′ = 0 (c) ab + a b′ ′ = 0 (d) ab − a b′ ′ = 0 94. If α + β =i tan −1 z z, = x + iy and α is constant, then the locus of z is (a) x2 + y2 + 2x cot 2α = 1 (b) (x2 + y2 ) cot 2α = 1 + x (c) x2 + y2 + 2y tan 2α = 1 (d) x2 + y2 + 2x sin 2α = 1 95. The equation | z + 1− i| = |z + i − 1|represents (a) a straight line (b) a circle (c) a parabola (d) a hyperbola 96. If the roots of (z − 1)n = i z( + 1)n are plotted in the argand plane, they as (a) on a parabola (b) concyclic (c) collinear (d) the vertices of a triangle 97. Locus of the point z satisfying the equation |iz − 1| + | z − i| = 2is (a) a straight line (b) a circle (c) an ellipse (d) a pair of straight lines 98. For x1 ,x2 , y1 , y2 ,∈R, if 0< x1 ,< x2 , y1 = y2 and z1 = x1 + iy1, z2 = x2 + iy2 and z z z , then z1 , z2 , z3 satisfy (a) |z1| = |z2| = |z3 | (b) |z1| < |z2| < |z3| (c) |z1| > |z2| > |z3| (d) |z1| < |z3| < |z2| 99. The equation not representing a circle is given by 1 + z (a) Re 0 1 − z (b) zz + iz − iz + 1 = 0 z − 1 π (c) arg = z + 1 2 z − 1 (d) = 1 z + 1 100. If z1 , z2 and z3 are three complex numbers in AP, then they lie on (a) a circle (b) a straight line (c) a parabola (d) an ellipse 101. If the area of the triangle on the argand plane formed by the complex numbers −z iz z, , − iz is 600 sq units, then | |z is equal to (a) 10 (b) 20 (c) 30 (d) None of these 102. The complex numbers given by1− 3 4i, + 3i and 3 + i represent the vertices of (a) a right angled triangle (b) an isosceles triangle (c) an equilateral triangle (d) None of the above z + 2i 103. If Re = 0,then z lies on a circle with centre 2
4 174 z + 4 (a) (− 2, −1) (b) (− 2 1, ) (c) (2, −1) (d) (2,1) z + 2i 104. If Im = 0, then z lies on the curve z + 2 (a) x2 + y2 + 2x + 2y = 0 (b) x2 + y2 − 2x = 0 (c) x + y + 2 = 0 (d) None of the above 105. In the argand plane, all the complex numbers satisfying |z − 4i| + |z + 4i| = 10lie on (a) a straight line (b) a circle (c) an ellipse (d) a parabola 106. If z = x + iy and a is a real number such that | z − ai| =|z + ai|, then locus of z is (a) X-axis (b) Y-axis (c) x = y (d) x2 + y2 = 1 107. If P represents z = x + iy in the argand plane and | z − 1|2 + |z + 1|2 = 4,then the locus of P is (a) x2 + y2 = 2 (b) x2 + y2 = 1 (c) x2 + y2 = 4 (d) x + y = 2 108. All the roots of the equation a z1 3 + a z2 2 + a z3 + a4 = 3, where |ai |≤ 1, i = 1, 2, 3, 4, lie outside the circle with centre origin and radius (a) 1 (b) (c) (d) None of the above 109. The region of argand diagram defined by | z − 1| + |z + 1|≤ 4 is (a) interior of an ellipse (b) exterior of a circle (c) interior and boundary of an ellipse (d) None of the above 110. The locus of the complex number z in an argand plane satisfying the inequality | z − 1| + 4 2 log1/2 3|z − 1| − 2 > 1 where,|z − 1|≠ 3 is (a) a circle (b) interior of a circle (c) exterior of a circle (d) None of the above 111. The closest distance of the origin from a curve given as az + az + aa = 0(a is a complex number), is |a | (a) 1 (b) 2 Re ( )a Im ( )a (c) (d) |a | |a | 112. Let a be a complex number such that | |a < 1 and z1 , z2 ,…, zn be the vertices of a polygon such that zk = +1 a + a2 +…ak , then the vertices of the polygon lie within the circle 113. If z1 , z2 , z3 and z4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such that z 1 − z 4 = z 2 − z 3 and z4 − z1 = π, then the quadrilateral is a amp z2 − z1 2 (a) square (b) rhombus (c) cyclic quadrilateral (d) None of the above 114. The locus of z satisfying Im (z2 ) = 4 is (a) a circle (b) a rectangular hyperbola (c) a pair of straight lines (d) None of the above 115. The curve represented by Re (z2 ) = 4 is (a) a parabola (b) an ellipse (c) a circle (d) a rectangular hyperbola ) a ( | | | | a z a = − − 1 1 ) (b | | | | z a − = − 1 1 1 c ( ) z a a − − = − 1 1 1 | |1 (d) None of these 1 3 2 3
175 116. A point z is equidistant from three distinct points z1 , z2 , z3 in argand plane. If z z, 1 and z2 are z3 − z1 collinear, then arg will be (z1 , z2 , z3 in z3 − z2 anti-clockwise sense) (a)(b) (c)(d) Type 2. More than One Correct Option 117. If z1 , z2 , z3 and z4 are roots of the equation a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 where, a0 , a1 , a2 , a3 and a4 are real, then (a) z 1, z 2, z 3, z 4 are also roots of equation (b) z1 is equal to atleast one of z1, z2, z3, z4 (c) −z1, − z2, − z3, − z4 are also roots of equation (d) None of the above 118. If z3 + (3 + 2i z) + −( 1+ ia) = 0 has one real root, then the value of a lies in the interval (a ∈R) (a) (−2 1, ) (b) (−1 0, ) (c) (0,1) (d) (−2 3, ) 119. The real value of θ for which the expression i + icosθ is a real number, is n n I n n I n n I n n I 120. Ifα is a complex constant such thatαz + z + a = 0has a real root, then (a) α + α = 1 (b) α + α = 0 (c) α + α = −1 (d) the absolute value of the real root is 1 121. If |z1 | = | z2 | = 1and arg z1 + arg z2 = 0,then (a) z z1 2 = 1 (b) z1 + z2 = 0 (c) z1 = z2 (d) None of the above 122. If |z1 | = 15and | z2 − 3 − 4i| = 5,then (a) |z1 − z2|min = 5 (b) |z1 − z2|min = 10 (c) |z1 − z2|max = 20 123. If |z − (1/ )|z = 1,then (d) |z1 − z2|max = 25 satisfying | z1 2 − z2 2 | = | z1 2 + z2 2 − 2z z1 2 |, then z1 is purely imaginary (b) z1 is purely real (a) z2 z2 (c) |arg z1 − arg z2| = π (d) |arg z1 − arg z2| = 125. If |z − 1| = 1,then (a) arg {(z − 1 − i z)/ }can be equal to −π /4 (b) (z − 2)/z is purely imaginary number (c) (z − 2)/z is purely real number (d) If arg (z) = θ, where z ≠ 0and θ is acute, then 1 − 2/ z = i tanθ 126. Let complex number z of the form x + iy satisfy 3z − 6 − 3i π arg and |z − 3 + i| = 3. Then, the 2z − 8 − 6i 4 ordered pairs (x y, )are 4 2 4 2 (a) 4 − ,1 + (b) 4 + ,1 − (c) (6, −1) (d) (0, −1) 127. If z = ω ω, 2 , whereω is a non-real complex cube root of unity, are two vertices of an equilateral triangle in the argand plane, then the third vertex may be represented by (a) z = 1 (b) z = 0 (c) z = − 2 (d) z = − 1 128. If amp (z z1 2 ) = 0and | z1 | = | z2 | = 1,then (a) z1 + z2 = 0 (b) z z1 2 = 1 (c) z1 = z2 (d) None of these π 2 π 6 π 4 2 3 π ) a ( || max z = + 5 1 2 (b) || min z = −1 5 2 (c) || max z = − 2 5 2 (d) || min z = − 5 1 2 If z z 1 2 and aretwocomplexnumbers ( ) z z 1 2 ≠ 5 5 5 5
4 176 129. If z is complex number satisfying z + z−1 = 1, then zn + z−n ,n ∈N has the value (a) 2( 1)− n , when nis a multiple of 3 (b) (−1)n − 1 , when nis not a multiple of 3 (c) (−1)n + 1 ,when nis a multiple of 3 (d) 0, when nis not a multiple of 3 130. Let z1 , z2 be two complex numbers represented by points on the circle | |z = 1and | z | = 2 respectively, then (a) max |2z1 + z2| = 4 (b) min |z1 − z2| = 1 (c) z≤ 3 (d) None of these 131. ABCD is a square, vertices being taken in the anti-clockwise sense. If A represents the complex number z and the intersection of the diagonals is the origin, then (a) B represents the complex number iz (b) D represents the complex number iz (c) B represents the complex number iz (d) D represents the complex number − iz 132. If z0 , z1 represent point P Q, on the locus | z − 1| = 1 and the line segment PQ subtend an angle at the point z = 1, then z1is equal to i (a) 1 + i z( 0 −1) (b) z0 − 1 (c)1 − i z( 0 −1) (d) i z( 0 −1) 133. If | z1 | = |z2 | = |z3 | = 1 and z1 , z2 and z3 are represented by the vertices of an equilateral triangle, then (a) z1 + z2 + z3 = 0 (b) z z z1 2 3 = 1 (c) z z1 2 + z z2 3 + z z3 1 = 0 (d) None of these Type 3. Assertion and Reason Direction (Q. Nos. 134-144) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II isa correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II isnot a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 134. Consider z1 = 1, z2 = 2and z Statement I If z 1 + 2 z 2 of z z2 3 + 8z z3 1 + 27z z1 2 Statement II z1 + z2 + z 135. Statement I 7+ >i 5 + i Statement II Cancellation laws does not hold true in complex numbers. 136. Statement I If z1and z2 are two complex numbers z1 = 0. such that |z1 | = | z2 | + | z1 − z2 |,then Im z2 Statement II arg ( )z = 0⇒ z is purely real. 1 137. Statement I If z + = 1(z ≠ 0 is a complex z 5 +1 number), then the maximum value of | |z is. 2 1 Statement II On the locus z + = 1, the farthest z 5 +1 distance from origin is. 2 138. Statement I The number of complex numbers z satisfying | |z 2 + a z| | + b = 0 (a b, ∈R)is atmost 2. Statement II A quadratic equation in which all the coefficients are non-zero can have atmost two roots. 139. Statement I Ifcos (1− i) = a + ib, where a b, ∈R and 1 1 1 1 i = −1, then a = e + cos1, b = e − sin .1 2 e 2 2 Statement II eiθ = cos θ + isin θ 140. Statement I The product of all values of (cos α + isin α)3 5/ is cos 3a + isin 3 .a StatementII The product of fifth roots of unity is 1. 141. Statement I The locus of the centre of a circle which touches the circles | z − z1 | = aand | z − z2 | = b externally (z z, 1 and z2 are complex numbers) will be hyperbola. Statement II | z − z1 | − |z − z2 |< | z2 − z1 | ⇒ z lies on hyperbola. 142. Statement I Consider an ellipse having its foci at A z( 1) and B z( 2) in the argand plane. If the eccentricity of the z 2 1 1 + 3 3 = . z3 3 6 + = , thenthevalue is36. z z z3 3 1 2 ≤ + +
177 ellipse is e and it is known that origin is an interior point of the ellipse, then | z1 + z2 | |z1 + z2 | e , | z1 | + |z2 | |z1 | + |z2 | Statement II If z0 is the point interior to curve |z − z1 | + |z − z2 | = λ | z0 − z1 | + |z0 − z2 |< λ 143. Statement I The equation | z − i| + | z + i| = k k, > 0, can represent an ellipse, if k > 2i. StatementII | z − z1 | + | z − z2 | = k, represents an ellipse, if | |k > | z1 − z2 |. 144. Statement I The equation zz + az + az + λ = 0, where a is a complex number represents a circle in argand plane, if λ is real. Statement II The radius of the circle zz + az + az + λ = 0is aa − λ. Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 145-147) Consider the complex 147. One of the possible argument of complex number numbers z1 and z2 satisfying the relation i z( 1 / z2 ) is z1 + z 1 (a) (c) 0 (d) None of these 145. Complex number z z1 2 is Passage II (Q. Nos. 148-150) If z satisfy the relation (a) purely real (b) purely imaginary 2 | z − {(α − 7α +11) + i}| =1,α ∈R. (c) zero (d) None of these π Also, arg ( )z ≥ is satisfied by atleast one z. 146. Complex number z1 / z2 is 2 (a) purely real 148. The values of α lies in the interval (b) purely imaginary (c) zero (a)[2, 6] (b) 1, 72 (c)[2, 5] (d)[− 4, −2] (d) None of the above 149. Maximum value of | z − i|is AB and points A andE lies in the opposite side of lineBC. If A B, and C are represent by the complex numbers 1, (a) (b) (c) (d) ω and ω2 respectively. 150. Maximum value of arg ( )z , for which | z − i| is151. Angle between AC and DE is equal to maximum, is (a) (b) (c) (d) (a) π − tan−1 4 (b) π − tan−1 9 9 4 (c) π − tan−1 1 (d) π − tan−1 4 9 5 PassageIII (Q. Nos. 151-153) On the sides AB and BC of a ∆ABC, squares are drawn with centres D and E such that points C and D lies on the same side of line AB Type 5. Match the Columns 152. The length of DE is (a) (b) 3 2 153. The length of AE is (a) (b) 2 2 (c) 2 (c) 3 − 3 (d) 6 (d) 3 + 3 3 3 − 3 3 + 3 z z 2 1 2 2 2 = + .
4 178 154. Match the following: 157. 155. For the equation z6 − 6z + 20 = 0, match the items of Column I with that of Column II. A. The number of the roots in the first p. 1 quadrant can be B.The number of the roots in the second q. 2 quadrant can be C.The number of the roots in the third r. 3 quadrant can be Type 6. Single Integer Answer Type Questions 158. If x = a + bi is a complex number such that x2 = 3 + 4i and x3 = 2 + 11i, where i = −1, then (a + b)equals _______. 159. If the complex numbers z is simultaneously satisfy 161. If z is a complex number satisfying z4 + z3 + 2z2 + z + 1= 0, then | |z is equal to _____ . 162. ABCD is a rhombus. Its diagonals AC and BD z − 12 5 z − 4 intersect at the point M and satisfy BD = 2AC. Its equations = , =1, then Re ( )z is ___. z − 8i 3 z − 8 points D and M represent the complex numbers 1+ i and 2 − i respectively. Then, the complex number 1 λ 160. If | |z ≥ 3,then the least value of z + is ,where λ represented by A is z,has Re ( )z = λ{ 1 , λ 2},then the z 3 value of λ1 + λ 2 is __________ . is __________ . C. Ifωk, k ∈ I; 0 ≤ k ≤ n − 1are the nth r. roots of unity, then the maximum value of ( !)n 1/n ω(n−1 2)/ is ω D. If z1, z2, z3 are the affixes points and A, s. B and C are lying on circle centred at origin. If the altitude is drawn from vertex A to base BC, such that it meets the circumcircle at P z( ), then zz1 + z z2 3 is ω2 − + ± ≠± = − − − − − = − π + + = ≠ ∈ − − − + + − = − − π Matchthefollowing: = − + = π π = + − = π π = + = + π π − = + = Match the following: ∞ −− −− = = = ∑ ω−
4 179 Entrances Gallery JEE Advanced/IIT JEE 1. Match the following : [2014] 2k 2k Let zk = cos 10 + i sin 10 ; k = 1 2, ,..., 9. Column I Column II A. For each zk, there exists a zj such zk ⋅ zj = 1 p. True B. There exists k ∈{1, 2, ..., 9}, such that z1 ⋅ z = zk has no solution z in the set of complex numbers q. False C.equals r. 1 10 Codes A B C D A B C D (a) p q s r (b) q p r s (c) p q r s (d) q p s r 2. Let complex numbers α and lie on the circles z0 = x0 + iy0 satisfies the equation 2 z = r2 + 2 , [2013] (a) (c) (d) 3. Let w = and P = {wn :n = 1, 2, 3,...}. Further 2 H1 = z ∈C : Re ( )z > 1 and 2 z ∈C :Re ( )z − 1 , where C is the set of all H2 2 complex numbers. Ifz1 ∈P∩ H1, z2 ∈P ∩ H2 and O represents the origin, then ∠z Oz1 2 equals [2013] (a) Passage (Q. Nos. 4-5) Let S = S1 ∩ S2 ∩ S3 where, S1 = {z ∈C : z < 4}, z −1+ 3 i S2 z ∈C :Im 0 1− 3 i and S3 {z ∈C :Re z > 0}. 4. The area of S equals [2013] z ∈S 6. Let z be a complex number such that the imaginary part of z is non-zero and a = z2 + z + 1is real. Then, a cannot take the value [2012] (a) − 1 (b) (c) (d) 7. If z is any complex number satisfying z − 3 − 2i ≤ 2, then the minimum value of 2z − 6 + 5i is________ .[2011] π 2i 8. Let ω = e 3 and a b c x y z, , , , , be non-zero complex numbers, such that a + b + c = x a, + bω + cω 2 = y and a + bω2 + cω = z. Then, the value of is ________ . [2011] 9. Let z1 andz2 be two distinct complex numbers and z = (1− t z) 1 + tz2 for some real number t with 0< t <1. If arg (w) denotes the principal argument of a non- zero complex number w, then [2010] a z − z1 + z − z2 = z1 − z2 b arg (z − z1) = arg (z − z2) z − z1 z − z1 c = 0 z2 − z1 z2 − z1 1 − z1 1 − z2 ... 1 − z9 (x − x0 )2 + ( y − y0 )2 = r2 and (x − x0 )2 + ( y − y0 )2 = 4r2 , respectively. If − = ∑ π 1 7 a) ( 10 3 π ) b ( 20 3 π ) c ( 16 3 π d ( ) 32 3 π min z i − −3 1 equals [2013] (a) 2 3 2 − ) b ( 2 3 2 + ) c ( 3 3 2 − d) ( 3 3 2 + x y z b a c 2 2 2 2 2 2 + + + + 0 2 1 3 then α equals 1 2 ( ) b 1 2 i + 3 1 3 3 4
180 2. 1 2 5 2 (d) 2 2 + 1 d arg (z − z1) = arg (z2 − z1) 10. Match the statement in Column I with those in Column II. Ø Here, z takes the values in the complex plane and Im z and Re z denote, respectively the imaginary part and the real part of z. [2010] JEE Main/AIEEE 11. A complex number z is said to be unimodular, if | z | = 1. Suppose z1 and z2 are complex numbers such z1 − 2z2 thatis unimodular and z2 is not 2 − z z1 2 unimodular. Then, the point z1 lies on a [2015] (a) straight line parallel to X-axis (b) straight line parallel to Y-axis (c) circle of radius 2 (d) circle of radius 12. If z is a complex number such that z ≥ 2, then the minimum value of z + [2014] (a) is equal to (b) lies in the interval (1, 2) (c) is strictly greater than than but less than (d) is strictly greater 13. If z is a complex number of unit modulus and 1+ z argumentθ, then arg equals [2013] 1+ z (a) − θ (b) − θ (c) θ (d) π − θ 2 z 14. If z ≠ 1and is real, then the point represented by z − 1 the complex number z lies [2012] (a) either on the real axis or on a circle passing through theorigin (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing throughthe origin (d) on the imaginary axis 15. Let α and β be real and z be a complex number. If z2 + α z + β = 0 has two distinct roots on the line Re ( )z = 1, then it is necessary that (a) β ∈ −( 1, 0) (b) β = 1 (c) β ∈(1, ∞) (d) β ∈(0,1) 16. If ω (≠ 1) is a cube root of unity and [2012] (1+ ω)7 = A + Bω. Then, (A B,) equals (a) (1,1) (b) (1, 0) (c) (−1 1, ) (d) (0,1) [2011] 17. The number of complex numbers z such that z − 1 =| z + 1|= z − i equals (a) 0 (b)1 (c) 2 (d) ∞ [2010] 18. If α and β are the roots of the equation x2 − x + 1= 0, thenα2009 + β2009 is equal to [2010] (a) − 2 (b) − 1 (c)1 (d) 2 4 19. If z − = 2, then the maximum value of z is z equal to [2009] (a) 3 + 1 (b) 5 + 1 (c) 2 20. The conjugate of a complex number is . Then, i − 1 that complex number is [2008] 1 1 1 1 (a) − (b) (c) − (d) i − 1 i + 1 i + 1 i − 1 21. If z + 4 ≤ 3, then the maximum value of z + 1 is [2007] (a) 4 (b) 10 (c) 6 (d) 0 22. The value of ∑10 sin 2kπ + icos 2k is [2006] k = 1 11 11 (a) 1 (b) − 1 (c) − i (d) i 23. If z2 + z + 1= 0, where z is complex number, then + − = + = − + = ω= =ω− ω ≤ ω= ω =ω+ ≤ ≤ 5 2 3 2
4 181 2 2 2 the value of z + 1z z2 + z12 z3 + z13 2 + ... z6 + z 1 6 is [2006] (a) 54 (b) 6 (c) 12 (d) 18 24. If the cube roots of unity are 1, ω ω 2 , then the roots of the equation (x − 1)3 + 8 = 0, are [2005] (a) −1,1 + 2ω 1 + 2ω 2 (b) −1,1 − 2ω 1 − 2ω 2 (c) −1, −1, −1 (d) −1, −1 + 2ω, −1 − 2ω2 25. If z1 and z2 are two non-zero complex numbers such that z + z , then arg (z1 ) − arg (z2 ) is equal to [2005] (a) − (c) −π (d) w = and w = 1, then 26. If z lies on [2005] (a) a parabola (b) a straight line (c) a circle (d) an ellipse 27. Let z and w be two complex numbers such that z + iw = 0and arg (zw)= π. Then, arg ( )z equals [2004] (a)(b) (c)(d) 28. If z = x − iy and z1/3 = p + iq, then x y 2 2 / ( p + q )is equal to [2004] p q (a) 1 (b) − 1 (c) 2 (d) − 2 29. If |z2 − 1| = | |z 2 + 1, then z lies on [2004] (a) the real axis (b) the imaginary axis (c) a circle (d) an ellipse 30. Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle. Then, [2003] (a) a2 = b (b) a2 = 2b (c) a2 = 3b (d) a2 = 4b 31. If z and w are two non-zero complex numbers such that zw = 1, and arg ( )z − arg (w) = , then z w( ) is equal to [2003] (a) 1(b) − 1 (c) i (d) − i Other Engineering Entrances 34. If α and β are two different complex numbers with β − α β = 1, thenis equal to [Karnataka CET 2014] 1−αβ (a) 1/2 (b) 0 (c) −1 (d)1 35. If z, then z is equal to [Kerala CEE 2014] (a) 8 (b) 2 (c) 5 (d) 4 (e) 10 36. Let w ≠ ± 1 be complex number. If w = 1 and w − 1 = , then Re ( z )z is equal to[Kerala CEE 2014] w + 1 1 (a) 1 (b)(d) 0 (e) w + w 2 37. The value of z is minimum, when z equals [WB JEE 2014] −i (b) 45 + 3i (a) 2 ii (c)1 +(d)1 − 3 3 38. Convert (i + 1)/ cos π − isin π in polar form. 4 4 [J&K CET 2014] 4 z z 1 2 1 2 + = π 2 (b) 0 z z i − 3 π 4 3 4 π π 2 5 4 π 1 w + (c) Re()w z z i 2 2 3 + − + − 2 3
182 1 2 39. Let z1 ≠ z2 and z1 = z2 . If z1 has positive real part z2 + z1 and z2 has negative imaginary part. Then, z1 − z2 may be [Manipal 2014] (a) 0 (b) real and positive (c) real and negative (d) None of these 40. If z = e2π /3 , then 1+ z + 3z2 + 2z3 + 2z4 + 3z5 is equal to [Kerala CEE 2014] (a) − 3eπ i / 3 (b) 3eπ i/ 3(c) 3e2π i/ 3 (d) − 3e2π i/ 3 (e) 0 x 1+ i 32. If = 1, then [2003] 1− i (a) x = 4n, where nis any positive integer (b) x = 2n, where nis any positive integer (c) x = 4n + 1, where nis any positive integer (d) x = 2n + 1, where nis any positive integer 33. If ω is an imaginary cube root of unity, then (1+ ω − ω 2 )7 equals [2002] (a) 128 ω (b) −128 ω (c) 128 ω2 (d) −128 ω2 41. If ω, ω 2 are the cube roots of unity, then roots of equation (x − 1)3 + 5 = 0are [ RPET 2014] (a) −5, − 5 ω − 5ω 2 (b) −4 1, − 5 ω 1 − 5ω 2 (c) 6 1, − 5 ω 1 + 5ω 2 (d) None of the above 42. The complex number z = x + iy, which satisfies the z − 3i equation= 1, lie on [BITSAT 2014] z + 3i (a) the X-axis (b) the straight line y = 3 (c) a circle passing through the origin (d) None of the above 43. If the complex numbers z1, z2 and z3 denote the vertices of an isosceles triangle, right angled at z1, then (z1 − z2 )2 + (z1 − z3 )2 is equal to [Kerala CEE 2014] (a) 0 (b) (z2 + z3)2 (c) 2 (d) 3 (e) (z2 − z3)2 44. Let z1 and z2 be two fixed complex numbers in the argand plane and z be an arbitrary point satisfying z − z1 + z − z2 = 2 z1 − z2 . Then, the locus of z will be [WB JEE 2014] (a) an ellipse (b) a straight line joining z1 and z2 (c) a parabola (d) a bisector of the line segment joining z1 and z2 45. Let z1 be a fixed point on the circle of radius 1 centred at the origin in the argand plane and z1 ≠ ± 1. Consider an equilateral triangle inscribed in the circle with z1, z2 and z3 as the vertices taken in the counter clockwise direction. Then, z1 z2 z3 is equal to [WB JEE 2014] (a) z1 2 (b) z1 3 (c) z1 4 (d) z1 46. Suppose that z1, z2 and z3 are three vertices of an equilateral triangle in the argand plane. Let α = ( 3 − i) and β be a non- zero complex numbers. The pointsαz1 + β α z2 +β α, z3 +β will be [WB JEE 2014] (a) the vertices of an equilateral triangle (b) the vertices of an isosceles triangle (c) collinear (d) the vertices of a scalene triangle 47. In the argand plane, the distinct roots of 1+ z + z3 + z4 = 0 (z is a complex number) represent vertices of (a) a square (b) an equilateral triangle (c) a rhombus (d) a rectangle [WB JEE 2014] 48. If z1 = 2 2 1( + i) and z2 = +1 i 3, then z z1 2 2 3 is equal to [Kerala CEE 2013] (a) 128i (b) 64i (d) −128i (e) 256 (c) −64i 49.The value of (1+ i)3 + (1− i)3 is equal to[J&K CET 2013] (a) 1 (b) −2 (c) 0 (d) −4
4 183 50. If z(2− i2 3)2 = i( 3 + i)4 , then amplitude of z is [UP SEE 2013] (a) (b) (c)(d) None of these 51. If z1, z2 and z3 are complex numbers such that + = 1, then [AMU 2013] (a) 3 (b) 1 (c) greater than 3 (d) less than 1 52. If ( 3 i + 1)100 = 299 (a + ib), then a2 + b2 is equal to [RPET 2013] (a) 4 (b) 3 (c) 2 (d) 0 53. The value of (1+ 3 )i 4 + (1− 3 )i 4 is [RPET 2013] (a) − 16(b) 16 (c) 14 (d) − 14 54. If the fourth roots of unity are z1 , z2 , z3 and z4 ,then z1 2 + z2 2 + z3 2 + z4 2 is equal to [Karnataka CET 2013] (a) 0 (b) 2 (c) 3 (d) None of these 55. Among the complex number z satisfying condition z + 1− i ≤ 1, the number having the least positive argument is [OJEE 2013] (a) 1 − i (b) 1 + i (c) −i (d) None of the above 56. If z1 and z2 are two complex numbers such that z1 = z 2 + z 1 − z 2 , then [Manipal 2012] (a) Im z1 = 0 (b) Re z1 = 0 z2 z2 z z (c) Re 1 = Im 1 (d) None of these z2 z2 57. If the conjugate of (x + iy)(1− 2i)is 1+ i, then [Karnataka CET 2012] (a) x − iy = (b) x + iy = (c) x = = − (d) x 58. If = 325 (x + iy), where xand yare real, 2 2 then the ordered pair (x y, )is [WB JEE 2012] (a) (−3 0, ) (b) (0, 3) 1 3 (c) (0, −3) (d) , 2 2 59. If z − z + z + z = 2, then z lies on [AMU 2012] (a) a circle (b) a square (c) an ellipse (d) a line 60. If 2x = 3 + 5i, then the value of 2x3 + 2x2 − 7x + 72is [MP PET 2011] (a) 4 (b) −4 (c) 8 (d) −8 2 2 2 z1 61. Ifz1 + z2 = z1 + z2 , then is [MP PET 2011] z2 (a) purely real (b) purely imaginary (c) zero of purely imaginary (d) neither real nor imaginary 62. The value of[Karnataka CET 2011] (a) 20 (c) 63. If z1 and z2 are two non-zero complex numbers such z1 thatz1 + z2 = z1 + z2 , then arg is z2 [Kerala CEE 2011] (a) 0 (b) −π (c) − (e) π 64. If ω ≠ 1is a cube root of unity, then the sum of the series S = 1+ 2ω + 3ω 2 + ... + 3nω 3n − 1 is z z 2 3 1 1 + −π 6 π 6 z z z z 1 2 3 1 1 = = = z z z 1 2 3 + + is i i − − 1 1 2 1 5 3 3 50 +i i i − − 1 1 2 1 5 1 3 1 1 1 2 + + + i i is (b) 9 (d) 4 5
184 [WB JEE 2011] 3n (a)(b) 3n (ω − 1) (c)(d) 0 3n [Kerala CEE 2010] (a) − 20(b) −60 (c) −120 (d) 60 (e)156 68. The modulus of the complex number z such that z + 3 − i = 1 and arg ( )z = π, is equal to [Kerala CEE 2010] 73. (a) 1 (b) 2 (c) 9 (d) 4 (e) 3 69. If z , z ,...., z are complex numbers such that 1 2 n = 1,then z2 + z2 + ... + zn is [Kerala CEE 2010] (b) z1 + z2 + ... + zn 74. (d) n 1− i of z) [WB JEE 2010] (a) 2(1 + i) (b) (1 + i) 2 4 (c) (d) 1 − i If − π < arg ( )z < − , then arg ( )z − arg(−z)is [WB JEE 2010] (a) π (b) −π (c) π /2 (d) −π /2 The value of is equal to [VITEEE 2010] (a) i 1 + 3i (c)(d) 2 2 4 1+ cos θ2 − isin θ2 4n 70. 65. is equal to [UP SEE 2011] 1+ cos θ 2 + isin θ 2 (a) cos nθ − i sin nθ (b) cos nθ + i sin θ (c) cos 2nθ − i sin 2nθ (d) cos 2nθ + i sin 2nθ − 3 + i 3 66. If x =is a complex number, then the value z z 2 71. If z = r (cos θ + isin θ),then the value of+ is of (x2 + 3x)2 (x2 + 3x + 1)is [UP SEE 2011] z z [Kerala CEE 2010] (a) − (b) 6 (a) cos 2 θ (b) 2 cos 2 θ (c) 2 cos θ (d) 2 sin θ (c) − 18 (d) 36 (e) 2 sin 2 θ 67. If (x + iy)1 3/ = 2 + 3i, then 3x + 2yis equal to 72. If z = 4 , then z is (where, z is complex conjugate 1 ω− ω−1 1 − i If i z1 2 4 4 + = sin cos π π and i z2 3 3 = + cos , sin π π 3 then zz 2 1 is 2010] KeralaCEE [ (b) 6 a) ( 2 c) ( 6 d ( ) 3 (e) 2 3 + z z zn 2 1 = = =... equalto ) (a z zzz n 123 ... ) c ( 1 1 1 2 1 z z zn + + + ... (e) n cos sin sin cos 30 30 60 60 ° °+ °− ° i i (b) −i 1 3 − i
185 Work Book Exercise 4.1 Answers 1. (b) 2. (a) 3. (c) 4. (b) 5. (a) 6. (d) 7. (b) 8. (d) 11. (d) 12. (d) 13. (c) Work Book Exercise 4.2 14. (c) 1. (a) 2. (b) 3. (a) 4. (c) 5. (a) 6. (b) 7. (c) 8. (b) 11. (a) 12. (a) 13. (d) Work Book Exercise 4.3 14. (b) 15. (b) 1. (c) 2. (b) 3. (b) 4. (c) 5. (a) 6. (d) 7. (c) 8. (b) 11. (a) 12. (b) 13. (b) 14. (c) 15. (b) 16. (c) 17. (c) 18. (a 21. (c) 22. (a) 23. (c) Work Book Exercise 4.4 24. (a) 25. (d) 26. (c) 27. (b) 1. (c) 2. (d) 3. (b) 4. (b) 5. (a) 6. (b) 7. (a) 8. (a) 11. (b) 12. (c) 13. (c) 14. (a) 15. (a) 16. (c) 17. (a) 18. (b 21. (a) 22. (d) Target Exercises 23. (a) 24. (b) 25. (a) 26. (b) 27. (b) 28. (a 1. (b) 2. (a) 3. (c) 4. (d) 5. (b) 6. (c) 7. (c) 8. (c) 11. (b) 12. (d) 13. (c) 14. (a) 15. (c) 16. (d) 17. (d) 18. (a) 21. (a) 22. (c) 23. (c) 24. (d) 25. (c) 26. (c) 27. (d) 28. (c) 31. (a) 32. (a) 33. (c) 34. (c) 35. (b) 36. (b) 37. (b) 38. (b) 41. (d) 42. (a) 43. (b) 44. (a) 45. (a) 46. (c) 47. (b) 48. (b) 51. (c) 52. (a) 53. (b) 54. (b) 55. (a) 56. (c) 57. (a) 58. (c) 61. (d) 62. (c) 63. (a) 64. (d) 65. (c) 66. (a) 67. (a) 68. (b) 71. (b) 72. (a) 73. (b) 74. (d) 75. (c) 76. (c) 77. (a) 78. (a) 81. (c) 82. (b) 83. (b) 84. (c) 85. (a) 86. (a) 87. (c) 88. (c) 91. (a) 92. (c) 93. (b) 94. (a) 95. (a) 96. (c) 97. (a) 98. (d) 101. (b) 102. (d) 103. (a) 104. (c) 105. (c) 106. (a) 107. (b) 108. (c) 111. (b) 112. (c) 113. (c) 114. (b) 115. (d) 116. (a) 117. (a,b) 118. (a,b,d) 121. (a,c) 122. (a,d) 123. (a,b) 124. (a,d) 125. (a,b,d) 126. (a,b) 127. (a,c) 128. (b,c) 131. (a,d) 132. (a,c) 133. (a,b) 134. (b) 135. (d) 136. (a) 137. (a) 138. (d) 141. (d) 142. (d) 143. (d) 144. (a) 145. (b) 146. (b) 147. (c) 148. (c) 151. (c) 152. (a) 153. (b) 154. (*) 155. (**) 156. (***) 157. (****) 158. (3) 161. (1) 162. (4) * A → r; B → p; C → q; D → s ** A → p, q; B → p, q; C → p, q; D → p, q *** A → s; B → r; C → p; D → q **** A → r,s; B → p; C → q; D → p Entrances Gallery 1. (c) 2. (c) 3. (c,d) 4. (b) 5. (c) 6. (d) 7. (5) 8. (3) 11. (c) 12. (b) 13. (c) 14. (a) 15. (c) 16. (a) 17. (b) 18. (c) 21. (c) 22. (c) 23. (c) 24. (b) 25. (b) 26. (b) 27. (c) 28. (d) 31. (d) 32. (a) 33. (d) 34. (d) 35. (b) 36. (d) 37. (c) 38. (d) 41. (b) 42. (a) 43. (a) 44. (a) 45. (b) 46. (a) 47. (b) 48. (d) 51. (b) 52. (a) 53. (a) 54. (a) 55. (d) 56. (a) 57. (b) 58. (d)
186 61. (b) 62. (d) 63. (a) 64. (a) 71. (b) 72. (d) 73. (a) 74. (a) * A → q; B → p, t; C → p; D → s,q 65. (c) 66. (c) 67. (c) 68. (e)
4 187 Explanations Target Exercises i 584 (i 8 + i 6 + i 4 + i 2 + 1) i 584 1. 574 8 6 4 2 − 1 = 574 − 1 i (i + i + i + i + 1) i = i10 − 1 = − 1− 1 = − 2 2. i 57 + i1251 = (i 4 14) i + (i 4 311) i = i + 1i = i − i = 0 3. We have, i n + i n + 1 + i n + 2 + i n + 3 = i n (1 + i + i 2 + i 3 ) = i n (1+ i − 1 − i ) [i 3 = i 2 ⋅ i = −(1)⋅ i = − i ] = i n (0) = 0 4. Given expression = 1 + i 2 + i 4 + i 6 + ... + i 2n = 1 − 1 + 1 − 1 + ... + (− 1)n which cannot be determined unless n is known. 5. We have, i 1i 25 2 i16 ⋅ i 3 + i 241⋅ i 2 19 2 1 2 2 i⋅ i + i = (− i − i ) = −( 2 )i 2 = 4i 2 = − 4 i 4n ⋅ i − i 4n 4 n n = [i= (i ) = ( )1 = 1] 2 i 7. (1 − i ) n 1 − 1 n = (1 − i ) n (1 + i ) n (1 + i ) n − 2 (1 + i ) n − 2 (1 − i ) n − 2 n − 1 = 2 (− i )n − 1 = 2 (− 1)2 n − 1 This is positive and real, if is even. 2 n − 1 Let = 2λ 2 ∴ n = 4λ + 1 n n n 9. 1 + i 1 + i × 1 + i 1 − 1 + 2i 1 − i 1 − i 1 + i 1 + 1 = i n = − 1, if n = 2 1 − i 1 − i 2 13. We have, 1 + i 2 1 − i 2 1 − i 1 + i 2 2 (1+ i ) (1 − i ) 2i − 2i = (1 − i )2 + (1 + i )2 = − 2i + 2i  (1 + i )2 = 1 + i 2 + 2i = 2i 2 2 and (1 − i ) = 1 + i − 2i = − 2i = − 1 − 1 = − 2 14. Re ( 13+−ii)2 = Re 2i 3 + i = Re 15. Given that, a2 + b2 = 1 1 + b + ia (1 + b + ia)(1 + b + ia) ∴ = (1+ b − ia (1 + b − ia) (1 + b + ia) (1 + b)2 − a2 + 2ia (1 + b) = 2 2 1 + b + 2b + a i = (1 + 1) n= 2n (1 − i ) n 8. = (1 − i ) n (1 − i ) n − 2 × 10. − = + ⇒ =− ∴ 11. + = + × + + = − + =− + ∴ + − 12. − − = − + − + = − − + =− 6. − + − ⋅ = − − = − = − − = − = − = − + − + + = − =−
188 = b + ia 16. We have, b + ic = (1 + a z) b + ic ib − c a + ib 1 + c 17. x + iy = 6 i (3 i 2 + 3) + 3 i (4 i + 20) + 1 12( − 60i ) = 0 − 12 + 60i + 12 − 60i = 0 + 0i ∴ x = 0, y = 0 18. 2i = 1 + i 2 + 2i = (1 + i )2 = 1 + i 19. Let z = x + iy, where x y, ∈R ∴ x2 + y2 ≠ 0 and x = 0 ∴ z = 0 + iy = iy, y ≠ 0 ⇒ z2 = − y2 ⇒ Im (z2 ) = 0 20. We have, (x + iy)1 3/ = a + ib On cubing both sides, we get x + iy = (a + ib)3 = a3 + 3 a2 (ib) + 3 a ib()2 + (ib)3 = a3 + 3 a2 bi + 3 ab i2 2 + i b3 3 = a3 + 3 a bi2 − 3 ab2 − ib3 [i 3 = i 2 ⋅ i = − i] = a a(2 − 3 b2 ) + ib (3 a2 − b2 ) ∴ x = a a( 2 − 3 b2 ) and y = b(3 a2 − b2 ) ⇒ x = a2 − 3 b2 and y = 3 a2 − b2 a b x y 2 2 ∴ + = 4(a − b ) a b 21. We have, 8 iz3 + 12 z2 − 18z + 27i = 0 ⇒ 4z2 (2iz + 3) + 9 i (2iz + 3) = 0 ⇒ (2iz + 3) (4z2 + 9 i ) = 0 ⇒ 2iz + 3 = 0 or 4z2 + 9 i = 0 ∴ z = 22. Multiplicative inverse of z2 1 − i 1 + i Hence, (x y, ) = (0, − 2). 24. Let z = (6 + 5 )i 2 = 36 + 2(6)(5 i ) + 25 i 2 = 36 + 60i − 25 = 11 + 60i 1 − i 1 + i ⇒ i 3 − −( i )3 = x + iy ⇒ − i + i 3 = x + iy ⇒ − i − i = x + iy x + iy = 0 − 2i ∴ = + ⇒ = + ⇒ + − = − + + − + + = + − + + − + × + + + + + + = + + + + + + = + + + + = = = + =− = 23. + − = =− ∴ + − − = + x
189 = − − = and ∴ Multiplicative inverse of − − i z = i 26. Let z = x + iy, where x < 0, y < 0 z x − iy (x − iy)2 x2 − y2 2 xy ∴ z = x + iy = (x2 + y2 ) = x2 + y2 − x2 + y2 i z 2 2 lies in III quadrant, if x − y < 0 and − 2 xy < 0 z x < 0, y < 0 ⇒ − 2 xy < 0 Also, x2 − y2 < 0, if (x + y) (x − y) < 0 If x − y > 0, if x > y [x y, < 0 ⇒ x + y < 0] ∴ lies in III quadrant, if y < x < 0. 27. Let z 2 − i − 2 + i 3 − 4 i ∴= × − 3 − 4 i 3 + 4 i 3 − 4 i i 29. (1 + ix) (1 − ix) 1 − x2 − 2ix a − ib ⇒ 1 + x2 = 1 1 − x2 2 x ⇒ 2 = a and 2 = b 1 + x 1 + x Now, x can be written as = = − −
4 190 32. We have, x + iy = u + iv …(i) u − iv We know that, when two complex numbers are equal their conjugates are also equal. u − iv ∴ x − iy = …(ii) u + iv On multiplying Eqs. (i) and (ii), we get u + iv u − iv (x + iy) (x − iy) = × u − iv u + iv 2 2 2 ⇒ x2 + y2 = u 2 − i v2 2 u − i v 2 2 u 2 + v2 ∴ x + y = 2 2 = 1 u + v 33. Taking modulus and squaring on both sides, we get 1 + i 2 ⋅ 1 + 2i 2 ⋅ 1 + 3 i 2 .... 1 + ni 2 = α + i β 2 ⇒ (1 + 1)⋅ (1 + 4)⋅ (1 + 9) ... (1 + n2 ) = α 2 + β2 ⇒ 2 ⋅ 5⋅ 10 ... (1 + n2 ) = α 2 + β2 12 36. z ⇒ x2 − 8x + 16 + y2 = x2 ⇒ y2 = 8 (x − 2) The given relation represents the part of parabola with focus (4, 0) lying above X-axis and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence, the greatest positive argument of z is . 42. Let a + bi be a square root of − 2 + 2 3 i. ∴ (a + bi )2 ⇒ a2 + 2abi + i b2 2 = + + − ⇒ = π π π π π − + = + − π 41. − = ⇒ x x − + = 0) (4, Y X X′ Y′ O
4 191 + =− =− + = − − − − −  − − = ⇒ (a2 − b2 ) + 2abi = − 2 + 2 ∴ a2 − b2 = − 2 …(i) and 2ab = 2 …(ii) Now, (a2 + b2 2) = (a2 − b2 2) + 4 a b22 = −(2)2 + (2 3)2 = 4 + 12 = 16 ∴ a2 + b2 = 16 = 4 …(iii) On solving Eqs. (i) and (iii), we get 2a2 = 2 or a2 = 1 ∴ a = ± 1 ⇒ 2b2 = 6 or b2 = 3 ∴ b = ± 3. From Eq. (ii), ab = 3, which is positive. Eithera = − 1, b = − i.e. ± (1 + Aliter From Eq. (i), we get| z2| ≥ 2ab ∴ | z|2 + a2 2 ≥ a2 + b2 + 2ab ⇒2 ⇒2 ⇒ 2 z ≥ a + b, as z is positive ∴ (a + b) 44. We have, 1 − z z − z − z =(1 − z z12 ) (1 − z z12 ) − (z1 − z2 ) (z1 − z2 ) [z z = z 2 ] and 1 = 1] =(1 − z z12 ) (1 − z z12 ) − (z1 − z2 ) (z1 − z2 ) [(z1) = z1] = 1 − z z1 2 − z z1 2 + z z z z1 1 2 2 − z z1 1 + z z1 2 + z z1 2 − z z2 2 = 1 + z1 2 z2 2 − z1 2 − z2 2 = (1 − z1 2 ) (1 − z2 2 ) ∴ k = 1 45. Let z = x + iy We have, z + α z − 1 + 2i = 0 ⇒ x + i (y + 2) + α (x − 1)2 + y2 = 0 On equating real and imaginary parts y + 2 = 0 ⇒ y = − 2 and x + α (x − 1)2 + 4 = 0 ∴ x2 = α 2 (x2 − 2 x + 5) or (1 −α 2 )x2 + 2α 2 x − 5α 2 = 0 Since, x is real. ∴ D = B2 − 4 AC ≥ 0 ⇒ 4α 4 + 20 α 2 (1 + α 2 ) ≥ 0 ⇒ − 4 α 4 + 5α 2 ≥ 0 ⇒ 4α 2 α 2 − 5 ≤ 0 4 − 5 5 ∴ ≤ α ≤ 2 2 46. Given, z = 2 = 10 Let) ⇒ ∴Locus of α, i.e. 5z − 1is the circle having centre at − 1 and radius 10. ∴ a = 10 Again, z1 2 + z2 2 − 2 z z12 cos θ = 0 z 2 z + + ≥ + + ≥ ≥ α= − ⇒ α+ = = α+ =
4 192 1 − 2 1 cos θ + 1 = 0 ⇒ ⇒ ∴ ⇒ ⇒ ⇒ ⇒ ⇒ = 0 ⇒ = 1 [z1 ≠ − z2 ] 49. Since, 1, ω ω 2, ...,ω n − 1 are the nth roots of unity. n − 1 n − 1 n − 1 = ∑ k = 0 n − 1 = ∑ k = 0 = n z ∴ ⇒ ⇒ ( ⇒ x2 + 4 = (1 + x) ⇒ 2 x = 3 or 3 ∴ z = x + iy = − 2i 2 5 1. ( 3 + i ) = (a + ib) (c + id ) On taking argument both sides, we get tan− 1 1 = tan− 1 b + tan− 1 d 3 a c
4 193 ⇒ tan− 1 b + tan− 1 d = nπ + π , n ∈Z a c 6 52. Let z − 1 = e iθ ⇒ z = 1 + cos θ + i sin θ [z − 1 = 1] θ θ θ θ = 2 cos cos + i sin 2 2 ∴ arg (z) = = arg (z − 1) or arg (z − 1) = 2 arg (z) 53. arg (z )arg (− 1) = = z − z2 (x + iy) − (6 + 4i ) (x − 6) + i (y − 4) z − z1 = π We have, arg z − z2 4 x + y − 14x − 8y + 64 ⇒ x y x y ⇒ ∴ z − (7 + 5 i ) = 2 55. arg (− z) = π − θ = π + (−θ) = π + arg (z) ∴ arg (− z) − arg (z) = π 56. Since, z − 1 = 1 Let z − 1 = cos θ + i sin θ Then, z − 2 = cos θ + i sin θ − 1 = − 2 sin2 θ + 2 i sin θ cos θ 2 2 2 = 2i sin cos + i sin …(i) 2 2 2 and z = 1 + cos θ + i sin θ = 2 cos2 θ + 2i sin θ cos θ 2 2 2 = 2 cos cos + i sin …(ii) 2 2 2 From Eqs. (i) and (ii), we get z − 2 z 2 arg z = 2 , from Eq. (ii) = i tan = i tan (arg z) 57. We have, C2 + S 2 = 1 Let C ∴ = 1 + C − iS 1 + cos θ − i sin θ cos cos + i sin 2 2 2 = cos cos − i sin 2 2 2 θ 2 cos + i sin 2 2 = cos θ + i sin θ 1 =C + iS 58. Let z − 1 = r (cos θ + i sin θ) = re iθ ∴ Given expression = re iθ ⋅ e −iα + 1 iθ e iα re = re i( θ − α) + 1 e −i( θ − α) r ⇒ = − − − π ⇒ − − − − − − − = x x π ⇒ − =
4 194 + + + − − 16 59. 8 8 cos π + i sin π 8 8 8 sin − i cos 8 8 61. We have, 1 + i 8 1 − i 8 π π 8 = cos 4 4 4 4 = cos 2π + i sin 2π + cos 2π − i sin 2π = 2 cos 2π = 2 (1) = 2 [using De-Moivre’s theorem] 10 = − i ∑ e 11 − 1 k = 1 = − i (Sum of 11th roots of unity − 1) = − i (0 − 1) = i 63. We have, 4 (cos 75° + i sin 75° ) 0 4. (cos 30° + i sin 30° ) 10 {cos (75° − 30° ) + i sin (75° − 30° )} = 2 2 cos 30° + sin 30° 10 = 10 (cos 45° + i sin 45° ) = (1 + i ) 2 64. Using De-Moivre’s theorem, the given expression (cos θ + i sin θ)− 14 (cos θ + i sin θ)− 15 = θ + i sin θ)48 (cos θ + i sin θ)− 30 (cos = (cos θ + i sin θ)− 47 = cos 47θ − i sin 47θ 65. We have, sin θ − i cos θ = − i 2 sin θ − i cos θ = − i (cos θ + i sin θ) ∴The given expression, using De-Moivre’s theorem = −(i )3 [cos (− 25 θ) + i sin (− 25 θ)] = i [cos 25 θ − i sin 25 θ] = sin 25 θ + i cos 25 θ 66. We have, z = (1 + cos θ + i sin θ)3 5 (cos θ + i sin θ) 2 θ θ 5 1 + 2 cos − 1 + 2 i sin cos = 2 2 2 cos 3 θ + i sin 3 θ 5 θ 5 32 cos cos + i sin = 2 2 2 cos 3 θ + i sin 3 θ = 32 cos5 cos 5 θ + i sin 5 {cos 3 θ − i sin 3 θ} 2 2 2 = 32 cos5 2 cos 5 2 θ − 3 θ + i sin 5 2 θ − 3 = 32 cos5 θ 2 cos θ 2 + i sin − θ 2 Hence, the modulus and argument of z are 32 cos5 θ 2 Since, imaginary part of given expression is zero, we have 1 r sin (θ − α ) − sin (θ − α ) = 0 r r 2 − 1 = 0 ⇒r 2 = 1 ⇒ r = 1 ⇒ z − 1 = 1 or sin (θ − α ) = 0 ⇒(θ − α ) = 0 ⇒ θ = α ⇒ π 8 sin + i cos arg (z − 1) = α = π− π= 60. + + + = π π + = π π π + =− π π ∴ + = − + = − π π 62. π π − = ∑ = − − = ∑ π π + =− =− = = ∑ ∑ π π π π
4 195 and , respectively. 2 67. We have, 1 1 = = cos θ − i sin θ z cos θ + i sin θ ∴ zn = (cos θ + i sin θ)n = cos nθ + i sin nθ , 1 n and n = (cos θ − i sin θ) = cos nθ − i sin nθ z Hence, zn + 1 n = 2 cos nθ and zn − 1 n = 2i sin nθ z z 68. We have, [using De-Moivre’s theorem] 69. We have, abcd = cos (2α + 2β + 2γ + 2δ) + i sin (2α + 2β + 2γ + 2δ) ∴ abcd = [cos (2α + 2β + 2γ + 2δ) +i sin (2α + 2β + 2γ + 2δ)]1 2/ or abcd = cos (α + β + γ + δ) + i sin (α + β + γ + δ)…(i) [using De-Moivre’s theorem] 1 ∴ = cos (α + β + γ + δ) abcd − i sin (α + β + γ + δ)…(ii) On adding Eqs. (i) and (ii), we get 1 abcd + = 2 cos (α + β + γ + δ) abcd 70. Let cot− 1 p = θ then p = cot θ ∴ e 2mi cot −1 p ⋅ pi + 1 m = e 2 miθ ⋅ i cot θ + 1 m pi − 1 i cot θ − 1 =e 2miθ. i (cot θ − i ) m = e 2miθ ⋅ cot θ − i m i (cot θ + i ) cot θ + i = e 2miθ ⋅ cos θ − i sin m = e 2miθ ⋅ e − iθ m 8 8 i 71. We have, α = cos i sin e 11 11 Re (α + α 2 + α 3 + α 4 + α 5 ) 1 + (1 + α + α 2 + α 3 + α 4 + α 5 2 3 4 5 = + α + α + α + α + α ) 2 − 1 + 0 =[sum of 11 and 11th roots of unity] 1 = − 2 72. We have, 1 + 1 + = 2ω2 = 2 1 ω 2 and a + ω2 b + ω2 c + ω2 = 2ω = ω ∴ ω and ω2 are roots of the equation 1 1 1 2 cos θ + i sin = e 2miθ (e − 2iθ )m = e 0 = 1 e iθ − + = + − + + θ θ) θ) θ = + − + + θ θ θ θ = − + − + − θ θ) θ θ θ θ+ = + + θ θ θ θ θ θ  =− = θ θ θ θ θ θ θ + + = + + + ω ω ω + +
4 196 + + = . a + x b + x c + x x When x = 1, 1 1 1 2 + + = = 2 a + 1 b + 1 c + 1 1 2 1 3 73. x − 2 x + 4 = 0 ⇒ x = 1 + 3 i = 2 ± i 2 2 = − 2ω − 2ω2 ⇒ α 6 + β6 = 128 1 74. z + = 1 z ⇒ z = − ω or − ω2 ∴ z99 + 1 = −( 1) + 1 = − 2 z99 (− 1) 75. (− 1 + − 3)62 + (− 1 − − 3)62 = 262 ω62 + 262 (ω2 62) = 262 [(ω3 20)ω2 + (ω3 41)ω] = 262 [(ω2 + ω] = − 262 76. Given expression = {3(1 + ω2 ) + 5ω}2 + {3 (1 + ω +) 5ω2}2 = (− 3ω + 5ω)2 + (− 3ω2 + 5ω2 2) = 4ω2 + 4ω = − 4 6 6 ω2 − 1 + i 3 2 2 z2 z2 3 25 2 2 2 z1 50 lies in I quadrant. z2 79.xyz = (a + b) (aω + bω2 ) (aω2 + bω) = (a + b) (a2 − ab + b2 ) = a3 + b3 80. (1 + ω)3 − (1 + ω2 3) = − ω( 2 3) − − ω)( 3 = − ω6 + ω3 = − 1 + 1 = 0 81. p1 3/ = −( p)1 3/ (− 1)1 3/ = − −(p)1 3/, − −(p)1 3/ω − −(p)1 3/ ω 2 = − q, − qω, − qω2, where q = −(p)1/ 3 Let α = − q, β = − qω and γ = − qω2 xα + yβ + z γ − q (x + yω + zω2 ) ∴ x y z = 2 = ω xω + yω + z 82. Given expression n = ∑ (k − 1)(k − ω) (k − ω2 ) k = 2 nn 2 2 2 n n(+ 1) n 2 83. Let α = ω and β = ω2 Now, 84. We have, z2 2 z 50 ⇒ 1 z1 2 25 = − 3 i 25 = 3 25 = − iω 77. ω ω 2 + = ω= ω = − − 78. = − ω +
4 197 (1 − ω + ω2 ) (1 − ω2 + ω4 )(1 − ω4 + ω8 )(1 − ω8 + ω16 ) = (1 + ω2 − ω) (1 − ω2 + ω) (1 − ω + ω2 ) (1 − ω2 + ω) [ω4 = ω3 ⋅ ω = ω;ω8 = (ω3 2) ⋅ ω2; ω16 = (ω3 5) ⋅ ω =ω andω3 = 1] = (− ω − ω) (− ω2 − ω2 ) (− ω − ω) (− ω2 − ω2 ) = (− 2ω) (− 2ω2 )(− 2ω) (− 2ω2 ) = 16⋅ω6 = 16 (ω3 2) = 16 1( )2 = 16 85. We have, x = ω − ω2 − 2 or x + 2 = ω − ω2 On squaring, x2 + 4x + 4 = ω2 + ω4 − 2ω3 = ω2 + ω3 ⋅ ω − 2ω3 = ω2 + ω − 2 [ω2 = 1] = − 1 − 2 = − 3 ⇒ x2 + 4x + 7 = 0 On dividing x4 + 3x3 + 2 x2 − 11x − 6 by x2 + 4x + 7, we get x4 + 3x3 + 2 x2 − 11x − 6 = (x2 + 4x + 7) (x2 − x − 1) + 1 = (0) (x2 − x − 1) + 1 = 0 + 1 = 1 86. We have, (1 + ω) (1 + ω)2 (1 + ω4 ) (1 + ω8 ) ... to 2n factors = (1 + ω) (1 + ω2 ) (1 + ω3 ⋅ ω) (1 + ω6 ⋅ω2 ) ... to 2n factors = (1 + ω) (1 + ω2 ) (1 + ω) (1 + ω2 ) ... to 2n factors [ω3 = ω6 = ... = 1] = [(1 + ω) (1 + ω ...)to n factors] [(1 + ω2 ) (1 + ω2 ) ... to n factors] = (1 + ω) n (1 + ω2 ) n = [(1 + ω) (1 + ω2 )]n = (1 + ω + ω2 + ω3 ) n = (0 + 1) n = 1 [1 + ω + ω2 = 0, ω3 = 1] 87. We have, α = ω and β = ω2 Then, xyz = (a + b + c ) (aω + bω2 + c ) (aω2 + bω + c ) = (a + b + c ) (a2 + b2 + c 2 − ab − bc − ca) = a3 + b3 + c 3 − 3 abc 88. We have, z3 + 2 z2 + 2 z + 1 = 0 ⇒ (z + 1) (z2 + z + 1) = 0 Its roots are − 1, ω and ω2. The root z = − 1 does not satisfy the equation, z1985 + z100 + 1 = 0 but z = ω and z = ω2 satisfy it. Hence, ω and ω2 are the common roots. 89. We have, (16)1 4/ = (24 1 4) / = 2 1( )1 4/ 1 1 0, 1, 2, 3 4 4 = 2 × 1, 2 × i, 2 × − 1, 2 × − i, = ± 2, ± 2i 90. We have, f(ω) = ω3p + ω3q + 1 + ω3r + 2 = ω3p + ω3q ⋅ ω + ω3r ⋅ ω2 = (ω3 ) p + (ω3 )q ⋅ ω + (ω3 ) r ⋅ ω2 = 1 + ω + ω2 = 0 [ω3 = 1] 91. Since 1, α, α 2, ..., α n − 1 are n and nth roots of unity. ∴ xn − 1 = (x − 1) (x − α) (x − α 2 ) ...(x − α n − 1) ⇒ log (xn − 1) = log (x − 1) + log (x − α ) + log (x − α 2 ) + ... + log (x − α n − 1) On differentiating both sides w.r.t. x, we get nxn − 1 1 1 1 1 = + + + ... + xn − 1 x − 1 x − α x − α 2 x − α n − 1 Putting x = 2, we get n2n − 1 1 1 1 1 = + + + ... + n n − 1 − 1 ∴ 1 2 − 1 i = 1 2 − α n − 1 n⋅2n − 1− 2n + 1 Hence, = n i i = 1 2 − 1 92. The vertices of the triangle are A (1, 0), B (1/ 2, 1/ 2 ) and C (0, 1). ∴ AB2 = 2 − 2 BC2 = 2 − 2
4 198 AC2 = 1 + 1 = 2 ∴ AB = BC 93. By definition, z1 = x1 + iy1, z2 = x2 + iy2 and collinear, if z3 = x3 + iy3 are x1 y1 1a x2 y2 1= 0 ⇒a′ x3 y3 1a − a′ ⇒ ab′ = a b′ 94. Let z = x + iy, where x y, ∈R b 1 b′ 1= 0 b − b′ 1 ∴ α + β =i tan− 1 (x + iy) ⇒ α − β =i tan− 1 (x − iy) ⇒ 2α = tan− 1 (x + iy) + tan− 1 (x − iy) = tan− 1 22x 2 1 − x − y ∴ x2 + y2 + 2 x cot 2α = 1 n z − 1 = z + 1 (x − 1)2 + y2 = (x + 1)2 + y2 ∴The roots lies on the Y-axis. 97. Let z = x + iy, where x y, ∈R We have, iz − 1 + z − i = 2 ⇒ ix − y − 1 + x + iy − i = 2 ⇒ (− y − 1)2 + x2 + x2 + (y − 1)2 = 2 ⇒ x2 = 0 or x = 0 ∴The locus of z is a straight line. 98. Clearly, z1 < z3 < z2 , as z3 is mid-point of z1 and z2. 99. (a) Re 1 + z = 0 ⇒ arg 1 + z = π 1 − z 1 − z 2 This represents a circle. (b) z z + iz − iz + 1 = 0 represent a circle. z − 1 π (c) arg z + 1 2 This represents a circle. (d) This represents a straight line. 100. We have, z2 = z + z The point representing z2 divided the line segment joining points representing z1 and z3 in the ratio 1: 1. ∴The points lies on a line. 101. Area of the triangle on the argand plane formed by 3 2 complex numbers − z iz z, , − iz is z . 2 3 2 ∴ z = 600 ⇒ z = 20 2 102. Let z1 = 1 − 3 i, z2 = 4 + 3 i and z3 = 3 + i. Then, z3 divides the line segment joining z1 and z2 in the ratio 2 : 1 internally. So, the points z1, z2 and z3 are collinear. 103. Let z = x + iy z + 2i z + 4 which represents a circle with centre (− 2, − 1). 104. Let z = x + iy z + 2i x + iy + 2i x + (y + 2)i Then, = = − + = ⇒ − = + 95. = − − 96. = − + ∴ = + − ⇒ + = − ⇒ = − + ⇒ ⇒ ⇒ x = ⇒ x = x x x x = + + + + = + + + + = x x x + + + − + + = + + + + + + + + x x x x + + = ⇒ x x = + + +
4 199 z + 2 x + iy + 2 (x + 2) + iy z + 2 which represents a straight line. 105. We have, z − 4 i + z + 4 i = 10 ⇒ z − (0 + 4 i ) + z − (0 − 4 i ) = 10 This represents an ellipse. 4 i + 4 i < 10 i.e. 10 > 8 106. We have, z − a i = z + a i ⇒ x + i (y − a) 2 = x + i (y + a) 2 ⇒ x2 + (y − a)2 = x2 + (y + a)2 ⇒ 4ay = 0; y = 0, which is X-axis. 107. We have, z − 1 2 + z + 1 2 = 4 ⇒ (x − 1) + iy 2 + (x + 1) + iy 2 = 4 ⇒ (x − 1)2 + y2 + (x + 1)2 + y2 = 4 ⇒ 2 (x2 + 1) + 2 y2 = 4 ∴The locus of P is x2 + y2 = 1. ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ∴ ⇒ ⇒ and i.e. A2 − B2 = − 4x …(ii) On dividing Eq. (ii) by Eq. (i), we get ≤ − x ⇒ ≤ 4 − x ⇒ 3x 2 ≤ 12 [squaring and simplifying] x2 y2 or + ≤ 1 4 3 which represents the interior and boundary of an ellipse. 110. We have, log1 2/ 3zz−−11+−42 > 1 = log1 2/ 21 z − 1 + 4 1 ⇒ < < 1 3 z − 1 − 2 2 [loga x is a decreasing function, if a < 1] ⇒ z − 1 + 4 < 3 z − 1 − 2 ⇒ 2 z − 1 > 6 ⇒ z − 1 > 3 which is an exterior of a circle. 111. The closest distance = length of the perpendicular from the origin on the line az + az + aa = 0 a (0) + a = = + + + − + + x x x = + + + + + + + + x x x x + = ⇒ x = + + x − + − x + + x − + + + = = + = + − +
4 200 112. We have, zk = 1 + a + a 1 − a 113. We have, z1 − z4 = z2 − z3 or z1 + z3 = z2 + z4 i.e. the 2 2 diagonals bisect each other. ∴It is a parallelogram. z4 − z1 = π Also, amp z2 − z1 2 ⇒ Angle at z1 is a right angle. ∴ It is a rectangle and hence, a cyclic quadrilateral. 114. We have, Im (z2 ) = 4 ⇒ Im [(x2 − y2 ) + 2 ixy] = 4 [putting z = x + iy] ⇒ 2 xy = 4 or xy = 2, which is a rectangular hyperbola. 115. We have, Re (z2 ) = 4 ⇒ Re [(x2 − y2 ) + 2 ixy] = 4 [putting z = x + iy] ⇒ x2 − y2 = 4 which is a rectangular hyperbola. 116. z − z1 = z − z2 = z − z3 ∴ z must be the mid-point of z1 and z2 and since z − z1 = z − z2 and z z, 1 and z2 are collinear. ⇒ z is circumcentre of ∆ formed by z1, z2 and z3. z3 − z1 = ± π [neglecting − ve value] ∴ arg z3 − z2 2 117. a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 ⇒ a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 [taking conjugate on both sides] ⇒ a0(z4 ) + a z1( )3 + a2(z)2 + a3(z)2 + a z3 + a4 = 0 ∴z is a root of the equation if z is a root. So, option (a) is correct. Also, if z1 is real, z1 = z1. If z1 is non-real complex, then z1 is also a root because imaginary root occurs in conjugate pairs. So, option (b) is also correct. 118. Let z =α be a real root. Then, α 3 + (3 + 2i )α + (− 1 + ia) = 0 ⇒ (α 3 + 3α − 1) − i (a + 2 α ) = 0 ⇒ α 3 + 3α − 1 = 0 and α = − a / 2 ⇒ ⇒ a3 + 12a + 8 = 0 Let f a() = a3 + 12a + 8 ∴ f(− 1) < 0, f(0) > 0, f(− 2) < 0, f(1) > 0 and f(3) = 0 119. We have, ⇒ θ = 2nπ ± where, n is an integer. 120. Let z = c be a real root. Then, αc 2 + c + α = 0 Let α = p + iq Then, (p + iq c) 2 + c + p − iq = 0 ⇒ pc 2 + c + p = 0 and qc 2 − q = 0 …(i) ⇒ c = ± 1 [q ≠ 0] From Eq. (i), we get α ± 1 + α = 0 Also, c = 1 121. Let z2 = r2 (cos θ2 + i sin θ2 ) ⇒ z2 = r2 Also, arg (z1) + arg (z2 ) = 0 ⇒ arg (z1) = − arg (z2 ) = − θ2 ∴ z1 = r2 [cos (− θ2 ) + i sin (− θ2 )] = r2 [cos θ2 − i sin θ2 ] ⇒ z1 = z2 1 ⇒ z1 = z2 ∴ z z12 = 1 + − = + + + − θ θ θ θ θ θ = − + + θ θ θ + − θ θ θ= ⇒ − − − − + ⇒ − − = − < − +  < ∴ lie within the circle − − = −
4 201 z = 15 = 25 ⇒ ⇒ From z ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ 1 lies on ⊥ bisector of 1 and − 1. z2 z1 lies on imaginary axis. ⇒ z2 z1 is purely is imaginary. ⇒ z2 122. = = − − − = = − = − − = + = + O C A B i= z2 –3–4 5 1
4 202 z1 = ± ⇒ arg z2 ∴ arg (z1) − arg(z2 125. Since, arg ((z − 1 − i ) / z) is the angle subtended by the chord joining the pointsO and 1 + i at the circumference Y z 2 z − 0 We have, 2 − z π z − 2 AP ⇒ arg ⇒ = i 0 − z 2 2 OP AP Now, in ∆OAP, tan θ = Thus, z − 2 = 0 OP 126. Given, z = x + iy 6x2 − 36x + 48 + 6y2 − 24y + 18 + i [6xy 4 a ⇒ 6x2 + 6y2 − 36x − 24y + 66 = 12 x − 12 y − 12 ⇒ 6x2 + 6y2 − 48x − 12 y + 78 = 8 ⇒ x2 + y2 − 8x − 2 y + 13 = 0 Again, z − 3 − i = 3 ⇒ x + iy − 3 + i = 3 ⇒ (x − 3)2 + (y + 1)2 = 9 …(i) ⇒ x2 + y2 − 6x + 2 y + 1 = 0 On subtracting Eq. (ii) from Eq. (i), W get − 2 x − 4y + 12 = 0 …(ii) ⇒ x = − 2 y + 6 Putting the value of x in Eq. (ii), we get (− 2 y + 6)2 + y2 − 6 (− 2 y + 6) + 2 y + 1= 0 ⇒ 5y2 − 10 y + 1 = 0 …(iii) ∴ y = ∴ x = − 2 y + 6 = 4 Then, 2 ⇒ x = 1, − 2 128. amp (z z1 2 ) = 0 ⇒ amp (z1) + amp (z2 ) = 0 ∴ amp (z1) = −amp (z2 ) =amp (z2 )  z1 = z2 , we get z1 = z2 . So, z1 = z2 Also, z z12 = z z2 2 = z2 2 = 1, because z2 = 1 129. z + z− 1 = 1⇒ z2 − z + 1 = 0 1 ± 3 i 2 ∴ z == − ω −ω = − 4 π/ − = + = ∴ − =± π ⇒ − Pz () O θ 2 A X 1 x x − − − − = + − − + − − = − − − − + − + − − x x x x − − − 8) = + + − − + − x x − + − − x = + + − − − + − x x x + x x − − − + − = + + = π ∴ π = ⇒ = ± = ±  ∴ x = = + ± +  x x = + + − π = π − + 127. . Let x = x x − = − = − ⇒ x + = + − + − − − = ⇒ x + =±
4 203 2 ∴ zn + z− n = −( ω) n + (− ω)− n or (− ω2 )n + (− ω2 )− n = −(1)nωn + (− 1)− n ⋅ 1 or = −(1) n ωn + = (− 1) n ⋅ (ωn + ω2n ) = −(1) n ⋅ (1 + 1) or (− 1) n ⋅ −(1) Dependening on whether n is a multiple of 3 or not. = 3 131. OB = OD = OA = z amp (z2 ) = θ + and z = z {cos θ + i sin θ} ∴ z1 = z cos θ − 2 + i sin θ − 2 = z {sin θ − i cos θ} = z (− i ) (cos θ + i sin θ) = − iz z2 = z cos θ + 2 + i sin θ + 2 iz ∴ 132.  = − − = − − = z –1=1 Y Pz ( ) 0 Qz ( ) 1 C 1 X ω + − =− − ω ω ω + =− ω ω = − + ω ω ω = =θ =θ− π π Dz ) ( 1 C Az () B ) (z2
4 204 z − 1 ⇒ The vertices are at equal distances from the origin z = 0. ∴The origin is at the centroid of the equilateral triangle. z1 + z2 + z3 = 0 ∴ 3 ∴ z1 + z2 + z3 = 0 with OA as real axis, z1 = 1, z2 = 1(cos 120° + i sin 120° ), z3 = 1(cos 120° − i sin 120° ) 135. Statement I is false, since there is no order relation in the set of complex numbers. Cancellation laws, a + c > b + c ⇒ a > b does not hold true in complex numbers, therefore Statement II is true. 136. Given that, arg (z) = 0 ⇒ z is purely real. ∴ Statement II is true. ⇒ arg(z1) − arg (z2 ) = 0 z1 = 0 ⇒ z1 is purely real. ⇒ arg z2 z2 z1 = 0 ⇒ Im z2 Hence, Statement I and Statement II are true and Statement II is correct explanation of Statement I. 137. We will show that Statement I is true and follows from Statement II. Indeed ⇒ ⇒ But as z > 0, we have ⇒ Maximum value of 138. Statement I is false, since z = 2 > 4b and z is a positive z = c ⇒ z = c (cos θ + i sin θ) ⇒ Infinite complex numbers satisfy the given equation. Statement II is true. A quadratic can have more than two roots, if all coefficients are zero. = + − ∴ − − = ⇒ + = − −θ θ = − + ⇒ θ θ = − ⇒θ θ = − = + − ≤ − ⇒ − = − π π ∴ − − = ⋅ ± + ± = − π π ∴ − − = − =− − 133. = = O 1 1 Bz ( ) 2 Cz ( ) 3 Az ( ) 1
4 205 139. e iθ = cos θ + i sin θ ⇒ e − iθ = cos θ − i sin θ e iθ + e − iθ e i(1 −i) + e − i(1−i) e(1 + i) + e − (1 + i) 2 e 2 e 140. We have, (cos θ + i sin θ)3 5/ = (cos 3θ + i sin 3θ)1 5/ = [cos (2rπ + 3θ) + i sin (2rπ + 3θ)]1 5/ where, r = 0, 1, 2, 3, 4 2πr + 3 i = e 5 , r = 0, 1, 2, 3, 4 Hence, product of all values of (cos θ + i sin θ)3 5/ 3θ i 2π + 3 i 6π + 3 i 8π + 3 i = e 4 e 5 e 5 e 5 = e i3θ + 4πi =e 4iπ e i3θ = cos 3θ + i sin 3θ Also, product of roots of the equation x5 − 1 = 0 is 1. Hence, Statement II is true, but it is not a correct explanation of Statement I. 141. Let A z( 1)and B z( 1)be the centres of the given circles and P be the centre of the variable circle, which touches given circles externally, then AP = a + r and BP = b + r where, r is the radius of the variable circle. On subtraction, we get AP − BP = a − b ⇒ AP − BP =|a − b is a constant. Hence, locus of P is (i) a right bisector of AB, if a = b (ii) a hyperbola if a − b < AB = z2 − z1 (iii) an empty set, if a − b > AB = z2 − z1 (iv) set of all points on line AB except those which lie between A and B, if a − b = AB ≠ 0 Thus, Statement I is false and Statement II is true. 142. If P z( ) is any point on the ellipse, then equation of the ellipse is z − z1 + z − z2 =…(i) e For P z( )to lie in ellipse, we have z1 − z2 z − z1 + z − z2 < e It is given that origin is an interior point of the ellipse. ⇒ 0 − z1 + 0 − z2 < e z1 − z2 ∴ e 0, z1 + z2 Hence, Statement I is false and Statement II is true. 143. As, we know z − z1 + z − z2 = k represents an ellipse, if k > z1 − z2 . Thus, z − i + z + i = k represents an ellipse, if k > i + i or k > 2. ∴ Statement I is false and Statement II is true. 144. The equation can be rewritten as zz + az + az + aa = aa − λ ⇒ (z + a) (z + a) = aa − λ ⇒ z + a = aa − λ Since, aa is real, 1 should be real. aa − λ represents radius of the circle. Hence, Statement I and Statement II both are true and Statement II is the correct explanation for Statement I. Solutions (Q. Nos. 145-147) Given that, | z1 + z2|2 =| z1|2 + | z2|2 ⇒| z1|2 + | z2|2 + z z1 2 + z z1 2 =| z1|2 + | z2|2 ⇒ z z1 2 + z z1 2 = 0 …(i) z1 + z1 = 0 [dividing by z z2 2 ] ⇒ z2 z2 z1 z1 = 0 …(ii) ⇒ z2 z2 z1 − z2 z1 − z2 = − = = + − − = + + − − = + + − ∴ + = = −
4 206 145. From Eq. (i),z z2 2 is purely imaginary. 146. From Eq. (ii), z1/ z2 is purely imaginary 147. Also, i z( 1/ z2 ) is purely real. Hence, its possible arguments are 0 and π. 148. Clearly, according to the least possibility α 2 − 7α + 11 ≤ 1 ⇒ α ∈[2, 5] X 151. zD = B z iA , zE = zB − z iC 1 − i 1 − i ∴Angle between AC and DE = arg C zA − arg zC − zA z − zE − zD zA − zC 1 − i π = i 4 A D B C E
4 207 z + 1 ⇒ z − z = 0, z + z = 0, y = 0, x = 0 Locus of z is portion of pair of lines xy = 0 z2 − 1  2 > 0 z + 1 B. Given,z − cos− 1 cos 12 − z − sin− 1 sin 12| = 8 (π − 3) Since, cos− 1 cos 12 − sin− 1 sin 12= 8(π − 3) ∴ Locus of z is portion of a line joining z1 and z2 except the segment between z1 and z2. C. z2 − i z1 2 = k2 − k1 ∴ x2 − y2 + 2ixy − λi 1 = λ 2 ⇒ x2 − y2 = λ2 and xy = ∴Locus of zis point of intersection of hyperbola. ⇒ Locus of z is the segment joining z1 and z2. 155. There are no real roots of the equation z6 − 6z + 20 = 0. If x + iy is a root, then x − iy is also a root. Let the roots be x1 ± iy1, x2 ± iy2, x3 ± iy3 Sum of the roots = 2 (x1 + x2 + x3 ) = 0 ⇒ x1 + x2 + x3 = 0 ⇒ One of x1, x2, x3 is negative and other two are positive or vice-versa. ⇒ The number of roots in each of the quadrant is either 1 or 2. 156. A. z4 − 1 = 0 ⇒ z4 = 1= cos 0 + i sin 0 ⇒ z = (cos 0 + i sin 0)1 4/ = cos 0 + i sin 0 B. z4 + 1 = 0 ⇒ z4 = − 1 = cos π + i sin π ⇒ z = (cos π + i sin π)1 4/ 8 8 D. iz4 − 1 = 0 ⇒ z = ω ω,2 B. Here, z1 + z2 + z3 + .... + z6 = 0 ... (i) z2 = z e1( i2π )/ n, if e( 2i π )/ n= α ⇒ z2 = z1 α Similarly, z3 = z1α 2, z4 = z2 α 3, z5 = z2 α 4, z6 = z1α 5 On squaring and adding and then using Eq. (i), we get z12 + z22 + z32 + z42 + z52 + z62 = 0 C. 1 + 2 ω + 3ω2 + 4ω3 + ... + nωn − 1 n !ω n n( 2+ 1) 1/ n n [AM ≥ GM] Now, E = 1 + 2 ω + 3ω2 + 4ω3 + ... + nωn − 1 +.... ω + 2 ω2 + 3ω3 + ... + (n − 1)ωn − 1 + nωn D. Let x2 + y2 = 1 where, z1 − i, z2 = 1, z3 = − 1 ∴ z = − i ⇒ zz1 + z z23 = 1 − 1 = 0 158. Consider, x3 2 + 11 i 3 − 4 i 50 + 25 i x = x2 = 3 + 4 i × 3 − 4 i = 25 = − + = − + = π π + + = ⇒ + = = π π ⇒ + = π π + = π π ⇒ − = =− π π/ ⇒ − = π π/ = π π − 157. − = ⇒ = + + ω ω) ω ω ω = + + − = + + − − ω ⇒ = − −ω ∴ = ω− − − − + + = − − π + 1 − = + − − π ∴ + + = − −
4 208 = 2 + i ∴ a + b = 2 + 1= 3 x − 4 + iy x − 8 + iy ⇒ y2 − 25y + 136 = 0 ⇒ 9 (36 + y2 ) = 25 [36 + (y − 8)2 ] ⇒ y = 17, 8 Thus, the required numbers are z = 6 + 17i, 6 + 8 i. Hence, the value of Re (z) is 6. Hence, λ is equal to 8. 161. Consider z4 + z3 + 2 z2 + z + 1 = 0 ⇒ z4 + z3 + z2 + z2 + z + 1 = 0 ⇒ z2 (z2 + z + 1) + (z2 + z + 1) = 0 ⇒ (z2 + z + 1) (z2 + 1) = 0 It is given that, BD = 2 AC ⇒ MD = 2 AM Also, DM is perpendicular to AM. ⇒ (1 − 2)2 + (1 + 1)2 = 4[(x − 2)2 + (y + 1)2 ] …(i) y + 1 1 + 1 and ⋅ = − 1 x − 2 1 − 2 ⇒ 2 (y + 1) = x − 2 With x − 2 = 2(y + 1) , Eq. (i) can be written as ⇒ y ⇒ x = 3, 1 ∴ λ1 + λ 2 = 4 Entrances Gallery 1. A. zk is 10th root of unity ⇒ zk will also be 10th root of unity. Take zj as zk. zk , we can always find z. B. z1 ≠ 0, take z = z1 C. z10 − 1 = (z − 1)(z − z1) (z − z9 ) ⇒ (z − z1)(z − z2 ) (z − sz9 ) = 1 + z + z2 + + z9, ∀ z ∈complex number Put z = 1 (1 − z1)(1 − z2) (1 − z9 ) = 10 D. 1 + z1 + z2 + + z9 = 0 ⇒ Re (1) + Re (z1) + + Re (z9 ) = 0 ⇒ Re (z1) + Re (z2)|+ + Re (z9) = − 1 | z0|2 + |α|2 − r 2 = |α | ⇒ 160. − ≥ + − = − = − ≥ x x − = x ≥ x x > + = ⇒ x = − + = = = ∴ = − ω ω = 162. M C A ( ) x,y B D 1) , (1 –1) (2, 159. − − = ⇒ = ⇒ x x = + − − + ⇒ x = x = − − = ∴ = − = ∑ π = α = = |α |α ∆ θ α α| = + − ∆ θ α |α = − + + − 2 D O A B()α 2r r C ) / (1 α θ z0 Y X
4 209 3 6 2 Possible position of z1 are A1, A2, A3, whereas of z2 are B1, B2, B3 (as shown in the figure). So, possible value of ∠z Oz1 2 according to the given options is . 4. Area of = 42 π 5. Distance of (1, − 3) from y + > 6. Given, is z2 + z + 1 − a = 0 Clearly, this equation do not have real roots, if D < 0 ⇒ 1 − 4 1( − a) < 0 ⇒ 4a < 3 ∴ a < ∴ Minimum value = 5 8. The expression may not attain integral value for all a, b,c. If we consider a = b = c, then x = 3 a y = a(1 + ω + ω2 ) = 0 z = a(1 + ω2 + ω) = 0 ∴ | x|2 + | y|2 + | z|2 = 9|a|2 | x|2 + | y|2 + | z|2 9 ∴ |a|2 + |b|2 + |c|2 = 3 = 3 As, x 2 + y 2 + z 2 = 3(|a|2 +|b|2 +|c|2 ) [using 1+ ω + ω2 = 0] | x|2 +| y|2 +| z|2 ∴ 2 2 2 = 3 |a| +|b| +|c| 9. Given, z = (1 − t z) 1 + tz2 z − z1 = t ⇒ z2 − z1 z − z1 = 0 ...(i) ⇒ arg z2 − z1 = ∩ ∩ = × + × π π + = π 60° x+y 2 2 <16 y+x= 0 3 √ Y X X′ Y′ = 0) (0, B (3 0) , C 2) , (3 A(3,–5/ 2) O Y Y¢ X¢ X α| α| ⇒ α|= = + = π = π π > π <− A A2 A1 B1 B2 B3 x 2 –1/ = x 2 / =1 π/6 O x = − + × > −
4 210 ⇒ ⇒ AP + PB = AB ⇒ | z − z1| + | z − z2| =| z1 − z2| z z 10. A. − i = + i , z ≠ 0 | z| | z| z is unimodular complex number | z| and lies on perpendicular bisector of i and − i z ⇒ = ± 1⇒ z = ± 1| z| ⇒ a is real number | z| ⇒ Im(z) = 0 B. | z + 4| + | z − 4| = 10 z lies on an ellipse, whose focus are (4, 0) and (− 4, 0) and length of major axis is 10. ⇒ 2ae = 8 and 2a = 10 ⇒ e = 4 / 5 |Re(z)| ≤ 5. C. ∴|ω| = 2 ⇒ w = 2(cos θ + i sin θ) x iy i i 2 x ⇒ 2 + (3 / 2) y = 1 2 (5 / 2) e ⇒ e D. |ω| = 1 ⇒ x + iy = cos + i sin θ + cos θ − i sin θ x + iy = 2 cos θ |Re(z)| ≤ 1,|Im(z)| = 0 11. Given, z2 is not unimodular i.e.| z2| ≠ 1 z1 − 2 z2 is unimodular. and − z z1 2 ⇒⇒| z1 − 2 z2|2 =|2 − z z1 2|2 ⇒ (z1 − 2 z 2 ) = (2 − z z [zz =| |z 2 ] ⇒ | z1|2 + 4| z2|2 − 2 z z12 − 2 z z1 2 = 4 + | z1| |2z2|2 − 2 z z12 − 2 z z1 2 ⇒ (| z2|2 − 1)(| z1|2 − 4) = 0  | z2| ≠ 1 ∴ | z1| = 2 Let z1 = x + iy ⇒ x2 + y2 = (2)2 ∴ Point z1 lies on a circle of radius 2. 12.| |z ≥ 2 2 2 13. Given,| |z = 1, arg z = θ ∴ z = e ie 1 But z = z 1 + z ∴ arg = arg(z) = θ 1 + 1 z z2 14. Given, a complex number (z ≠ 1) is purely real. z − 1 To find the locus of the complex number z. z2 Since, (z ≠ 1) is purely real. z − 1 z2 z2 ∴ = z − 1 z − 1 ⇒ z2(z − 1) = z2(z − 1) ⇒ z z2 − z2 = z z2 − z2 ⇒ zzz − z2 = zzz − z2 ⇒ z z| |2 − z2 = z z| |2 − z2 ∴ ≥ + − ≥ − ≥ − ⇒ − = − − − = − − − − − − = Pz () Az ( ) 1 Bz ( ) 2 − − = − −
4 211 On rearranging the terms, we get z z| |2 − z| |z 2 = z2 − z2 ⇒ | z|2 (z − z) = (z − z)(z + z) ⇒ | |z 2 (z − z) − (z − z)(z + z) = 0 ⇒ (z − z)(| z|2 − (z+ z)) = 0 Either (z − z) = 0 or[| z|2 − (z+ z)] = 0 Now, z = z ⇒ Locus of z is real axis and (| z|2 − (z + z)) = 0 ⇒ zz − (z + z) = 0 Locus of z is a circle passing through the origin. Aliter Put z = x + iy, then z2 (x + iy)2 (x2 − y2 ) + i(2 xy) = = z − 1 (x + iy) − 1 (x − 1) + iy 2 2 (x − y ) + i(2 xy) (x − 1) − iy = × (x − 1) + iy (x − 1) − iy z2 Since, (z ≠ 1) is purely real, hence its imaginary z − 1 part should be equal to zero. ⇒ (x2 − y2 )(− y) + (2 xy)(x − 1) = 0 ⇒ y (x2 − y2 + 2 x − 2 x2 ) = 0 ⇒ y x( 2 + y2 − 2 x) = 0 Either y = 0 or x2 + y2 − 2 x = 0 Now, y = 0, locus of z is real axis, and x2 + y2 − 2 x = 0, locus of z is a circle passing through the origin. Locus of z is either real axis or a circle passing through the origin. 15. Let z = x + iy, given Re (z) = 1 ∴ x = 1⇒ z = 1 + iy Since, complex roots are conjugate of each other. ∴z = 1 + iy and 1 − iy are two roots of z2 + α z + β = 0 Product of roots = β ⇒ (1 + iy)(1 − iy) = β ∴ β = 1 + y2 ≥ 1 ⇒ β ∈(1, ∞) 16. (1 + ω)7 = A + Bω, we know 1 + ω + ω2 = 0 ∴ 1 + ω = − ω2 ⇒ (− ω2 7)= A + Bω ⇒ − ω14 = A + Bω [ω14 = ω12 ⋅ω2 = ω2 ] ⇒ − ω2 = A + Bω ⇒ 1 + ω = A + Bω On comparing A = 1, B = 1 17. We have,| z − 1| =| z + 1| =| z − i| Clearly, z is the circumcentre of the triangle formed by the vertices (1, 0), (0, 1) and (− 1, 0), which is unique. Hence, the number of complex number z is one. 18. Since, α and β are roots of the equation x2 − x + 1 = 0. ⇒ α + β = 1, αβ = 1 ⇒ x = 1 + 3i 1 − ⇒ x =or 2 ⇒ x = − ω or Thus, α = − ω2, then β = − ω or α = − ω , then β = − ω2 [where,ω3 = 1] Hence, α 2009 + β2009 = −(ω)2009 + (− ω2 2009) = − [(ω3 669)⋅ω2 + (ω3 1337)⋅ω] = − [ω2 + ω] = − −(1) = 1 19. | z| z − 4 + z ⇒ ⇒ ⇒ | z|2 ⇒ ∴Maximum value of| z| is Ø For minimum value of |z|, we take 4 4 4 4 z − − z − z z z z from which we get | |z ≥5 − 1 O (0, 1) (–1, 0) (1, 0) z i + ± ≤ + − ≤ + ≤ − − ≤ + +
4 212 1 20. Let z = 1 1 1 ∴ z = = − i − 1 − i − 1 i + 1 21. From the argand diagram, maximum value of| z + 1| is 6. Aliter | z + 1| =| z + 4 − 3| ≤| z + 4| + −|3| ≤ 6 Thus, maximum value Y¢ of| z + 1| is 6. 10 22. ∑ sin 2 kπ + i cos 2k k = 1 11 11 10 = i ∑ k = 1 = i ∑ e 11 k = 1 = i ∑10 e − 211kπ − 1 = − i k = 0 23. Given, equation is z2 + z + 1 = 0 ⇒ z = ω ω,2 1 2 Now, z + z2 + 12 2 z3 + z13 2 z z 2 2 2 z4 + 14 z5 + z15 z6 + z16 z = (ω + ω2 2)+ (ω2 + ω)2 + (ω3 + ω−3 2) + (ω + ω2 2)+ (ω2 + ω)2 + (ω6 + ω−6 2) = −(1)2 + (− 1)2 + (1 + 1)2 + (− 1)2 + (− 1)2 + (1 + 1)2 = 1 + 1 + 4 + 1 + 1 + 4 = 12 24. Given that, (x − 1)3 + 8 = 0 −2 x − 1 1 3/ ⇒ ( )1 −2 x − 1 2 Cube roots of are 1, ω, ω . −2 Cube roots of (x − 1) are − 2, − 2ω and − 2ω2. ∴Cube roots of x are − 1, 1 − 2ω and 1 − 2ω2. 25. Let z1 = x1 + iy1 and z2 = x2 + iy2 Given,| z1 + z2| =| z1| + | z2| ∴ On squaring both sides, = x12 + x22 + 2 x x12 + y12 + y22 + 2 y y1 2 = x12 + y12 + x22 + y22 + 2 (x12 + y12 )(x22 + y22 ) ⇒ x x12 + y y12 = (x12 + y12 )(x22 + y22 ) Again, squaring both sides, we get ⇒ = x x1222 + y y1222 + 2 x x y y1 2 1 2 = x x1222 + y y1222 + x y1222 + y x12 22 ⇒ (x y12 − y x12 )2 = 0 y1 = y2 ⇒ x1 x2 ⇒ tan−1 y1 = tan−1 y2 x1 x2 ⇒ arg (z1) = arg(z2 ) ⇒ (x − 1)3 = −( 2)3 ⇒ x − 1 3 = 1 x x + + + + = + + x x − X X¢ Y (–7, 0) (–4,0)(–1, 0) O − π π π
4 213 ∴ arg(z1) − arg(z2 ) = 0 Aliter Given, | z1 + z2| =| z1| + | z2| =| z1|2 + | z2|2 + 2 Re(z z12 ) [squaring both sides] =| z1|2 + | z2|2 + 2| z1|| z2| ⇒ Re (z z12 ) =| z1|| z2| ⇒ | z1|| z2|cos(θ1 − θ2 ) =| z1|| z2| ⇒ θ1 − θ2 = 0 ∴ arg (z1) − arg(z2 ) = 0 z 26. Given, w = and|w| = 1 ⇒ = 1 ⇒ | |z = z − 1 ⇒ z lies on perpendicular bisector of (0, 0) and 0, . ⇒ − + 2 arg(z) = π arg(− i ) = − 2 ∴ arg(z) = 28. Since, z1/ 3 = p + iq ∴ z = (p + iq)3 = p3 − iq3 + 3ip q2 − 3pq2 Given that, z = x − iy ∴ x − iy = p3 − 3pq2 + i(3p q2 − q3 ) ⇒ x = p3 − 3pq2 and − y = 3p q2 − q3 ⇒ x 2 2 = p − 3q p and= − 3p2 + q2 ∴= − 2 p2 − 2q2 ⇒ (p2 + q2 ) p q = − 2 29. Using the relation, if | z1 + z2| =| z1| + | z2| Then, arg(z1) = arg(z2 ) Since, | z2 + (− 1)| =| z2| + −|1| Then, arg(z2 ) = arg(− 1) ⇒ 2 arg(z) = π [arg(− 1) = π] ⇒ arg(z) = ⇒ z lies on Y-axis (imaginary axis). 30. Since, origin, z1 and z2 are the vertices of an equilateral triangle, then z12 + z22 = z z1 2 ⇒ (z1 + z2 )2 = 3z z1 2 ...(i) Again, z1, z2 are the roots of the equation z2 + az + b = 0 Then, z1 + z2 = − a and z z12 = b On putting these values in Eq. (i), we get (− a)2 = 3b ⇒ a2 = 3b 31. Let z = r e1 iθ ⇒ w = r e2 iφ ⇒ z = r e1 iθ Given, | zw| = 1 ⇒ |r e1iθ ⋅ r e2iφ| = 1 ⇒ r r1 2 = 1 ...(i) and arg(z) − arg(w) = ⇒ ...(ii) Now, zw = r e1−i θ ⋅ r e2i φ Hence, z lies on a straight line. 27. Given, z + i w = 0 ⇒ z = − iw ⇒ z = iw ⇒ w = − iz and arg(zw) = π ⇒ arg(− iz2 ) = π ⇒ arg(− i ) + 2 arg(z) = π 3 − − x + x +
4 214 = r r e1 2−i( θ − φ ) = 1⋅e iπ / 2 [from Eqs. (i) and (ii)] ∴ zw = − i 32. Now, 11+− ii x ((11+− ii)()(11++ ii)) x (1 + i )2 x 1 − 1 + 2i x = = 34. Given,|β| = 1 β − α β − α ∴ = [1 = ββ] 1 − αβ ββ − αβ 1 β − α 1 = = × 1 = 1 [| z| =| z|] |β| β − α ( )1 35. Given, z ∴ | |z = | z1| = | z1| | z2| | z2| =[| zn| =| |z n ] (2) (3 5)2 102 ⋅2 =2 = 2 = 2 (10) (10) 36. Given,|w | = 1and z = w + 1 ⇒ z w( + 1) = w − 1 ⇒ w z(− 1) = − 1 − z 1 + z 1 + z ⇒ w =⇒|w | = 1 − z 1 − z |1 + z| ⇒ 1 =⇒ |1 − z| =|1 + z| |1 − z| On squaring both sides, we get 12 + | |z 2 − 2 Re (z) = 12 + | |z 2 − 2 Re (z) ⇒ 4Re (z) = 0 ∴ Re (z) = 0 37. Let z = x + iy ∴ | |z 2 + | z − 3|2 + | z − i|2 =| x + iy|2 + |(x − 3) + iy|2 + | x + i y(− 1)|2 = x2 + y2 + (x − 3)2 + y2 + x2 + (y − 1)2 = x2 + y2 + x2 − 6x + 9 + y2 + x2 + y2 + 1 − 2 y = 3x2 + 3y2 − 6x − 2 y + 10 = 3(x2 − 2 x + 1) + 3 y2 −y + 1 + 10 − 3 − 9 2 = 3(x − 1)2 + 3 y − 1 + 3 It is minimum, when x − 1 = 0 and y − = 0 − i 2 2 1 1 + i x ⇒ = ( )i x = 1 1 − i ⇒ ( )i x = ( )i 4n where, n is any positive integer. ∴ x = 4n [given] 33. Now, (1 + ω − ω2 7)= − ω(2 − ω2 7) = −(2ω2 7) = − 27 ⋅ω14 = − 128(ω3 4) ω2 [1 + ω + ω2 = 0] = − 128ω2 [ω3 = 1] + + + + + + + + + + + + = −
4 215 π = −iπ / 4 e = 2e iπ / 4 + iπ / 4 = 2e 2iπ / 4 = 2e iπ / 2 π = 2 cos 2 + i sin 2 39. Let z1 = x + iy and z2 = p − iq, where x q, > 0 Given,| z1| =| z2| ⇒ x2 + y2 = p2 + q2 ...(i) z1 + z2 = ∴ z1 − z2 =...(ii) If xq + yp ≠ 0, then it is purely imaginary and if x y xq + yp = 0 or= − = λ [say] p q ⇒ x = pλ, y = − qλ From Eq. (i), we get p2 + q2 = λ2(p2 + q2 ) ⇒ λ2 = 1 ⇒ λ = ± 1 For λ = − 1 and z1 ≠ z2, but| z1| =| z2| In this case, Eq. (ii) is zero. 40. Given, z = e 2π / 3 ∴ 1 + z + 3z2 + 2 z3 + 2 z4 + 2 z4 + 3z5 = 1 + e 2π / 3 + 3(e 2π / 3 2)+ 2(e 2π / 3 3) + 2(e 2π / 3 4)+ 3(e 2π / 3 5) 2 4 = 1 cos 3 + i sin 3 4 4 + 3 cos 3 + i sin 3 8π 8 + 2[cos 2π + i sin 2π ] + 2 cos 3 + i sin 3 10 10 + 3 cos 3 + i sin 3 = 1 + [cos 120° + i sin 120° ] + 3[cos 240° + i sin 240° ] + 2[cos 360° + i sin 360° ] + 2[cos 480° + i sin 480° ]+ 3[cos 600° + i sin 600°] 1 3 1 3 = 1 2 +2i + 3 2 − 2 i + 2[1 + 0] + 2[cos 120° + i sin 120° ] + 3[cos 120° − i sin 120° ] 41. Given, (x − 1)3 = − 5 (x − 1) = − 5, − 5ω, − 5ω2 ⇒ x = 1 − 5 1,− 5ω, 1 − 5ω2 ⇒ x = − 4 1,− 5ω, 1 − 5ω2 42. Given, z − 3 i = 1 z + 3 i ⇒ | z − 3 i| =| z + 3 i| [if| z − z1| =| z − z2|, then it is perpendicular bisector of z1 and z2] Hence, perpendicular bisector of (0, 3) and (0, − 3) is X-axis. 43. (z3 − z1) = (z2 − z1)(cos 90° + i sin 90° ) ⇒ (z3 − z1) = i z( 2 − z1) ⇒ (z3 − z1)2 = − (z2 − z1)2 ⇒ (z3 − z1)2 + (z2 − z1)2 = 0 Aliter Let z1 = 0, z2 = 1 − i and z3 = 1 + i Now, (z1 − z2 )2 + (z1 − z3 )2 = [0 − (1 − i )]2 + [0 − (1 + i )]2 = (1 − i )2 + (1 + i )2 = 12 + i 2 − 2i + 12 + i 2 + 2i = 1 − 1 + 1− 1 = 0 44. We know that, ∴ x = = ∴ = + + − = + −π + π π x x + + − − + + − + + − + x
4 216 | z − z1| + | z − z2| = k will represent an ellipse, if | z1 − z2| < k. Hence, the equation | z − z1| + | z − z2| = 2| z1 − z2|, represents an ellipse. 45. Given, z1 ≠ ± 1 Since, z2 and z3 can be obtained by rotating vector representing through and , respectively. ∴ z2 = z1ω and z3 = z1ω2 ∴ z z z1 2 3 = z1 × z1ω × z1ω2 = z13ω3 = z13 [ω3 = 1] 46. Since, z1, z2 and z3 are the vertices of an equilateral triangle, therefore | z1 − z2| =| z2 − z3| =| z3 − z1| = k [say] Also, α = 3 + i ) × 2 = 1 ⇒ |α =| Let A = αz1 + β, B = αz2 + β and C = αz3 + β Now, | AB| = α|z2 + β − α( z1 + β)| =|α(z2 − z1)| =|α|| z2 − z1| =|1|| z2 − z1| = 1| z2 − z1| =| z2 − z1| = k Similarly, BC = CA = k Hence, the points αz1 + β α, z2 + β and αz3 + β are the vertices of an equilateral triangle. 47. Given, equation is 1 + z + z3 + z4 = 0. ⇒ (1 + z) + z3(1 + z) = 0 ⇒ (1 + z)(1 + z3 ) = 1 ⇒ z = − 1, z3 = − 1 ⇒ z = − 1 ∴ z = − 1, − ω, − ω2 Hence, roots are in cubic roots of unity. Hence, these roots are the vertices of an equilateral triangle. 48. Given, z1 = 2 2 1( + i ) and z2 = 1 + i 3 (− 1 − 3 i ) (− 1 + i 3) i(− 2 − 2i 3) 3 1 = − i 4 2 2 ∴ Amplitude 51.| z1| =| z2| =| z3| = 1 ⇒ | z1|2 =| z2|2 =| z3|2 = 1 ∴ z z1 1 = z z22 = z z33 = 1 1 1 ⇒ z1 = , z2 = z1 z2 1 and z3 = z3 ∴ | z1 + z2 + z3| = = 1 = 1 52. Given, ( 3 i + 1)100 = 299(a + ib) 1001 100 99 2 ∴ 2 i + 2 = 2 (a + ib) ω= 2 ⇒ 2(− ω2 100) = a + ib ⇒ 2 × ω200 = a + ib ⇒ 2 × (ω3 66) × ω2 = a + ib 66 − 1 − 3 i ⇒ 2 × ( )1 a + ib 2 ⇒ − 1 − 3i = a + ib ∴ a2 + b2 = 4 53. (1 + 3 i )4 + (1 − 3 i )4 = −(2ω2 4) + (− 2ω)4 1 1 1 + + z1 z2 z3 = =− − + + − + + − − = + + − + = + − = =− = + − = + − − = + − − − = − + × − + + = − −
4 217 − 1 + 3 i 2 − 1 − 3 i ω =and ω 2 2 = 16ω8 + 16ω4 = 16(ω3 2) ω2 + 16 (ω3 )ω = 16(ω2 + ω) = 16(− 1) = − 16 54.11/ 4 = (cos θ + i sin θ)1/ 4 [1 + ω + ω2 = 0] = (cos 2πr + i sin 2πr )1 4/ πr πr = cos + i sin 2 2 where, r = 0, 1, 2, 3 ∴ 11/ 4 = 1, i, − 1, − i ∴Required value = 12 + i 2 + (− 1)2 + (− i )2 = 1 − 1 + 1 − 1 = 0 55. From figure, it is clear that argument ≥ 90°. The given equation | z − −( 1 + i ) ≤ 1|represents the points inside and on the boundary of the circle, centred at (− 1 1, ) and whose radius is 1. It lies in 2nd quadrant. The point (0, 1), i.e. i lies on it such that, it has least argument. ∴The required complex number is i. 56. (| z1| −| z2|)2 =| z1 − z2|2 ⇒ | z1|2 + | z2|2 − 2| z1|| z2| =| z1|2 + | z2|2 − 2| z1|| z2|cos(θ1 − θ2 ) ⇒ cos(θ1 − θ2 ) = 1 ⇒ θ1 = 2nπ + θ2 Thus, if z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ1 + i sin θ2 ) ⇒ z1 = r1(cos θ2 + i sin θ2 ) and z2 = r2(cos θ2 + i sin θ2 ) z r ⇒ 1 = 1 z2 r2 z ⇒ Im 1 = 0 z2 57. Gievn, (x + iy)(1 − 2i ) = 1 + i ⇒ (x + iy)(1 − 2i ) = 1 + i ⇒ (x − iy)(1 + 2i ) = 1 + i 1 + i ⇒ x − iy = ...(i) 1 + 2i 1 + i 1 − i ⇒ x − iy ...(ii) i + 2i 1 − 2i 1 − i ⇒ x + iy = 1 − 2i 50 ⇒ ⇒ ∴ 59. Locus of z is a circle passing through origin. 60. Given, 2 x = 3 + 5 i ⇒ 8x3 = 27 − 125 i + 27 + 5i − 25 × 9 ⇒ 2 x3 = − 198 + 10i 4 ...(i) Also, 4x2 = 9 − 25 + 30i P1 P(0, 1) X Y X' Y' Q G (–1,1) P1
4 218 i ...(ii) [form Eqs. (i) and (ii)] 61. We have,| z1 + z2|2 =| z1|2 + | z2|2 ⇒| z1|2 + | z2|2 + 2Re (z z12 ) =| z1|2 + | z2|2 ⇒ Re (z z12 ) = 0 ⇒ z z1 2 + z z2 1 = 0 z1 = − z1 ⇒ z2 z2 z1 z2 ⇒ z1 is purely imaginary. z2 62. Let z 1i++i2 32 = (1 +3i+ 34)(i 2i ) i + 1 63. We have, | z1 + z2|2 = (| z1|2 + | z2|2 + 2| z1|| z2|) ⇒ =| z1|2 + | z2|2 + 2| z1|| z2|cos(θ1 − θ2 ) =| z1|2 + | z2|2 + 2| z1|| z2| ⇒ cos(θ1 − θ2 ) = 1 ⇒ θ1 = θ2 ⇒ arg (z1) = arg (z2 ) ∴ z arg 1 = 0 z2 64. Given, S = 1 + 2 ω + 3ω2 ++ 3 nω3n − 1 ∴ Sω = ω + 2ω2 + 2ω2 ++ (3 n − 1)ω3n − 1 + 3 nω2n ⇒ S(1 − ω =)1 + ω + ω2 ++ ω3n − 1 − 3 nω3n = 0 − 3 n 3 n 3 n ⇒ S = − = 1 − ω ω − 1 2 4n 65. Let z 2 cos 4 − i2 sin 4 cos 4 2 cos2 2i sin cos 4 4 4 4n cos 4 − i sin 4 cos + i sin θ 4 4 cos nθ − i sin nθ (cos θ + i sin θ)−n = cos nθ + i sin nθ = (cos θ + i sin θ)n = (cos θ + i sin θ)−2n = cos 2nθ − i sin 2nθ 66. Given, x = − 3 + i 3 2 − 1 + i 3 ⇒ x =− 1 = ω − 1 2 ∴ (x2 + 3x) (2 x2 + 3x + 1) = [(ω − 1)2 + 3(ω − 1)] [(2ω − 1)2 + 3(ω − 1) + 1] = (ω2 + ω − 2) (2 ω2 + ω − 1) = −(1 − 2) (2 − 1 − 1) = − 18 [1 + ω + ω2 = 0] 67. Given, (x + iy)1 3/ = 2 + 3 i ⇒ x + iy = (2 + 3 i )3 = 8 + 36 i + 54 i 2 + 27i 3 = − 46 + 9 i On equating real and imaginary parts from both sides, we get x = − 46, y = 9 ∴ 3x + 2 y = − 138 + 18 = − 120 68. Let z = x + iy ∴ | z + 3 − i| =|(x + 3) + i y(− 1)| = 1 ⇒ (x + 3)2 + (y − 1)2 = 1 ...(i) = + − x = + − + = = = − + ⇒ = − + = × × =
4 219  arg z = π ⇒ tan−1 y = π x y ⇒ = tan π = 0 ⇒ y = 0 ...(ii) x From Eqs. (i) and (ii), we get x = − 3, y = 0 ∴ z = − 3 ⇒ | |z = −|3| = 3 69. Given,| z1| =| z2| = =| zn| = 1 n ⇒ arg (z) = π − θ arg (−z) = − θ ∴ arg (z) − arg (− z) = π − θ − −(θ) = π − θ + θ = π 74. cos 30° + i sin 30° cos 60° − i sin 60° = (cos 30° + i sin 30° )(cos 60° + i sin 60° ) = cos 90° + i sin 90° = i ⇒ = = =  ⇒ = = = =  ⇒ = = = + = + +  + + + = + + + =   = + + +  from Eq. (i  = = ∴ = = + = + = = + − − − θ θ θ θ θ θ θ = + + = = + ∴ = − =−π+θ
5 Inequalities and Quadratic Equation Inequality An equation having signs >, <, ≥ or ≤ at the place of sign of equality ( )= is known as inequality or inequation. or Let a and b be two real numbers. If a − b is negative, we say that a is less than b a( < b) and if a − b is positive, then a is greater than b a( > b). Properties of Inequalities We shall learn about some elementary properties of inequalities which will be used in the subsequent discussion. Chapter Snapshot ● Inequality ● Generalised Method of Intervals for Solving Inequalities by Wavy Curve Method (Line Rule) ● Absolute Value of a Real Number ● Logarithms ● Arithmetico-Geometric Mean Inequality ● Quadratic Equation with Real Coefficients ● Formation of a Polynomial Equation from Given Roots ● Symmetric Function of the Roots ● Transformation of Equations ● Common Roots ● Quadratic Expression and its Graph ● Maximum and Minimum Values of Rational Expression ● Location of the Roots of a Quadratic Equation ● Algebraic Interpretation of Rolle’s Theorem ● Condition for Resolution into Linear Factors ● Some Application of Graphs to Find the Roots of Equations
5 221 i. ii. iii. iv. v. vi. vii. If a > b and b > c, then a > c. Generally, if a1 > a2, a2 > a3,...,an −1 > an , then a1 > an If a > b, then a ± c > b ± c, ∀ ∈c R a b b a If a > b, then for m > 0, am > bm, > and for m < 0, bm > am, > m m m m If a > b > 0, then (a) a 2 > b2 (b) | |a > | |b (c) 1 < 1 a b and if a < b < 0 , then (a) a 2 > b2 (b) | |a > | |b (c) 1 > 1 a b If a < 0 < b, then (a) a 2 > b2 , if | |a > | |b (b) a 2 < b2 , if | |a < | |b If a < x < b and a, b are positive real numbers, then a 2 < x2 < b2 If a < x < b and a is negative number and b is positive number, then (a) 0 ≤ x2 < b2 , if | |b > | |a (b) 0 ≤ x2 ≤ a 2 , if |a | > | |b
5 X 222 1 1 1 x b a, , if a and b have same sign If x ∈[a b, ] ⇒ 1 x ∈ − ∞ , 1 a 1 b, , if a and b have opposite signs a If > 0, then b (a) a > 0, if b > 0 (b) a < 0, if b < 0 If ai > bi > 0, where i = 1, 2, 3, …, n, then a a a1 2 3an > b b b1 2 3bn If ai > bi , where i = 1, 2, 3, …, n, then a1 + a2 + a3 ++ an > b1 + b2 ++ bn If 0 < a <1and n is a positive rational number, then (a) 0 < a n <1 (b) a −n >1 If a and b are positive real numbers such that a < b and if n is any positive rational number, then (a) a n < bn (b) a −n > b−n (c) a1/n < b1/n If a >1and n is any positive rational number, then (a) a n >1 (b) 0 < a −n <1 If 0 < a <1and m, n are positive rational numbers, then (a) m > n ⇒ a m < a n (b) m < n ⇒ a m > a n
5 223 and a2 < b2 < c2 ⇒ 3a2 < a2 + b2 + c2 < 3c2 …(ii) From Eqs. (i) and (ii), we get 2 2 2 2 + b + c 2 a < < viii. ⇒ c a + b + c a X Example 2. The value of x for which ix.12x − 6 < 0,12 − 3x < 0, is (a) φ (b) R (c) R − {0} (d) None of these x.Sol. (a) 12 x − 6 < 0 ⇒ 12 x < 6 ⇒ x < xi.and 12 − 3x < 0 …(i) ⇒ 12 < 3x ⇒ 4 < x xii.⇒ x > 4 …(ii) no real number x satisfying both the inequalities (i) and (ii). So, there is Hence, the given system of inequalities has no solution. xiii.X Example 3. The value of x for which x − 3 x −1 x − 2 − x < − , 2 − x > 2x − 8 10 (a) 1, 3 (b) 1, 3 xiv.(c) R (d) None of these Sol. (b) We have, x − 3 − x < x − 1 − x − 2 4 2 3 ⇒ 3x − 9 − 12 x < 6x − 6 − 4x + 8 xv.⇒ −11x < 11 ⇒ − x < 1 ⇒ x > − 1 …(i) and 2 − x > 2 x − 8 ⇒ −3x > − 10 3 3c a + b + c 3aa1 < a2 < a3 << an −1 < an If a >1and m n, are positive rational numbers, then (a) m > n ⇒ a m > a n (b) m < n ⇒ a m < a n + + < + + < 4 2 3 10
5 X 224 ⇒ x < …(ii) xvi.From Eqs. (i) and (ii), the solution of the given system of inequalities is given by x ∈ − 1, 10 . 3 Generalised Method of Intervals X Example 1. If a, b and c are positive real for Solving Inequalities by Wavy numbers such that a < b < c, then show that Curve Method a 2 a 2 + b2 + c2 c2 < < . (Line Rule) c a + b + c a Sol. We have, c > b > a > 0 Let F x( ) = (x − a1 )k1 (x − a2 )k2(x − an −1 )kn−1 ⇒ c2 > b2 > a2 > 0 (x − an )kn Now, a < b < c ⇒ 3a < a + b + c < 3c where, k1, k2, …, kn ∈Z and a1, a2, …, an are ⇒ 1 < 1 < 1 …(i) fixed real numbers satisfying the condition For solving F x( ) > 0 or F x( ) < 0, consider the following algorithm: ■ We mark the numbers a1, a2,…, an on the number axis and put the plus sign in the interval on the right of the largestof thesenumbers, i.e. on the right of an. ■ Then, we put the plus sign in the interval on the left of an, if kn is an even number and the minus sign, if kn is an odd number. In the next interval, we put a sign according to the following rule: ■ When passing through the point an − 1 the polynomial F x( ) changes sign, if kn − 1 is an odd number.Then, we consider the next interval and put a sign in it using the same rule. ■ Thus, we consider all the intervals. The solution of the inequality F x( )> 0is the union of all intervals in which we have put the plus sign and the solution of the inequality F x( )< 0is the union of all intervals in which we have put the minus sign. Solution of Rational Algebraic Inequation If P x( ) and Q x( ) are polynomial in x, then the inequation and are known as rational algebraic inequations. To solve these inequations, we use the sign method as explained in the following algorithm : Algorithm Step I Obtain P x( ) and Q x( ). Step II Factorise P x( ) and Q x( ) into linear factors. Step III Make the coefficient of x positive in all factors. Step IV Obtain critical points by equating all factors to zero. Step V Plot the critical points on the number line. If there are n critical points, then they divide the number line into (n +1) regions. Step VI In the right most region the expression P x( ) bears positive sign and in other Q x( ) regions the expression bears positive and negative signs depending on the exponents of the factors. Example 4. Solve (x −1)(x − 2 1)( − 2x) > 0. Sol. We have, (x − 1)(x − 2)(1 − 2 x)> 0 ⇒ −(x − 1)(x − 2)(2 x − 1)> 0 ⇒ (x − 1)(x − 2 2)( x − 1)< 0 On number line mark x = , 1, 2 – + – +
5 225 1/2 1 2 When x > 2, all factors (x − 1), (2 x − 1) and (x − 2) are positive. Hence, (x − 1)(x − 2)(2 x − 1)> 0 for x > 2. Now, put positive and negative sign alternatively as shown in figure. Hence, solution set ofor (x − 1)(x − 2)(2 x − 1)< 0 is , ∪ (1 2, ). 2 X Example 5. Solve . Sol. ⇒ ⇒ ⇒ ⇒ ⇒ (7 x − 12)(3x − 5)≤ 0 Sign scheme of (7 x − 12) (3x − 5) is as follows : + – + ⇒ 3 7 Ø x = is not included in the solutions as at x = denominator becomes zero. X Example 6. The value of x for which (x − 2) (3 x − 3) < 0, is (a) (2, 3) (b) [2, 3) Sol. (a) (x − 2)3 (x − 3)< 0 ⇒ (x − 2)(x − 3)< 0 (c) (0, 3) (d) (2, 3] [as (x − 2)2 is positive for all real values of x ≠ 2] –∞ + – + ∞ 2 3 Here, interval is open as x = 2, 3 do not satisfy inequality. i.e. 2 < x < 3 or x ∈(2, 3) Work Book Exercise 5.1 1 The value of x satisfying the inequalities 1 x + ≥ 2 hold x (0, ∞) R φ [0, ∞) x2 2 ≥ 0 x − 1 (1, ∞) [1, ∞) {0} ∪ (1, ∞) None of these 3 (x − 2) (4 x − 3) (3 x − 4) (2 1 − x) ≤ 0 (1, 3) (− ∞, 1) ∪ (3, ∞) (−∞, 1] ∪ [3, ∞) None of the above 4 If c < d, x2 + (c + d x) + cd < 0 (−d, − c] (−d, − c) R φ 2 1, 2 R φ 1, 2 2 Absolute Value of a Real Number The absolute value of a real number ‘x’ is denoted x, if x ≥ 0 by | |x and defined by | |x x, if x < 0 Ø ● (i) |x| is also defined as (ii) If x is positive, then (iii) If x is negative, then e.g. or x ∈ 5/3 12/7 = =− = − = = =±
5 X 226 ( ) b ∞ − − ∪ ∞ [ ) , , 3 2 3 ● (i) a≤ |a| (ii) |ab| = |a||b| (iii) a = |a| b |b| (iv) |a + b|≤ |a| + |b| [triangle inequality] (v) |a − b|≥ |a| − |b| [triangle inequality] (vi)|a + b| = |a| + |b| , iff ab ≥ 0 (vii) |a + b| = |a| − |b|,iff ab ≤ 0 Inequations Containing Absolute Values By definition, | |x < a ⇒ −a < x < a a( > 0) | |x ≤ a ⇒ −a ≤ x ≤ a | |x > a ⇒ x < − a and x > a | |x ≥ a ⇒ x ≤ − a and x ≥ a a ≤ | |x ≤ b ⇒ x ∈ −[ b, − a] ∪ [a b, ] where, a, b > 0 Forms of the Inequations Containing Absolute Values Form 1 The inequation of the form f (| |x ) < g x( ) is equivalent to the collection of systems f x( ) < g x( ), if x ≥ 0 f (−x) < g x( ), if x < 0 Form 2 The inequation of the form | f x( )| < g x( ), is equivalent to the systems f x( ) < g x( ), if g x( ) > 0 f x( ) < g x( ), if g x( ) > 0 Form 3 The inequation of the form | f x( )| > g x( ), is equivalent to the systems f x( ) > g x( ), if g x( ) < 0 f x( ) > g x( ), if g x( ) < 0 Form 4 The inequation of the form | f (| |)|x≥ g x( ) is equivalent to the collection of systems | f x( )| ≥ g x( ), if x ≥ 0 | f (−x)| ≥ g x( ), if x < 0 Form 5 The inequation of the form | f x( )| ≥ | ( )|g x is equivalent to the collection of system f 2 ( )x ≥ g 2 ( )x Form 6 The inequation of the form h x( , | f x( )|) ≥ g x( ) is equivalent to the collection of systems h x f x{ , ( )} ≥ g x( ), if f x( ) ≥ 0 h x{ , − f x( )} ≥ g x( ), if f x( ) < 0 X Example 7. The solution set of |3 − 4x| ≥ 9 is 3 (a) , 2 ∪ [3, ∞) (c) (−∞, )2 ∪ [2, ∞) (d) None of the above Sol. (b) We have, |3 − 4x|≥ 9 ⇒ 3 − 4x ≤ − 9 or 3 − 4x ≥ 9 [since,| |x ≥ a ⇒ x ≤ − a or x ≥ a] ⇒ −4x ≤ − 12 or −4x ≥ 6 ⇒ x ≥ 3 or x [dividing both sides by −4] ⇒ x ∈ −∞ , − 2 3 ∪ [3, ∞) |x + 3| + x X Example 8. The solution set of >1is x + 2 (a)[−5, − 2] ∪ −[ 1, ∞) (b)[−5, − 2) ∪ −[ 1, ∞) (c) (−5, − 2) ∪ −( 1, ∞) (d) None of these have, |x + 3| + x > 1 Sol. (c) We ⇒ − 1> 0 ⇒> 0 Case I When x + 3 ≥ 0, i.e. x ≥ − 3 0 ∴ |x + 3| − 2 > ⇒> 0 x + 1 ⇒> 0 x + 2 ⇒ {x + 1> 0 and x + 2 > 0} or {x + 1< 0 and x + 2 < 0} ⇒ {x > − 1 and x > − 2} or {x < − 1 and x < − 2} ⇒ x > − 1 or x < − 2 ⇒ x ∈ −( 1, ∞) or x ∈ −∞ −( , 2) ⇒ x ∈ −( 3, − 2) ∪ −( 1, ∞) [x ≥ − 3] …(i) Case II When x + 3 < 0, i.e. x < − 3 x + x x x + + + x x + − + x + x x + − +
5 227 ∴ |x + 3| − 2 > 0 x + 2 ⇒ − x − 3 − 2 > 0 x + 2 ⇒ −(x + 5) > 0 x + 2 ⇒ (x + 5) < 0 x + 2 ⇒ {x + 5 < 0 and x + 2 > 0} or {x + 5 > 0 and x + 2 < 0} ⇒ (x < − 5 and x > − 2), which is not possible or {x > − 5 and x < − 2} ⇒ x ∈ −( 5, − 2) …(ii) From Eqs. (i) and (ii), we get x ∈ −( 5, − 2) ∪ −( 1, ∞) Example 9. Solve the inequation 1 − | |x 1. 1 + | |x 2 Sol. The given inequation is equivalent to the collection of systems. ⇒ 1 − 1 +x x 21, if x ≥ 0 |1 +1 x| ≥ 21, if x ≥ 0 ⇒ 1 + 1 −x x 21, if x < 0 |1 −1 x| ≥ 21, if x < 0 1 ≥ 1 , if x ≥ 0 ⇒ 1+1 x 21 ≥ , if x < 0 1− x 2 1 − x ≥ 0, if x ≥ 0 x − 1 ≤ 0, if x ≥ 0 ⇒ 1 + x ⇒ x + 1 1 + x ≥ 0, if x < 0 x + 1 ≤ 0, if x < 0 1 − x x − 1 From Eqs. (i) and (ii), the solution of the given equation is x ∈ −[ 1, 1]. X Example 10. The value of x, |x + 3| > |2x −1| is −2 2 (a) , 4 (b) , 3 3 (c) (0, 1) (d) None of these Sol. (a) Given,|x + 3|>|2 x − 1| On squaring both sides, we get |x + 3|2 >|2 x − 1|2 ⇒ {(x + 3) − (2 x − 1)} {(x + 3) + (2 x − 1)} > 0 ⇒ {(− x + 4)(3x + 2)} > 0 – + – ⇒ 3 Logarithms Let there be a number a > 0, ≠1. A number p is called the logarithm of a number x to the base a, if a p = x and is written as p = log a x. Obviously x must be positive. Ø ● If x < 0, loga x is imaginary and if x = 0, loga x does not exist . ● loga x exists if and only if x, a> 0 and a ≠ 1 . x x x − + ≤ ≥ ≤ ≤ x x x x + − ≤ < ∴ < ≤ − x –1 + + 1 – –1 + + 1 – x ∈ − –2/3 4
5 X 228 Properties of Logarithms i.x. ii. iii. xi. iv. v. X Example 11. The value of x, log (e x − 3) <1is (a) (0, 3) (b) (0, e) (c) (0, e + 3) (d) (3, 3 + e) vi.Sol. (d) From definition of logarithms x − 3 > 0 ⇒ x > 3 …(i) Also, e > 1 given inequality may written as x − 3 < ( )e 1 ⇒ x < 3 + e ⇒ x ∈(3, 3 + e) vii.X Example 12. The value of x, log1 2/ x ≥ log1 3/ x is (a) (0,1] (b) (0,1) (c)[0,1) (d) None of these Sol. (a) Case I When x ≠ 1and x > 0 log1 2/ x ≥ log1 3/ x viii.⇒ log x 2 ≥ log x 3, when x ≠ 1 which is possible, only if 0 < x < 1. Case II When x = 1 ix.log1 2/ x = log1 3/ x, i.e. equality holds. Combining the above cases, 0 < x ≤ 1 or x ∈(0, 1] Work Book Exercise 5.2 If a >1, then (a) log a x > p ⇒ x > a p (b) 0 < log a x < p ⇒ 0 < x < a p If 0 < a <1, then (a) log a x > p ⇒0 < x < a p (b) 0 < log a x < p ⇒ a p < x <1 a loga x = x; a ≠ 0, ±1, x > 0 a logb x = xlogb a ; a > 0, b > 0, ≠1, x > 0 log a a =1, log a 1 = 0; a > 0, ≠1 1 log a x = ; x, a > 0, ≠1 log x a log x log a x = log b x ⋅ log a b = b ; a, b > 0, ≠1, log b a x > 0 For m, n > 0, a > 0, ≠1 (a) log (a m n⋅ ) = log a m + log a n m (b) log a n = log a m − log a n (c) log (a mx ) = x log a m For x > 0, a > 0, ≠1 1 (a) log n ( )x log a x a n (b) log n (xm ) m log a x a n (a) log a x > 0, iff x >1, a >1or 0 < x <1, 0 < a <1 (b) log a x < 0, iff x >1, 0 < a <1or 0 <x <1, a >1 For x > y > 0, (a) log a x > log a y, if a >1 (b) log a x < log a y, if 0 < a <1
5 229 Solve for x, 1 x + 4 > 5 (−∞, 1) R − −[ 9, 1] 2 | x + 2| < 4 (−6 2,) 3 log2 x > 0 (0, ∞) 4 log4 x > 1 (0, ∞) (−6 0,) (− ∞, 0) (4, ∞) (−∞, 9) None of these (−6 2,] (0, 2) (1, ∞) (4, ∞) (−4, ∞) (−∞, 4) 7 (x − 1)2 > 9 (−∞, − 2) ∪ (4, ∞) (4, ∞) (−∞, 2) (−∞, − 2] ∪ (4, ∞) 8 (x + 1)2 < 25 (−6 4,] (4, 6) 9 log x(x + 7) < 0 (−6 4,) (−4 6,) (−1 0,) (−1 1, ) (0, 1) (−∞, 1) 10 x2 + 6 5 log0.2(x + 5) > 0≥ 1 5x (−5, − 4) (−5 4,) (0, 4) (0, 5) 6 log x 0 5. > 2 (−∞, − 3) (−∞, − 3) ∪ (3, ∞) R , 1 (−∞, 1) (−1 0, ) (−1 1, ) (−∞, − 3] ∪ −[ 2 0,) ∪ (0 2,] ∪ [3, ∞) Arithmetico-Geometric Mean Inequality If a1, a2, …, an are n distinct positive real numbers, then a1 + a2 ++ an 1/n > (a a12an ) n i.e. AM > GM Ø If a1 = a2 == an, then AM = GM. Thus, equality occurs when all quantities are equal. X Example 13. If a, b and c are three distinct positive real numbers, then 1 1 1 (a) (a + b + c) + + 9 a b c 1 1 1 (b) (a + b + c) + + 9 a b c 1 1 1 (c) (a + b + c) + + 9 a b c 1 1 1 (d) (a + b + c) + + 9 a b c Sol. (a) We have, AM > GM 1 1 1 + + 1 3/ a + b + c 1 3/ a b c 1 ∴ > (abc) and 3 3 abc ⇒ a + b + c > 3 (abc)1 3/and a1 + b1 + c1 > (abc3)1 3/ ⇒ (a + b + c) 1 + 1 + 1 > 9 a b c Weighted AM and GM Inequalities If a1, a2,, an are n positive real numbers and m1, m2, …, mn are n positive rational numbers, then m a1 1 + m a2 2 ++ m an n m1 + m2 +... + mn 1 > (a1m1 ⋅ a2m2anmn ) m1 + m2 ++ mn i.e. weighted AM > weighted GM X Example 14. If a, b and c are distinct positive integers, then axb − c + bxc − a + cxa − b is (a) > 2(a + b + c) (b) R (c) > (a + b + c) (d) < (a + b + c)
5 X 230 Sol. (c) We have, a⋅ x b − c + b⋅ x c − a + c ⋅ x a − b > a + b + c {(x b − c ) (a x c − a b) (x a − b ) }c 1/ a + b + c ax b − c + bx c − a + cx a − b ⇒ > a + b + c {xa b( − c ) + b c( − a) + c a( − b)}1/ a + b + c ⇒ ax b − c + bx c − a + cx a − b > (a + b + c) Example 15. If the equation x4 − 4x3 + ax2 + bx +1 = 0 has four positive roots, then a b, are (a) a = 4, b = 6 (b) a = − 4, b = 6 (c) a = 2, b = 3 (d) a = 6, b = − 4 Sol. (d) Let α β γ and δ be four roots of the given equation. Then, α + β + γ + δ = 4 and αβγδ = 1 ⇒ AM of α β γ δ = GM of αβ γ δ ⇒α =β =γ =δ ⇒ α = β = γ = δ = 1 [α + β + γ + δ = 4, also αβ + αγ + αδ + βγ + βδ + γδ = a, αβγ + αβδ + βγδ + αγδ = − b] ∴ x4 − 4x3 + ax2 + bx + 1 = x4 − 4x3 + 6x2 − 4x + 1 ⇒ a = 6 and b = − 4
5 231 Two Important Inequalities
5 X 232 i. ii. X Example 16. If m >1and n ∈ N, then 1m + 2m + 3m ++ nm n +1 m (a) n 2 m m m m m 1 + 2 + 3 ++ n n +1 (b) n 2 1m + 2m + 3m ++ nm (c)≥1 (d)≤1 n Sol. (a) We know that for m > 1, AM of mth power > mth power of AM m m m m m ∴ 1 + 2 + 3 + + n 1 + 2 + 3 + + n ⇒ n 2 If a1, a2, ..., an are n positive distinct real numbers, then a1m + a2m + anm a1 + a2 ++ an m (a) , n n if m < 0 or m >1 a1m + a2m +...+ anm a1 + a2 ++ an m (b) n n if 0 < m <1. i.e. the arithmetic mean of the mth power of n positive quantities is greater than mth power of their arithmetic mean except when m is a positive proper fraction known as mth power mean. (c) If a1, a2 an and b1, b2 bn are rational numbers and m is a rational number, then b a1 1m + b a2 2m ++ b an nm b1 + b2 ++ bn b a + b a + ... + b a m 1 1 2 2 n n , b1 + b2 ++ bn if m < 0 or m >1and b a1 1m + b a2 2m ++ b an nm b1 + b2 +... + bn m b a + b a + + b a 1 1 2 2  n n , if 0 < m <1 b1 + b2 + + bn If a 1, a 2, a 3, …, a n are distinct positive real numbers and p, q, r are natural numbers, then a 1p + q + r + a 2p + q + r ++ a np + q + r n a1p + a2p ++ anp a1q + a2q ++ anq n n a1r + a2r ++ anr n n 1 2 3 m m m m n + + + +  + + + + > + 
5 233 Some Other Standard Inequalities i. an an ii. ) ) iii. X Example 17. If none of b1, b2, , bn is zero, a1 a2 an 2 then + ++ is b1 b2 bn Weierstrass Inequality (a) If a1, a2,, an are n positive real numbers, then for n ≥ 2 , (1 + a1 )(1 + a2 )(1 + an ) >1 + a1 + a2 + + (b) If a1, a2,, an are positive numbers less than unity, then (1− a1 )(1− a2 )(1− an ) >1 − a1 − a2 − − Cauchy-Schwartz Inequality If a1, a2,, an and b1, b2,, bn are 2n real numbers, then (a b1 1 + a b2 2 +...+ a bn n )2 ≤ (a12 + a22 ++ an2 (b12 + b22 ++ bn2 with the equality holding if and only if a1 a2 an = = = b1 b2 bn Tchebychef’s Inequality If a1, a2,, an and b1, b2,, bn are real numbers such that a1 ≤ a2 ≤ a3 ≤≤ an and b1 ≤ b2 ≤≤ bn , then n a b( 1 1 + a b2 2 ++ a bn n ) a b a b a b or n a1 + a2 ++ an b1 + b2 ++ bn n n
5 X 234 Applications of Inequalities to Find the Greatest and Least Values i. ii. iii. X Example 18. The greatest value of x y2 3 when 3x + 4y = 5, is (a) (b) (c) (d) Sol. (b) Let P = x y2 3. Clearly, P is the product of 5 factors such that two of them are equal to x and the remaining 3 are equal to y. Now, 3x + 4y = 5 ⇒ 2 3x + 3 4y = 5 2 3 ⇒ 5 ⇒ 16 x y23 5 ⇒ x y2 3 ≤ 3 3 5 16 or maximum of x y2 3 = 3 16/ If x1, x2,, xn are n positive variables such that x1 + x2 ++ xn = c(constant), then the product x x1 2 xn is greatest when c x1 = x2 == xn = and the greatest value is n c n . n If x1, x2,, xn are positive variables such that x x1 2 xn = c(constant), then the sum x1 + x2 ++ xn is least when x1 = x2 == xn = c1/n and the least value of the sum is n c( 1/n ). If x 1, x 2,, x n are variables and m 1, m 2,, m n are positive real numbers such that x1 + x2 +… + xn = c(constant), then x1 1 ⋅ x2 m2 … xn mn is greatest, when m x1 x2 xn x1 + x2 +… + xn = =… = = m1 m2 mn m1 + m2 +… + mn
5 235 + < λ λ Work Book Exercise 5.3 1 If a, b and c are all positive real numbers, which one of the following is true? (b + c)(c + a)(a + b) ≥ 8abc (b + c)(c + a)(a + b) < 8abc ea + b + + f + gc < 9 (a + b) 1 a b 2 If a 1, a 2, , a n and b 1, b 2, , b n are any two sets of positive real numbers, then a12 + a22 +... + an2 2 b1 b2 bn ≤ (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) ≥ (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) < (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) > (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) 3 If a and b are two positive quantities whose sum is 1 1 λ, then the minimum value of 1 + 1 + is a b λ − λ − 1 + 1 + 4 If a x2 4 + b y2 4 = c 6, then the greatest value of xy is c3 c3 c3 c3 2ab 4ab 5 Which of the following is true, where a, b, c > 0? 2(a3 + b3 + c3) ≥ bc b( + c) + ca c( + a) + ab a( + b) a3 + b3 + c3 (a + b + c) (⋅ a2 + b2 + c2 ) 11 Quadratic Equation with Real Coefficients An equation of the form ax2 + bx + c = 0 …(i) Ø where, a ≠ 0, a, b, c∈R is called a quadratic equation with real coefficients. The quantity D = b2 − 4acis known as the discriminant of the quadratic equation in (i) whose roots are given by −b + b2 − 4ac −b − b2 − 4ac α =and β = 2a 2a The nature of the roots is as given below: (i) If a quadratic equation in x has more than two roots, then it is an identity in x that is a = b = c = 0. (ii) If a =1and b, c∈I and the roots are rational numbers, then these roots must be integers. X (iii) The roots are of the form p ± q p q( , ∈Q), iff a, b and care rational and D is not a perfect square. (iv) The roots are real and distinct, iff D > 0. (v) The roots are real and equal, iff D = 0. (vi) The roots are complex with non-zero imaginarypart, iff D < 0. (vii) The roots are rational, iff a, b and care rational and D is a perfect square. ● If α is a root of the equation f( )x = 0, then the polynomial f( )xis exactly divisible by (x −α) or (x −α) is a factor of f( )x and conversely. ● Every equation of nth degree (n≥ 1) has exactly n roots and if the equation has more than n roots, it is an identity. ● If the coefficients of the equation f( )x = 0 are all real and α + βi is its root, then α − βi is also a root, i.e. imaginary roots occur in conjugate pairs. ● If the coefficients in the equation are all rational and α + β is one of its roots, then α − β is also a root, where α, β ∈Q and β is not a perfect square. > + + + + + + < + + + + + + < + +
5 236 ● If there is any two real numbers ‘a’ and ‘b’ such that f( )a and f b( ) are of opposite signs, then f( )x = 0 must have atleast one real root between‘a’ and ‘b’. ● Every equation f( )x = 0 of odd degree has atleast one real root of a sign opposite to that of its last term. Example 19. The number of values of a for which (a 2 − 3a + 2)x2 + (a 2 − 5a + 6)x + a 2 − 4 = 0 is an identity in x, is (a)0 (b)2 (c)1 (d)3 Sol. (b) It is an identity in x, if a2 − 3 a + 2 = 0, a2 − 5a + 6 = 0, a2 − 4 = 0 ⇒ a = 1, 2 and a = 2, 3 and a = 2, −2 ∴ Equation is identity, if a = 2

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mathematics part-2.docx

  • 1. 3 90 ⇒ β4 − β2 − 39β2 + 39 = 0 ⇒ (β2 − 1) (β2 − 39) = 0 Since, numbers are integers. ∴ β2 ≠ 39 ∴ β2 − 1 = 0 ⇒ β = ± 1 Hence, the four integers are 3, 5, 7 and 9. = 3 n2 − 3(n − 1)2 = 6 n − 3 On putting n = 1, 2, 3, ..., the sequence is 3, 9, 15, 21, ... which is clearly an AP. 27. The first two digit number which when divided by 4 leaves remainder 1 is 4⋅ 3 + 1 = 13 and last is 4 24⋅ + 1 = 97. Thus, we have to find the sum of the series 13 + 17 + 21 + + 97 which is an AP. ∴ 97 = 13 + (n − 1 4) ⇒ n = 22 n and Sn = [a + l ] = 1113[+ 97] 2 = 11 × 110 = 1210 28. The least and the greatest numbers of three digits divisible by 7 are 105 and 994, respectively. So, it is required to find the sum of the series 105 + 112 + 119 + + 994 Here, a = 105, d = 7, an = 994 ⇒ n − 1 = ∴ n − 1 = 127 ⇒ n = 127 + 1 = 128 ∴ Sum =[2a + (n − 1)d ] =[2 × 105 + (128 − 1) × 7] = 64(210 + 889) = 64 × 1099 = 70336 29. The odd numbers of four digits which are divisible by 9 are 1017, 1035, …, 9999. These are in AP with common difference 18. a = 1017, d = 18and l = 9999 ∴ nth term, an = a + (n − 1)d ⇒ 9999 = 1017 + (n − 1) × 18 ⇒ 18 n = 9999 − 999 = 9000 ⇒ n = 500 ∴ Sn = (a1 + an ) =(1017 + 9999) = 250 × 11016 = 2754000 3 30. Here, a = 10 and d = − Then, tn = 10 + (n − 1) 7 Now, tn is positive, if 3 10 + (n − 1) 0 7 ⇒ 70 − 3(n − 1) ≥ 0 ⇒ 73 ≥ 3 n ⇒ n So, first 24 terms are positive. Now, sum of the positive terms, 24 3 S24 = 2 × 10 + 23 7 2 69 852 = 12 20 − 7 =7 31. Let nth term be the first negative term. Then, nth term, tn < 0 ∴ 40 + (n − 1) (−2) < 0 ⇒ 42 − 2n < 0 ⇒ 2n > 42 ⇒ n > 21 ∴ Least value of n = 22 Hence, first 21 terms of an AP are non-negative. Since, sum will be maximum, if no negative terms is taken. ∴ Maximum sum, S 26. Given, Sn = 3 n2 ∴ Tn = Sn − Sn−1 Now, an = a + (n − 1)d ⇒ 994 = 105 + (n − 1) × 7 ⇒ 994 − 105 = 7(n − 1) ⇒ 889 = 7 (n − 1)
  • 2. 3 91 32. Let the first instalment be a and common difference of an AP be d. Given, 3600 = Sum of 40 terms ⇒ 3600 = [2a + (40 − 1)d ] ⇒ 3600 = 20 [2a + 39d ] ⇒ 180 = 2a + 39d …(i) After 30 instalments, one-third of the debt is unpaid. Hence, = 1200 is unpaid and 2400 is paid. Now, 2400 = [2a + (30 − 1)d ] ⇒ 160 = 2a + 29d …(ii) On subtracting Eq. (ii) from Eq. (i), we get 20 = 10 d ∴ d = 2 From Eq. (i), 180 = 2a + 39 2⋅ ⇒ 2a = 180 − 78 = 102 ∴ a = 51 Now, the value of 8th instalment = a + (8 − 1)d = 51 + 7 ⋅2 = ` 65 33. Let a be the length of the smallest side and d be the common difference. Here, n = 25 and S25 = 2100 Now, Sn =[2a + (n − 1)d ] ⇒ 2100 =[2a + (25 − 1)d ] ⇒ a + 12d = 84 ...(i) The largest side = 25th side = a + (25 − 1)d = a + 24d ∴ a + 24d = 20a [given] ...(ii) On solving Eqs. (i) and (ii), we get a = 8 , d = 6 n 34. Sn = [2a + (n − 1)d ] 2 Here, a = 1st term = 7 yr, d = 3 months = yr Sn = 250 yr n 1 ⇒ 250 = 2 2 × 7 + (n − 1) × 4 n n + 55 ⇒ 250 = 2 4 ⇒ 2000 = n2 + 55 n ⇒ n2 + 55 n − 2000 = 0 ⇒ (n − 25) (n + 80) = 0 ⇒ n = 25 ∴ Number of members in the club = 25 n 35. Sn = [2a + (n − 1)d ] 2 n ⇒ 60100 = [4 + (n − 1)(3)] 2 n ⇒ [3 n + 1] = 60100 2 ⇒ n = 200 36. Tn = Sn − Sn−1 = [3 n2 + 5 n] − [3(n − 1)2 + 5(n − 1)] = [3 n2 + 5 n] − [3(n2 − 2n + 1) + 5(n − 1)] = 6 n + 2 Let Tn = 164 Then, 164 = 6n + 2 ⇒ 6 n = 162 ⇒ n = 27 Hence, 164 is 27th term. 37. Here, a = 1, l = 11and Sn = 36 n  Sn n ∴ n ⇒ n = 6 38. Given series is 2 + 2 2 + 3 2 + 4 2 + ... = 2 [1 + 2 + 3 + 4 + upto n terms] n n( + 1) n n( + 1) = 2 (∑ n) = 2 2 ∑ n = 2 n n(+ 1) ∴ Sn = 1 1
  • 3. 3 92 2n + 1 − 1 = 2 40. Two digit odd numbers are 11, 13, ..., 99. These clearly form an AP with a = 11, d = 2 and n = 45. ∴ Required sum n 45 = (a + l ) = (11 + 99) = 2475 2 2 41. Sp − Sq = 2a p(− q) + (p2 − q2 − p + q d) = 2(q − p) ⇒ 2a + (p + q − 1)d = − 2 ∴ Sp+q = [2a + (p + q − 1)d ] = = − (p + q) 42. Given, Sn = 3 n2 − n If tn is the nth term of an AP, then tn = Sn − Sn−1 = (3 n2 − n) − [3(n − 1)2 − (n − 1)] = 6 n − 4 ∴ t1 = 6 − 4 = 2 n 43. We have, Sn = [2 p + (n − 1)Q] 2 By definition, d = Q 44. Given, Sn = 5 n2 + 2n ⇒ S1 = 5 + 2 = 7 and S2 = 5 4() + 2 2() = 24 ∴ T2 = S2 − S1 = 24 − 7 = 17 n 45. Given, tn = + y x 1 2 ⇒ t1 = + y and t2 = + y x x 1 ⇒ d = t2 − t1 = x where, d is common difference of given AP. r ∴ Sr = [2a1 + (r − 1)d ] 2 r r(+ 1) 1 = 2 x + ry a1 = x + y 46. Sn+ 3 − 3Sn+ 2 + 3Sn+1 − Sn = (Sn+ 3 − Sn+ 2 ) − 2(Sn+ 2 − Sn+1) + (Sn+1 − Sn ) = Tn+ 3 − 2Tn+ 2 + Tn+1 = (Tn+ 3 − Tn+ 2 ) − (Tn+ 2 − Tn+1) = d − d = 0 47. a + be y = b + ceyy = c + deyy y a − be b − ce c − de Applying componendo and dividendo rule, we get b c d = = a b c ⇒ a b c, , and d are in GP. 48. Given, Tm+ n = p and Tm n− = q ⇒ ar m+ n−1 = p …(i) and ar m n− −1 = q …(ii) On multiplying Eqs. (i) and (ii), we get a r2 2m−2 = pq ⇒ (ar m−1 2) = pq ⇒ ar m−1 = pq ⇒ Tm = pq 49. Since,1/2 cosec2θ, 2 cot θ and sec θ are in GP. ∴ ⇒ ⇒ 50. Let a and r be the first term and the common ratio respectively of GP. 51. Let A and R be the first term and the common ratio respectively of given GP. Then, T4 = p ⇒ AR3 = p T7 = q ⇒ AR6 = q and T10 = r ⇒ AR9 = r Now, (AR3 ) (AR9 ) = A R2 12 = (AR6 2) ⇒ pr = q2 Then, T10 = 9 ⇒ ar 9 = 9 and T4 = 4 ⇒ ar 3 = 4 ∴ T 7 = ar 6 = (ar 9 ⋅ ar 3 1 2) /= (9 × 4)1 2/ = 6 + − +
  • 4. 3 93 52. Let a and r be the first term and the common ratio respectively of GP. x1 y1 1 1 1 1 =x r1 y r1r r 1= 0 x r1 2 y r1 2 1 r 2 r 2 1 So, points are collinear, i.e. lie on a straight line. 54. Let ax = by = c z = k ⇒ a = k1/ x, b = k1/ y and c = k1/ z Since, a b, and c are in GP. ∴ b2 = ac ⇒ (k1/ y )2 = k1/ xk1/ z ⇒ k 2/ y = k1/ x + 1/ z ⇒ 2 = 1 + 1 y x z 1 1 1 ⇒ , and are in AP. x y z ⇒ x y, and z are in HP. 55. Let d be the common difference of an AP and R (≠ 0)be the common ratio of a GP. Then, q = p + d, r = p + 2d and y = xR, z = xR2  q − r = −d, r − p = 2d and p − q = −d ∴ xq−r ⋅ yr −p ⋅ zp − q = x− d ⋅(xR)2d ⋅(xR2 )− d = (x− d ⋅ x2d ⋅ x−d ) (R2d ⋅ R−2d ) = (x− d + 2d −d ) (⋅ R2d −2d ) = x0 ⋅ R0 = 1 × 1 = 1 56. Let the three numbers in GP be a, ar and ar 2. ∴ a + ar + ar 2 = 56 [given]…(i) On subtracting 1, 7, 21 from the numbers, we get a − 1, ar − 7 , ar 2 − 21, which are given to be in AP. ∴ (ar − 7) − (a − 1) = (ar 2 − 21) − (ar − 7) ⇒ ar − a − 6 = ar 2 − ar − 14 ⇒ a − 2ar + ar 2 = 8 …(ii) On subtracting Eq. (ii) from Eq. (i), we get 16 3 ar = 48 ⇒ a = …(iii) r 16 On substituting a = in Eq. (i), we get r 16 + 16 + 16 r = 56 r ⇒ 16 r 2 − 40r + 16 = 0 ⇒ 2r 2 − 5 r + 2 = 0 ⇒ (r − 2) (2r − 1) = 0 ∴ r = 2, If r = 2, then from Eq. (iii), a = 8 and the numbers are 8, 16, 32. If r = , then from Eq. (iii), a = 32 and the numbers are 32, 16, 8. 57. Let the sides of the right angled triangle be a, ar and ar 2, out of which ar 2 is the hypotenuse, then r > 1. Now, a r24 = a2 + a r2 2 ⇒ r 4 − r 2 − 1 = 0 ⇒ r 2 = 1 ± 5 2  r > 1 ∴ r 2 > 1 ⇒ r 2 = 1 + 5 2 Let ∠C be the greater acute angle. a 1 2 Now, cos C = ar 2 = r 2 = 1 + 5 58. Let the three digits be a, ar and ar 2. According to the hypothesis, 100a + 10ar + ar 2 − 792 = 100ar 2 + 10ar + a ⇒ a(1 − r 2 ) = 8 ...(i) and a, ar + 2, ar 2 are in AP. ∴ 2(ar + 2) = a + ar 2 ⇒ a r( 2 − 2r + 1) = 4 ...(ii) On dividing Eq. (i) by Eq. (ii), we get a(1 − r 2 ) 8 = a r( 2 − 2r + 1) 4 Then, a6 = 32 ⇒ ar 5 = 32 …(i) and a8 = 128 ⇒ ar 7 = 128 On dividing Eq. (ii) by Eq. (i), we get r 2 = 4 ⇒ r = ± 2 53. Let r be the common ratio of a GP, then Area formed by three points …(ii) =
  • 5. 3 94 (1 + r )(1 − r ) r + 1 ⇒ 2 = 2 ⇒ 1 − r = 2 (r − 1) ∴ r = 1/ 3 From Eq. (i), a = 9 Thus, digits are 9, 3, 1 and so the required number is 931. 59. Given a, b and c are in AP. ⇒ 2b = a + c …(i) Also, given a + 1, b, c are in GP and a, b, c + 2 are also in GP. b2 = (a + 1)c…(ii) and b2 = a c( + 2) …(iii) On solving Eqs. (i), (ii) and (iii), we get a = 8, b = 12 and c = 16 Hence, b = 12 60. The given series is a GP with a = 0.4. r = 40 a ∴ S = S∞ = ∴ =− x < 1, for x > 0] S 62. We have, = 255 [r = 2] r − 1 2 − 1 ⇒ 256 − a = 255 ⇒ a = 1 (x + 2)n − (x + 1)n 63. Clearly, (x + 2) − (x + 1) = (x + 2)n − 1 + (x + 2)n − 2(x + 1) + (x + 2)n − 3(x + 1)2 + … + (x + 1)n − 1 ∴Required sum = (x + 2)n − (x + 1)n [(x + 2) − (x + 1) = 1] 64. Here, c = ar, e = ar 2, d = bs, f = bs2 a b 1 1 2 c d 1 ∴ Area of triangle = e f 1 = ab s(− r )(s − 1)(r − 1) 65. Let S denotes the sum of all terms and S1 denotes the sum of odd terms. a S1 = (i.e. sum of odd terms) 1 1 Given, S = 5 ⋅S1 ⇒ 1 − r 5 1 − r 2 ⇒ ⇒ r 66. The given series is a GP with a = 2, r = 3 > 1 a r( 10 − 1) 2[( 3)10 − 1] ∴ S10 = = r − 1 3 − 1 2 = (242)( 3 + 1) = 121( 6 + 2 ) 2 n = 1 = = 212 .16 68. Let a and r be the first term and the common ratio respectively of given infinite GP. Then, ar = 2 …(i) a Also,= 8 …(ii) 2 From Eq. (i), r = a On putting the value of r in Eq. (ii), we get a = 4 69. Let a and r be the first term and the common ratio respectively of given GP. Then, n n =1 a(1 − r n ) 1 − r 99 61. The given series is a GP with 1 − r = x = − x x x = − =− − − − x x x − = + = + − −  − − = ⇒ − − = ⋅ −
  • 6. 3 95 ⇒ Sn = 1 − r ⇒ β = 2 1 − r ⇒ r 70. Given, f x( + y) = f x f y( ) ( ), ∀ x, y ∈N For any x ∈N, f x() = [ ( )]f 1x = 3x n ∴ ∑ f x() = 120 x = 1 n ⇒ ∑ 3x = 120 x = 1 ⇒ 31 + 32 + 33 + + 3n = 120 ⇒ 3n − 1 = 80 [f(1) = 3] ⇒ 3n = 81 = 34 ⇒ n = 4 71. 0.423 = 0.4 + 0.023 + 0.00023 +  72. Given, x = 1 + y + y2 +  1 1 ⇒ x = ⇒ 1 − y = 1 − y x 1 x − 1 ⇒ y = 1 − = x x 73. Let the GP be a, ar, ar 2 and ar 3. Then, a (1 + r + r 2 + r 3 ) [given] ⇒ a = 2 74. Sum of areas of all the squares 2 a2 a2 75. Let a be the first term and r (| r | < 1)be the common ratio of infinite GP. Then, a a2 = 3 and 2 = 3 …(i) 1 − r Also, 2 = 9 …(ii) (1 − r ) × = − − = − = + − + = − + ⇒ − + = − + = + + +  = − = −
  • 7. 3 + 96 = ∀ ≥ On solving Eqs. (i) and (ii), we get 3 a = and r = 2 Hence, the sequence corresponding to the series will 3 3 3 be , , ,  2 4 8 1 1 1 76. Given product = x 2 4 8 = x 2 = x 77.Length of a side ofSn =Length of a diagonal ofSn + 1 ⇒ Length of a side ofSn = 2(Length of a side ofSn + 1) Length of a side ofSn + 1 1 ⇒ ⇒ Side of S1,S2, ,Sn form a GP with common ratio and first term 10. n − 1 1 ∴ Side of Sn = 10 2 10 = n − 1 2 2 ⇒ Area of Sn = (Sn )2 = 100n − 1 2 Area ofSn < 1 [given] ⇒ ⇒ 78. Let r be the common ratio of the given GP Then, terms of GP are α α α, r r 2 and αr 3. According to the question, α(1 + r ) = 1, α 2r = p, α r 2 (1 + r ) = 4, α 2r 5 = q On solving, we get (p q, ) = −( 2, − 32) 79. Consider, (1 + x) + (1 + x + x2 ) + (1 + x + x2 + x3 ) + upto n terms 1 − x2 1 − x3 1 − x4 = + + + upto n terms 1 − x 1 − x 1 − x 1 = [(1 + 1 + 1 + to n terms) 1 − x − (x2 + x3 + x4 + upto n terms)] 1 x2(1 − xn ) = n − 1 − x 1 − x 80. We have, 9 + 99 + 999 + upto n terms = (10 − 1) + (102 − 1) + (103 − 1) + upto n terms = (10 + 102 + 103 + upto n terms) − n =− n =− n = − 9n − 10) 81. Let Sn = 0.5 0.55 + 0.555 + upto n terms upto n terms] 82. Let the three digits be a, ar and ar 2. Given, a + ar 2 = 2ar + 1 ⇒ a r( 2 − 2r + 1) = 1 ⇒ a r(− 1)2 = 1 …(i) Also, given a + ar = (ar + ar 2 ) ⇒ 3 a(1 + r ) = 2ra (1 + r ) ⇒ (1 + r ) (3 − 2r ) = 0 [a ≠ 0] 3 ∴ r = , − 1 2 1 1 If r =, then from Eq. (i), a = 2 = 2 = 4 (r − 1) 3 − 1 2 ⇒ n − 1 ≥ 7 ⇒ n ≥ 8 + + + − − = − + − + +  + + + − + + = +   − = ⋅ − − = − × − − − = − − − +
  • 8. + 3 97 If r = − 1, then from Eq. (i), a = , which is not possible, as a is an integer. Hence, a and ar 83. Let the two numbers be a and b, then G = ab or G2 = ab Also, p and q are two AM’s between a and b. ∴ a, p, q and b are in AP. Now, p − a = q − p and q − p = b − q ∴ a = 2 p − q and b = 2q − p Hence, G2 = ab = (2 p − q) (2q − p) 84. Let the 3 n terms of a GP be a, ar, ar 2, …, ar n − 1, ar n, ar n + 1, ar n + 2, …, ar 2n − 1, ar 2n, ar 2n + 1, ar 2n + 2,…, ar 3n − 1. Then, S1 = a + ar + ar 2 + + ar n − 1 a(1 − r n ) = 1 − r S2 = ar n + ar n + 1 + ar n + 2 + + ar 2n − 1 ar n(1 − r n ) = 1 − r S3 = ar 2n + ar 2n + 1 + ar 2n + 2 + + ar 3n − 1 2n n ar (1 − r ) = 1 − r Now, (S2 )2 = a r2 2n a(1 − r n ) 2n (1 − r n ) = ⋅ ar = S S1 3 1 − r 1 − r Hence, S1, S2 and S3 are in GP. 1 85. (32)(32)1 6/ (32)1 36/ = 321 + 61 + 361 + = 32 1−1 6 = 326/ 5 = (25 6) / 5 = 26 = 64 4 On subtracting, we get S = 1 + ++ +  7 5 35 ⇒ 4 4 16 87. Let S = 1 + 2 x + 3x2 + 4x3 + x S = x + 2 x2 + 3x3 +  On subtracting, we get (1 − x S) = 1 + x + x2 + x3 +  1 1 ⇒ (1− x S)= 1 − x ⇒S = ( 1 − x)2 88. Let S = 1 + 2 ⋅2 + 3 2⋅ 2 + 4 2⋅ 3 + + 100 ⋅299 2S = 1 2⋅ + 2 ⋅22 + 3 2⋅ 3 + 4 2⋅ 4 + + 99 2⋅99 + 100 ⋅2100 On subtracting, we get − S = 1 + (1 2⋅+ 1 2⋅2 + 1 2⋅3 +to 99 terms − 100 ⋅2100 ) = 1 + 2 2(99 − 1) − 100 ⋅2100 ⇒ S = 99 2⋅ 100 + 1 n2(n + 1)2 89. Here, Tn n n ⇒ S k 4 k = 1 91. We can write S as S = (1 − 2)(1 + 2) + (3 − 4)(3 + 4) + + (2001 − 2002)(2001 + 2002) + 20032 = − [1 + 2 + 3 + 4 + + 2002] + 20032 = − (2002)(2003) + (2003)2 = 2007006 92. According to the given condition, 1 1984 2() (− 1)n (704)(2) 1 − 2n = 3 1 − 2n 86. + + + + =  ⇒ + + + =  = 90. = + ⋅ = + + Σ = + = − + ∴ − = + = + = × =
  • 9. 3 + 98 2112 (−1)n ⇒ 128 = n − 1984 n 2 2 If n is odd, then we get 2n = 32 ⇒ n = 5 128 If n is even, then we get 128 = n ⇒ n = 0 2 93. We have, 2n + 10 = 2 ⋅22 3 2⋅ 3 + 4 2⋅ 4 + + n⋅2n ⇒ 2 2( n + 10 ) = 2 ⋅23 + 3 2⋅ 4 + + (n − 1)⋅2n + n⋅2n + 1 On subtracting, we get ⇒ ⇒ n = 513 94. Let S = (1)(2003) + (2)(2002) + (3)(2001) + + (2003)(1) and K = 12 + 22 + 32 + + 20032 On adding, we get S + K = (2004)[1 + 2 + 3 + + 2003] ⇒ (2003)(334)(x) + (2003)(4007)(334) = (2004)(1002)(2003) ⇒ x = 2005 95. We can write the given equation as 1 1 1 1 1 + + + + +  log 2 x 2 4 8 16 = 4 − (a b1 1 + a b22 + + a bm m)2 = (a b1 2 − a b2 1)2 + (a b1 3 − a b3 1)2 ++ (am − 1bm − a bm m − 1)2 Thus, − (x x1 2 + x x2 3 + + xn−1xn )2 ≤ 0 ⇒ (x x1 3 − x x2 2 )2 + (x x1 4 − x x3 3 )2 + + (xn − 2 xn − xn − 1xn − 1)2 ≤ 0 As x1, x2, …, xn are real, this is possible if and only if x x1 3 − x22 = x x2 4 − x32 = = xn − 2 xn − xn2 − 1 = 0 x1 = x2 = x3 = = xn ⇒ x2 x3 x4 xn − 1 ⇒ x1, x2, …, xn are in GP. 97. We have, r = = 1 = 1 98. We have, 2 2 = 4 2 = 1 + (2r + 1)(2r − 1) tan−1 21r 2 = tan−1 1(2+r(2r1+) 1) ((22rr −11)) = tan−1 (2r + 1) − tan−1 (2r − 1) n n ⇒ r∑= 1tan−1 2 1 r 2 = r∑= 1{ tan−1 (2r + 1) − tan−1 (2r − 1)} n ⇒ lim→ ∞ r∑= 1tan−1 1 2 = nlim→ ∞ tan−1 (2n + 1) − π 4 n 2r 99. 13 = 1⋅(1 − 1) + 1 23 = (2 ⋅1 + 1) + 5 , 33 = (3 2⋅+ 1) + 9 + 11 , 43 = (4⋅ 3 + 1) + 15 + 17 + 19, etc ∴ n3 = {n⋅(n − 1) + 1} + , in which next term being 2 more than the previous ∴ n3 = (n2 − n + 1) + (n2 − n + 3) + + (n2 + n − 1) 100. ∑n r2 − r − 1 = ∑n r − 1 − r = −n r = 1 (r + 1 )! r = 1 r! (r + 1)! (n + 1)! n ⇒ log2(x2 ) = 4 ⇒ x2 = 24 ⇒ 96. We shall make use of the identity (a12 + a22 + + am2 )(b12 + b22 + + bm2 ) x = 4
  • 10. + 3 99 101. Applying AM ≥ GM Since, AM =GM ∴ ( 102. As, D < 0 2 )x = ( 2 − 2 )x i.e. for x = 0 [AM = GM iff a = b] ⇒ (a + b + c )2 − 4(ab + bc + ca) < 0 ⇒ (a − b + c )2 < 4 ac ⇒ − 2 ac < a − b + c ⇒ ( a + c ) > b 1 103. Given , a and b are in GP. 16 ⇒ a2 = b …(i) 16 Also, a, b and are in HP. ⇒ b = …(ii) On solving Eqs. (i) and (ii), we get a and b , 1 104. Since, a b( − c ) + b c( − a) + c a( − b) = 0 ∴ x = 1is a root of the given equation. Since, both the roots are equal, therefore the other root is also 1. ∴Their sum = 1 + 1 = 2 −b c( − a) ⇒ = 2 a b( − c ) ⇒ −bc + ab = 2ab − 2ac ⇒ 2ac = b a( + c ) ∴ 2ac b = a + c Hence, a, b and c are in HP. 105. Let two numbers be a and b. Then, a + b = 5 ⇒ a + b = 10 Also, ab = 4 ⇒ ab = 16 2ab 2 16( ) 16 ∴HM between a and b = = = a + b 10 5 106. Given, log(a c ), log (c − a) and log (a − 2b + c ) are in AP. ⇒ 2 log(c − a) = log (a + c ) + log (a − 2b + c ) ⇒ (c − a)2 = (a + c ) (a − 2b + c ) ⇒ (c − a)2 = (a + c )2 − 2b a( + c ) ⇒ 2b a( + c ) = (a + c )2 − (c − a)2 = 4 ac ⇒ 2ac b = a + c ⇒ a b, and c are in HP. 107. Given, cos (x − y), cos x and cos (x + y) are in HP. 2 cos(x − y) cos (x + y) ⇒ cos x = cos (x − y) + cos (x + y) 2 2 2 (cos x − sin y) ⇒ cos x = 2 cos x cos y ⇒ cos2 x = 1 + cos y = 2 cos2 y 2 ⇒ cos2 x sec2 = 2 ⇒ cos x sec 108. log2 6 = log2(3 × 2) = log2 3 + log2 2 = 1+ log2 3 and log2 12 = log2(22 × 3) = log2 3 + 2 log2 2 = 2 + log2 3 Since, log2 3, 1 + log2 3, 2 + log2 3 are in AP. ⇒ log2 3, log2 6, log2 12 are in AP. ⇒ log3 2, log6 2, log12 2 are in HP. 109. Given a, b and c are in HP ⇒ b is HM of a and c a + c ⇒ > b 2 ⇒ a2 + c 2 > 2b2 110.AM> GM> HM + × × + = + ⇒ >  > + > + + =±
  • 11. 3 + 100 = ∴ x > y > z ⇒ 111. Given a, b and c are in HP. 1 1 1 ⇒ , , are in AP. a b c z < y < x ⇒ = + ⇒ Hence, the straight line = 0 passes through a b c a fixed point (1, − 2). 112. Given, a10 = 3 ⇒ a1 + 9d = 3 ⇒ 2 + 9d = 3 [a1 = 2 ] ⇒ d = 7 ∴ a4 = a1 + 3d = 2 + = 3 1 h10 = 3 ⇒ h10 3 − + = x + +
  • 12. 3 101 ⇒ D = − ∴ ⇒ ∴ a h4 7 = 6 3 7 113. Since, x, y and z are in HP. ∴ y = ⇒ x − 2 y + = x + z ⇒ (x + z) (x − 2 y + z) = (z − x)2 ⇒ log(x + z) + log(x − 2 y + z) = 2 log(z − x) 1 1 114. Let− = K Hi + 1 Hi ∴ (− 1) ∑2=n1i HHii +− HHii ++ 11 i 2n (− 1)i 1 1 = i∑= 1 Hi + 1 + Hi = 2n K 115. Let the given AP be a1, a1 + d, a1 + 2d, … By substituting the value, we can find that only options (b) and (d) are correct. 116. Let the AP be a, a + d, a + 2d, a + 3d,  Given that, 3 a + 3d = 9 ⇒ a + d = 3 and a2 + (a + d )2 + (a + 2d )2 = 35 ⇒ a2 + (3)2 + (3 + d )2 = 35 ⇒ (3 − d )2 + 32 + (3 + d )2 = 35 ⇒ d = ± 2 and a = 1, 5 n ∴ Sn = [2a + (n − 1) d ] 2 where d = 2, a = 1 ⇒ Sn = n2 When d = − 2, a = 5 , then Sn = n(6 − n) 117. cot−1 1a+2 −a a1 2a1 + cot−1 1a+3 −a a2a23 + cot−1 1a+4 −a a3a34 + + cot−1 1 a+n −a anann−−11 = cot−1 a1 − cot−1 a2 + cot−1 a2 − cot−1 a3 + + cot−1 an − 1 − cot−1 an = cot−1 a1 − cot−1 an = tan−1 an − tan−1 a1 118. If a, bandc are pth,qth and rth terms of an AP, then b − c a − b is always a rational number. 119. Let the roots be a1, a2, a3, a4, a5, then 1 1 1 1 1 + + + + = 10 a1 a2 a3 a4 a5 Σ a a a a1 2 3 4 = 10 ⇒ a a a a a1 2 3 4 5 ⇒ 10S = Σa a a a1 2 3 4 Since, a1, a2, a3, a4, a5 are in GP, then consider a1, a2, a3, a4 and a5 as a, ar, ar 2, ar 3, ar 4 ⇒ 10S = 10a r5 10 = a r36 (a + ar + ar 2 + ar 3 + ar 4 ) ⇒ 10a r5 10 = a r36(40) [sum of roots = 40] ⇒ a r24 = 4 ⇒ ar 2 = ± 2 ⇒ S = a r36(4) = ±(2)3 (4) = ± 32 120. Since, a1, a2, a3, …, an, … are in AP and b1, b2, … bn, …are in GP. ⇒ 1, (1 + d ) , …, [1 + (n − 1)d ] … are in AP ⇒ 1, r, r 2, …, r n − 1, … are in GP. Also, a9 = b9 ⇒ 1 + 8d = r 8 = + − = = = × = x + x x − = + + − + + x x = − x
  • 13. 3 102 ∴ S9 = [2 + (9 − 1)d ] = 369 ⇒ 2 + 8d = 41 × 2 ⇒ d = 10 ⇒ r 8 = 81 ⇒ r 8 ⇒ ∴6 = 9 × 3 = 27 and 121. Given, sin β = sin α cos α 4 or 2 cos2 π + α 4 122. Since, ais AM between 1st and (2n + 1)th terms, bis GM between 1st and (2n + 1) th terms and c is HM between 1st and (2n + 1) th terms. ⇒ AM ≥ GM ≥ HM and AM × HM = (GM)2 123. If each of x, y and z is less than 1, then Statement I is obviously true. Also,1 − 2 x + 1 − 2 y + 1 − 2 z = 3 − 2(x + y + z ) = 1, The sum of the three given numbers is positive also at must one of x, y and z can be more than 1. If one of x, y and z is more than or equal to , then their product is less than equal to zero, hence still remains true. Statement II is always true but it does not explain Statement I. 124. Since, ax2 + bx + c = 0 and a x1 2 + b x1 + c1 = 0 have common root. (a c1 − ac1)2 = 4(ab1 − ba1) (bc1 − b c1) …(i) a c11 On putting these values in Eq. (i), we get c a1 1 = b12 125. Statement I is false, since each term of the series +++ + is smaller than 10−5 but its 6 6 6 6 sum upto infinity is infinity. n Statement II is true, since lim is not finite as n→ ∞ 105 n → ∞ 126. Statement I can be proved by taking the intersection of the inequalities. a > 0, ar > 0, ar 2 > 0 at a + ar > ar 2, ar + ar 2 > a, ar 2 + a > ar The inequalities follow from reason. 127. If we could show that Statement II is true, then Statement I will be false. Indeed if Statement II is false, then 2 − 3 = (p − q d) ...(i) and 3 − 5 = (q − r d) ...(ii) On dividing Eq. (i) by Eq. (ii) we get 2 − 3 p − q = 3 − 5 q − r ⇒ Rational =Irrational Hence, Statement I is false and Statement II is true. 128. Statement I is true, since for any x > 0, we can choose xn sufficiently larger n such that is small. Statement II is n ! (n !)2 false, since contains n ! in the denominator but n ! ⇒ sin2 β = sin α cos α Now, cos 2β = 1 − 2 sin2 β = 1 − 2 sin α cos α = (sin α − cos α )2 = 2 sin2 − = − = − = = = = =
  • 14. 3 103 diverges to ∞. 129. We can show that Statement I is true and follows from Statement II. Indeed a1 + a2 ++ an + an + 1 a1 + a2 + + an Sn + 1 − S = − n n(+ 1) [an + 1 > a1, an + 1 > a2, , etc. ] Solutions (Q. Nos. 130-132) Given, a b(− c x)2 + b c( − a x)+ c a(− b) = 0 ...(i) Since, a b( − c ) + b c( − a) + c a( − b) = 0, therefore x = 1 is a root of Eq. (i). Let other root be α, then c a( − b) 1 × α = a b(− c ) 2ac c a − a + c == 1 2ac a − c a + c ∴ α = 1 Hence, both roots of Eq. (i) are 1, 1. Then, roots of x2 − Px + Q = 0 are also 1, 1. Now, 1 + 1 = P, 1 1⋅= Q ∴ P = 2, Q = 1 ∴ [P] = [2] = 2 Now, [2P − Q] = [4 − 1] = 3 Roots of Eq. (i) are 1, 1, then a X′ ,0) = 4 an ⋅ 2 x 32 an/2 an 3 0 an/2 = 8 an anan − 3 2 2 2 2 ⋅ 8 an = 4 2 an2 sq units = 3 2⋅ 2 3 134. We have, y2 = 4x and y2 = 4 − 4x ⇒ 8x = 4 ⇒ x and y = ± 135. For an = 1, An =sq units 3 136. a b ca b c = (aa a times) (bb b times) (cc c times) Applying AM-GM inequality, (a b ca b c )1/ n a + a + + a times) + (b + b + + b times) ≤ + (c + c + + c times) n a2 + b2 + c 2 = n Similarly, (a b cb c a )1/ n ≤ ab + bc + ca ≤ a2 + b2 + c 2 + = − + + + + +  = − + − + + − > + + +  =− + − − ⇒ − − =− 133. = + − ∫ ∫ X ′ − ⋅ − x −
  • 15. 3 104 Also, (a b cc a b)1/ n ≤ n n So, (a b ca b c )1/ n + (a b cb c a )1/ n + (a b cc a b)1/ n (a + b + c )2  and G = H + …(ii)  G2 = AH ⇒ G2 = G + 3 G − 6 [from Eqs. (i) and (ii)] 2 5 ∴ G = 6, A = [from Eq. (i)] Since, a and b are the roots of x2 − 15x + 36 = 0. On solving, we get a = 12, b = 3 or a = 3, b = 12 ∴ α = 15, β = 9 ⇒ α + β2 = 15 + 81 = 96 and α 2 + β = 225 + 9 = 234 B.  A = G + 2 …(i) and H b [if b > a] 2 bA  G = AH = 5 ⇒ 5 ab = bA ⇒ A = 5 a a + b ⇒ ⇒ ab + 2 From Eq. (i), ⇒ 5 a = 3 a + 2 ∴ a = 1and b = 9 ⇒ α = a + b = 10 and β =|a − b| = 8 ∴ α + β2 = 10 + 64 = 74 C.  H = 4 ⇒ 2 A + G2 = 27 ⇒ 2 A + AH = 27 [G2 = AH] ⇒ 6A = 27 ⇒ A and G Since, a and b are the roots of x2 − 9x + 18 = 0. ∴ a = 6, b = 3 or a = 3, b = 6 Now, α = a + b = 9, β =|a − b| = 3 ∴ α 2 + β = 81 + 3 = 84 2 3 4 5 and 1 = 1 + 3 + 32 + 33 + 34 + 35 = 364 138. Let us add one more number an+1 to the given sequence. The number an+1 is such that | an+1| =| an + 1|. On squaring all the numbers, we have a12 = 0 a22 = a12 + 2a1 + 1 a32 = a22 + 2a2 + 1 a42 = a32 + 2a3 + 1     an2 = an2−1 + 2an−1 + 1 an2+1 = an2 + 2an + 1 On adding the above equalities, we get a a a a = a12 + a22 + + an2 + 2(a1 + a2 + + an ) + n ⇒ a a a n a n a1 + a2 + + an ≥ − 1 = − λ ⇒ µ ≤ = n n Further, a < n ⇒ aa < an ∴ (a b ca b c ) < (abc )n⇒(a b ca b c )1/ n < (abc ) 3 137. A. A = G + …(i) + + ≤ + + = = =
  • 16. 3 105 139. Let abe the first term andd be the common difference of an AP. Again, let x, y, z be the (m + 1) th, (n + 1) th and (r + 1) th terms of an AP. Then, x = a + md, y = a + nd 140. Let f x( ) = x4 + ax3 + bx2 + cx + 1 As a b, and c are non-negative, no root of the equation f x( ) = 0 can be positive. Further as f(0) ≠ 0, all the roots of the equation, say x1, x2, x3 and x4 are negative. We have, ∑ x1 = − a, ∑ x x12 = b, ∑ x x x1 23 = −c and x x x x1 2 3 4 = 1 Using AM ≥ GM for positive numbers −x1, −x2, −x3 and −x4, we get a Using AM ≥ GM for positive numbers x x1 2, x x1 3, x x1 4, x x2 3, x x2 4 and x x3 4, we get b Finally, using AM ≥ GM for positive numbers −x x x1 23, −x x x124, −x x x1 34 and −x x x2 3 4, we get ≥⇒ c ≥ 4 ∴ HCF of {a b c, , } = HCF of {4, 6, 4} = 2 Entrances Gallery b c 1. = = (integer) a b a a 2b b2 6 ⇒ 1 − a + a2 = a b 2 6 Let the three consecutive terms be a − d, a, a + d, where d > 0 Then, a2 − 2ad + d 2 = 36 + K a2 = 300 + K and a2 + 2ad + d 2 = 596 + K On subtracting Eq. (i) from Eq. (ii) we get d(2a − d ) = 264 On subtracting Eq. (ii) from Eq. (iii), we get d (2a + d ) = 296 …(i) …(ii) …(iii) …(iv) …(v) Again, subtracting Eq. (iv) from Eq. (v), we get 2d 2 = 32 ⇒ d 2 = 16 ⇒ d = 4 [d = − 4, rejected] From Eq. (iv), 4(2a − 4) = 264 ⇒ 2a − 4 = 66 ⇒ 2a = 70 ⇒ a = 35 ∴ K = 352 − 300 = 1225 − 300 = 925 ⇒ K − 920 = 5 ⇒ b2 = ac ⇒ c = b2 a ...(i) Also, a + b + c = b + 2 3 ⇒ a + b + c = 3 b + 6 ⇒ a − 2b + c = 6 ⇒ 2 2 b b a − 2b + = 6 from Eq. (i), c = + = ∴ = + + + = = − + − ⇒ = + ⇒ = + = + − − = − − = − =− λ ⇒ λ=
  • 17. 3 106 − 1 = = cot ∑ cot−1 (1 + 2 + 4 + 6 + 8 + ... + 2n) n = 1 23 = cot ∑ cot−1 {1 + n n(+ 1)} n = 1 = cot n∑23= tan−1 1 + n n(1 + 1) 1 = cot ∑23 tan 1 1 n+ +n n(1 −+n1) − n = 1 = cot ∑23 [tan−1(n + 1) − tan−1 n] n = 1 = cot [(tan−1 2 − tan−1 1) + (tan−1 3 − tan−1 2) + (tan−1 4 − tan−1 3) + + (tan−1 24 − tan−1 23)] = cot{tan−1 24 − tan−1 1} −1 24 − 1 = cot tan−1 23 = cot tan + ⋅ 25 1 24 ( )1 = cot cot−1 25 23 25 = 23 k k( + 1) 4n 2 3. Sn = ∑ (− 1) ⋅ k 2 k = 1 = − (1)2 − 22 + 32 + 42 − 52 − 62 + 72 + 82 +  = (32 − 1 )2+ (42 − 22 ) + (72 − 52 ) + (82 − 62 ) +  = 2{(4 + 12 + 20 + ... ) + (6 + 14 + 22 + ... )}   n terms n terms n n = 2 2 {2 × 4 + (n − 1 8) } + 2 {2 × 6 + (n − 1 8) } = 2[n(4 + 4 n − 4) + n(6 + 4 n − 4)] = 2 4[ n2 + 4 n2 + 2n] = 4 n(4n + 1) Here, 1056 = 32 × 33, 1088 = 32 × 34, 1120 = 32 × 35, 1332 = 36 × 37 Hence, 1056 and 1332 are possible answers. 4. Here, a1 = 5, a20 = 25 for HP ∴ = 5 and ⇒ ⇒ ∴ a + (n − 1)d ⇒+ (n − 1)d < 0 1 ⇒ (n − 1) < 0 ⇒ (n − 1) > a a ⇒ a = 6 only 2 a + a − 14 ⇒ = 4 a + 1 23 n 2. cot n∑= 1cot−1 1 + k∑= 12k 23 + = + = = − =− = − ×  < ⇒ < −
  • 18. 3 107 5 19 × 25 ⇒ n > 1 + ⇒ n > 24.75 Hence, the least positive value of n is 25. 5. Given, a1 = 3, m = 5 n and a1, a2, ..., a100 are in AP. Sm = S5n is independent of n. Also, Sn Sn 5 n [2 × 3 + (5 n − 1)d ] Sm = 2 Consider Sn n [2 × 3 + (n − 1)d ] 2 5{(6 − d ) + 5 nd} = (6 − d ) + nd For independent of n, put 6 − d = 0 ⇒ d = 6 ∴ a2 = a1 + d = 3 + 6 = 9 Sm is independent of n. Also, if d = 0, then Sn ∴ a2 = 3 6. Using AM ≥ GM, a−5 + a−4 + a−3 + a−3 + a−3 + 1 + a8 + a10 ⇒ a−5 + a−4 + 3 a−3 + 1 + a8 + a10 ≥ 8 1⋅ Hence, the minimum value is 8. k − 1 7. We have, Sk = 1 − k ! 1 = (k − 1 1)! k Now, (k 2 − 3k + 1)Sk = {(k − 2)(k − 1) − 1} × 1 (k − 1)! 1 1 = − (k − 3)! (k − 1)! ⇒ |(kk )Sk| k = 1 1 1 = 1 + 1 + 2 − 99! 98! ⇒ 100 ! k = 1 8. Since, ak = 2ak − 1 − ak − 2 So, a1, a2, a11 are in AP. a d a d ⇒ 225 + 35d 2 + 150 d = 90 ⇒ 35d 2 + 150 d + 135 = 0 ⇒ d Given, a ∴ d = − 3 and d a1 + a2 + + a11 = 1 [30 − 10 × 3] = 0 ⇒ 11 2 9. Given, m is the AM of l and n. ∴ l + n = 2m …(i) and G1, G2, G3 are geometric means between l and n. ∴ l, G1, G2, G3, nare in GP. Let r be the common ratio of this GP. ∴ G1 = lr G2 = lr 2 G3 = lr 3 n = lr 4 n ⇒ r l Now, G1 4 + 2G2 4 + G3 4 = (lr )4 + 2 (lr 2 4)+ (lr 3 4) = l 4 × r 4(1 + 2r 4 + r 8 ) = l 4 × r 4(r 4 + 1) = l 4 × n n + l 2 l l = ln × 4 lm2 = 4 lm n2
  • 19. 3 108 − = = 10. Given, series is Let Tn be the nth term of the given series. 13 + 23 + 33 + ... n3 ∴ Tn = 1 + 3 + 5 + ... + to n terms n n(+ 1) 2 2 (n + 1)2 = n2 = 4 9 2 (n + 1) 1 2 2 2 2 2 9 ∑ ∴ S = = [(2 + 3 + … + 10) + 1 − 1 ] n = 1 4 4 1 10 10(+ 1)(20 + 1) 384 = − 1 = = 96 11. Given, α and β are roots of px2 + qx + r = 0, p ≠ 0. −q r ∴ α + β = ,αβ = ...(i) p p Since, p q, and r are in AP. ∴ 2q = p + r ...(ii) Also, + = 4 ⇒ ⇒ α + β = 4αβ [from Eq. (i)] p p ⇒ q = − 4r On putting the value of q in Eq. (ii), we get 2(− 4 )r= p + r ⇒ p = − 9r −q 4r 4 Now, α + β = = = = − p 9 r and αβ = p − 9r − 9 ∴ 2 ⇒⇒ |α − β| = 13 9 12. Given, k ⋅109 = 109 + 2 11( ) (1 10)8 + 3 11( ) (2 10)7 + + 10 11()9 11 11 2 11 9 ⇒ k = 1 + 2 3 + + 10 ...(i) 10 10 10 11 11 11 2 11 9 11 10 k = 1 + 2 + 9 10 10 10 10 10 10 ...(ii) On subtracting Eq. (ii) from Eq. (i), we get 11 11 11 2 11 9 11 10 k 1 − 1  10 10 10 10 10 10 11 10 1 10 − 1 ⇒ k 10 − 11 = − 10 11 10 10 11 10 1 10 n a r( − 1) [in GP, sum of n terms =, when r > 1] r − 1 11 10 11 10 ⇒ − k = 10 10 10 − 10 − 10 10 ∴ k = 100 13. Let a ar, and ar 2 be in GP (r > 1). On multiplying middle term by 2, then the numbers a, 2ar and ar 2 are in AP. 4 6 4 α β α β αβ + = ⇒ − =
  • 20. 3 109 = [(1 + 1 + upto 20 terms) 1 1 1 − 10 + 102 + 103 + upto 20 terms 1 − = 7 1 101 20 20 − 9 10 1 − 1 10 7 110 1 − 101 20 = 20 − × 9 10 9 = 7 20 − 109(1 − 10−20 ) = 81 7 (180 − 1 + 10−20 ) 9 = (179 + 10−20 ) 15. Here, T100 = a + (100 − 1)d = a + 99d T50 = a + (50 − 1)d = a + 49d T150 = a + (150 − 1)d = a + 149d Now, according to the given condition, 100 × T100 = 50 × T50 ⇒ 100(a + 99d ) = 50(a + 49d ) ⇒ 2(a + 99d ) = (a + 49d ) ⇒ a + 149d = 0 ⇒ T150 = 0 16. Statement I Let S = (1) + (1 + 2 + 4) + (4 + 6 + 9) +... + (361 + 380 + 400) S = (0 + 0 + 1) + (1 + 2 + 4) + (4 + 6 + 9) + + (361 + 380 + 400) Now, we can clearly observe the first elements in each bracket. In second bracket, the first element is 1 = 12 In third bracket, the first element is 4 = 22 In fourth bracket, the first element is 9 = 32 In last bracket, the first element is 361 = 192 Hence, we can conclude that there are 20 brackets in all. Also, in each of the bracket, there are 3 terms out of which the first and last terms are perfect squares of consecutive integers and the middle term is their product. Now, the general term of the series is Tr = (r − 1)2 + (r − 1)r + (r )2 So, the sum of n terms of the series is n Sn = ∑ {(r − 1)2 + (r − 1)r + (r ) }2 r = 1 n 3 3 r − (r − 1) r = 1 n Now, let Sn = ∑ {k 3 − (k − 1) }3 k = 1 On substituting the value of k and rearranging the terms, we get Sn = − 03 + (13 − 13 ) + + [(n − 1)3 − (n − 1)3 ] + n3 ⇒ Sn = n3 Since, the number of terms is 20, hence substituting n = 20, we get S20 = 8000 Statement II We have already proved in the ⇒ Sn = ∑ r = 1 r − (r − 1) ⇒ n Sn = ∑ {r 3 − (r − 1) }3 ⇒ + = ⇒ = + − ⇒ = ± − = ± ⇒ + =  14. + = + +  + + + =  − = − + − + + 
  • 21. 3 110 Statement I that n Sn = ∑ {k 3 − (k − 1) }3 = n3 k = 1 17. Let the time taken to save ` 11040 be (n + 3) months. For first 3 months, he save `s 200 each month. In (n + 3) months, his total savings is n 3 × 200 + [ (2 240) + (n − 1) × 40] = 11040 2 ⇒ 600 + 20n n(+ 11) = 11040 ⇒ 20n n(+ 11) = 10440 ⇒ n n(+ 11) = 522 ⇒ n2 + 11n − 522 = 0 ⇒ n2 + 29n − 18n − 522 = 0 ⇒ n n(+ 29) − 18(n + 29) = 0 ⇒ n = 18 or n = − 29 ∴ n = 18 [neglecting n = − 29 ] Hence, total time = n + 3 = 18 + 3 = 21months 18. Number of notes that the person counts in 10 min = 10 × 150 = 1500 We have, a10, a11, a12, ... are in AP with common difference − 2. Let n be the time taken to count remaining 3000 notes, then n [2 × 148 + (n − 1) × −(2)] = 3000 2 ⇒ n2 − 149 n + 3000 = 0 ⇒ (n − 24)(n − 125) = 0 ⇒ n = 24,125 Hence, total time taken by the person to count all notes 19. Let S ⇒ S ...(i) S − 1 2 6 10 14 ⇒ = 2 + 3 + 4 + 5 +  3 3 3 3 3 On subtracting Eq. (ii) from Eq. (i), we get ...(ii) S = 2 ⇒ + 20. Since, a + ar = a (1 + r ) = 12 …(i) and ar 2 + ar 3 = ar 2 (1 + r ) = 48 …(ii) From Eqs. (i) and (ii), we get r 2 = 4 ⇒ r = − 2 On putting the value of r in Eq. (i), we get a = − 12 21. Since, each term is equal to the sum of the next two terms. ∴ ar n − 1 = ar n + ar n + 1 ⇒ 1 = r + r 2 ⇒ r 2 + r − 1 = 0 5 − 1 − 5 − 1 r = 2 r ≠ 2 a1 + a2 + + ap p2 22. Since, a1 + a2 + + aq = q2 p ∴ 2q [[22aa11 ++ ((qp−−11))dd]] = qp22 2 (2a1 − d ) + pd = p ⇒ (2a1 − d ) + qd q ⇒ (2a1 − d ) (p − q) = 0 + 5d 2 23. Since, a1, a2, a3, ..., an are in HP. 1 1 1 1 + + + = − + ⇒ − + + = + + − = + = ⇒ =  ≠ = + + = + = =
  • 22. 3 111 ∴ , , , , are in AP. a1 a2 a3 an Let d be the common difference of AP. 1 1 ∴ − = d a2 a1 ⇒ a1 − a2 = a a d1 2 Similarly, a2 − a3 = a a d2 3    an − 1 − an = an − 1a dn On adding all the equations, we get a1 − an = d {a a1 2 + a a2 3 + ...+ an − 1an} …(i) 1 1 Also, = + (n − 1)d an a1 a1 − an ⇒ d = a a1 n (n − 1) On putting the value of d in Eq. (i), we get a1 − an {a a1 2 + a a23 + + an − 1 an} a1 − an = a a1 n(n − 1) ⇒ 24. Given that, x = ∑ an, y = ∑ bn, z = ∑ c n n = 0 n = 0 n = 0 1 ...(i) Similarly, y ⇒ x = 1 + a + a2 + = =…(ii) and z =...(iii) 1 − c Now, a b, and c are in AP. ⇒ − a, − band − c are in AP. ⇒ 1 − a, 1 − band 1 − c are also in AP. 1 1 1 ⇒ , and are in HP. 1 − a 1 − b 1 − c ⇒ x y z, , are in HP. Aliter From Eqs. (i), (ii) and (iii), 1 1 1 x = , y = and z = 1 − a 1 − b 1 − c x − 1 y − 1 z − 1 ⇒ a = , b = and c = x y z Since, a b, and c are in AP. ∴ 2b = a + c ⇒ 1 1 ⇒ Hence, x y, and z are in HP. 25. We know that, e x + e − x x2 x4 x6  putting x On in both sides, we get 1 26. Given, Tm = n 1 ⇒ a + (m − 1)d =...(i) n 1 and Tn = m 1 ⇒ a + (n − 1)d =...(ii) m On solving Eqs. (i) and (ii), we get 1 a = d = mn ∴ a − d = 0 27. Given that the sum of n terms of given series is n n( + 1)2 , if n is even. 2 Let n be odd, i.e. n = 2m + 1 Then, S2m + 1 = S2m + (2m + 1)th term (n − 1)n2 = + nth term 2 + + = + + −  ⇒ + + = ⋅ + ⋅ + ⋅ +  − − ⇒ x x − = − + − − = − + − x x = +
  • 23. 3 112 (n − 1) n2 2 = + n 2 = n2 n − 1 + 2 2 (n + 1) n2 = 2 28. We know that, 1 1 1 1 e = 1 + + + + + ... ...(i) 1! 2 ! 3! 4! and e − 1 = 1 − 1 + 1 − 1 + 1 − ... ...(ii) 1! 2 ! 3! 4! On adding Eqs. (i) and (ii), we get e e e 2 + 1 2 2 ⇒ 11 ⇒ + ... e 2 ! 4! ⇒ = + + ... 2 2 ! 4! 29. Consider, 1 1 1 1 1 1 − − − ... 2 2 3 3 4 1 1 1 = 2 1 − + − + ... − 1 2 3 4 e 30. Since, 1, log3 31 − x + 2 and log3 (4⋅ 3x − 1) are in AP. ∴3x = t] t ⇒ 12 t 2 − 5t − 3 = 0 ⇒ (3t + 1) (4t − 3) = 0 ⇒ t ⇒ x cannot be negative] 3 ⇒ log3 4 = x ⇒ x = 1 − log3 4 31. Let S = 21/ 4 ⋅ 41/ 8 ⋅ 81/16... = 21/ 4 ⋅ 22/ 8 ⋅ 23/16... 1 2 3 1 + + + ... = 2 4 2 2 2 = 2 where, S It is an infinite arithmetico-geometric progression. a d ⋅ r ∴ S1 = 1 − r + (1 − r )2 = 1 1 2 = + = 1 − 1 − 2 2 ∴ S 32. Since, 5th term of a GP = 2 ∴ ar 4 = 2 …(i) where, a and r are the first term and common ratio respectively of a GP. Now, required product = a × ar × ar 2 × ar 3 × ar 4 × ar 5 × ar 6 × ar 7 × ar 8 = a r936 = (ar 4 9) = 29 = 512 [from Eq. (i)] (loge x )n loge x (loge x)2 n = 0 n ! =eloge x = x 34. e e x 35. Vr = Sum of first r terms of an AP whose first term is r and common difference(2r − 1) r − = + + + − + =
  • 24. 3 113 r 2 = [2r + 2r − 3r + 1] 2 r 2 = [2r − r + 1] n n n n 1 3 2 1 36. Since, a b, and c are in GP. ∴ b2 = ac Given equation is (loge a x) 2 − (2 loge b x) + (loge c ) = 0 ...(i) On putting x = 1in Eq. (i), we get loge a − 2 loge b + loge c = 0 ⇒ 2 loge b = loge a + loge c ⇒ loge b2 = loge ac ⇒ b2 = ac, which is true. Hence, one of the roots of given equation is 1. Let another root be α. 2 loge b = loge b2 ∴ Sum of roots, 1 + α = loge a loge a ⇒ α = loge ac − 1 = (loge a + loge c ) − 1 loge a loge a = loge c = loga c loge a Hence, roots are 1 and loga c. = x + = 37. Given, f x( ) ∴ f(2 x) = 2 x + f(2 x) = and f(4x) = 4x + ⇒ f(4x) = Since, f x( ), f(2 x) and f(4x) are in HP. 1 1 1 ⇒ (2 x + 1) (8x + 1) = (5x + 1) (4x + 1) ⇒ 16x2 + 10 x + 1 = 20 x2 + 9x + 1 ⇒ 4x2 − x = 0 ⇒ x (4x − 1) = 0 ⇒ x = [x ≠ 0] Hence, one real value of x for which the three unequal terms are in HP. 38.AM ≥ GM ⇒ M ≤ 1 ⇒ M ≤ 1 Also, (p + q) (r + s) > 0 [p q r s, , , > 0] ∴ M > 0 Hence, 0 < M ≤ 1 39. Case I Let α = ω and β = ω2 n ∴ S = − n = 0 ω n = 0 = 1 − ω2 + ω4 − ω6 + ω8 −ω10 + ω12 + ... + ω600 − ω602 + ω604 = 1 − ω2 + ω − 1 + ω2 − ω + 1 + ... + 1 − ω2 + ω = + − + + + = + − + + − = + + + ∴ x x ⇒ x x = + ⇒ x x x + = + + + ⇒ x x x x + = + + + ∴ + + + ≥ + + ⇒ ≥ = = = = ∑ ∑∑∑ + − = = ⋅ + − + + + + = + − + + + + x + ⇒ x + x +
  • 25. 3 114 = 0 + ... + 1 − ω2 + ω = − ω2 − ω2 = − 2ω2 [1 + ω + ω2 = 0] Case II Let α = ω2 and β = ω 302 n 2 n 302 n ω4 n ω ∴ S = ∑ (− 1) ∑ (− 1) 3 n = 0 n = 0 n n = 0 = 1 − ω + ω2 − ω3 + ω4 − ω5 + ω6 − ... + ω300 − ω301 + ω302 = 1 − ω + ω2 − 1 + ω − ω2 + 1 − ... + 1 − ω + ω2 = 0 + ... + 1 + ω2 −ω = − ω − ω = − 2ω α 2 + β2 2 α 3 + β3 3 x − Now, α + β = p and αβ = q α 2 + β2 2 α 3 + β3 3 Hence, (α + β) x − x x +  2 3 = log (1 + px + qx2 ) 41. Given, 1 1 1 1 = 1 − 1 − 2 + 3 − 4 + 5 − ... = 1 − loge 2 42. We have, e x = (1 − x) (B0 + B x1 + B x2 2 + ... + B x + B xn n + .... ) By the expansion of e x, we get x x2 xn 1 + + + ... + + ... 1! 2 ! n ! = (1 − x) (B0 + B x1+ B x2 2 + ... + Bn − 1xn − 1 + B xn n + ... ) On equating the coefficient of xn both sides, we get 1 Bn − Bn − 1 = n ! ∞ 3n ∞ 3n − 2 x x 43. We have, a = ∑ , b = ∑ n = 0(3 n)! n = 1(3 n − 2)! and c 1 ∞ 3n ∞ 3n − 2 ∞ 3n − 1 x x x Now, a + b + c = ∑ + ∑ + ∑ (3 n)! (3 n − 2)! (3 n − 1)! n = 0 n = 1 n = 1 x2 x3 x e a b c x x and a + bω2 + cω = e ω2x, ω is an imaginary cube root of unity. Now, a3 + b3 + c 3 − 3 abc = (a + b + c ) (a + bω + cω ) (2a + bω2 + cω) =e x ⋅ e ωx ⋅ e ω2x = e x (1 + ω + ω2) = e 0⋅x = 1 x 3 2 7 3 15 4 44. Given, f x( ) = + x + x + x + ... 1! 2 ! 3! 4! + ... − + + + + ... 1! 2 ! 3! 4! 40. β α − + + + x x x  − + α αx x + β β β x x x − + − = +β + +α x x + + + = β α αβ x x
  • 26. 3 115 x x2 x3 x4 + ... − 1 + + + + + ... 1! 2 ! 3! 4! ⇒ f x() = e 2 x − e x Put f x() = 0, we get e 2 x − e x = 0 ⇒ e x(e x − 1) = 0 ⇒ e x = 0 or e x = 1 ⇒ x = 0 Hence, exactly one real solution exists. 8 21 40 65 45. We have, S = 1 + + + + + ... 2 ! 3! 4! 5! Let S1 = 1 + 8 + 21 + 40 + 65 + .... + Tn ...(i) and S1 = 1 + 8 + 21 + 40 + ...+ Tn − 1 + Tn On subtracting Eq. (ii) from Eq. (i), we get ...(ii) 0 = 1 + 7 + 13 + 19 + 25 + ... − Tn Tn = 1 + 7 + 13 + 19 + 25 + ... upto nterms n = [ ( )2 1 + (n − 1 6) ] 2 = n [1+ 3 (n − 1)] = n (3 n − 2) n(3 n − 2) ∴ S = Σ = Σ = Σ ⇒ S = Σ = 3e + e = 4e (n − 2)! (n − 1)! 1 1 e = 1 + 1! + 2 ! + ... We know that, 2 < e < 3 ∴ 8 < 4e < 12 ⇒ 8 < S < 12 46. Since, a1, a2, ..., an are in AP, therefore a2 − a1 = a3 − a2 = ... = a2k − a2k − 1 = d [say] Now, a1 2 − a2 2 = (a1 − a2 ) (a1 + a2 ) = − d (a1 + a2 ) a32 − a42 = − d (a3 + a4 ) a22k −1 − a22k = − d (a2k−1 + a2k ) On adding, we get S = − d (a1 + a2 + ... + a2k ) 2k = − d 2(a1 + a2k ) = − a( + a dk ) = (a2 − a2 2 k ) = [(a1 − a2 ) + (a2 − a3 ) + (a3 − a4 ) + ...+ (a2k−2 − a2k−1) + (a2k−1 − a2k )] k 2 2 == (a1 − a2k ) (− d ) (2k − 1) 2k − 1 47. Given, log x (ax), log x (bx)and log x (cx) are in AP. ⇒ 1 + log x a, 1 + log x b, 1 + log x c are in AP. ⇒ log x a, log x b, log x care in AP. − − )! − + − )! + Σ − − − −
  • 27. 3 116 log a log c log b ∴ + = 2 log x log x log x ⇒ log a + log c = 2 log b ⇒ ac = b2 48. Let the six numbers in AP be a − 5d a, − 3d a, − d a, + d a, + 3d a, + 5d ∴ a − 5d + a − 3d + a − d + a + d + a + 3d + a + 5 d = 3[sum = 3] ⇒ 6 a = 3 ⇒ a = Also, T1 = 4T3, where T T1, 3 are respectively first and third terms of an AP. ⇒ a − 5d = 4 (a − d ) ⇒ d = − 3 a = − So, the fifth term = a + 3d 1 3 1 9 = + 3 − = − 4 2 2 2 2 49. Since,log10 2, log10 (2 x − 1) and log10 (2 x + 3) are in AP. ∴ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ Now, let 2 x = t ⇒ t 2 − 4t − 5 = 0 ⇒ (t − 5) (t + 1) = 0 ⇒ t or t = − 1 50. Sn = 0.2 + 0.22 + 0.222 + ... upto n terms = 2 [0.1 + 0.11 + 0.111 + ... upto n terms] 2 =[0 9. + 0 99.+0 999. + ... upto n terms] = [(1− 0 1. ) + {1 − (0 1. ) }2+ {1 − (0 1. ) }3 +...upto n terms] upto n terms}] = 2 n − (0 1. ) {11−−(0 10 1. ) }. n = 29 n − 19 (1 − 10−n ) 9 2 2 2 3 n 1 1 1 = 1 2 − 2 − ... + 2 − 2 3 n [1 + {2 + 2 +... + 2}] 1 1 1 = (n− 1) times − 2 + 3 + ... + n ⇒ 2 = 5 [neglecting 2 x = − 1as 2 x is always positive] ⇒ x log2 2 = log2 5 ⇒ x = log2 5 +… + + = ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + + = + + + + + + − = = = − 52. = + + + = + + + + −
  • 28. 3 117 1 1 1 = [1 + 2 (n − 1)] − 1 + 2 + 3 + ... + n + 1 1 1 1 = 2n − 1 + 2 + 3 + ... + n = 2n − Hn 53. Let the numbers be a and b. a + b 2ab ∴ = 27 and = 12 2 a + b ⇒ a + b = 54 and 2ab = 12 (a + b) ⇒ 2ab = 12 (54) ⇒ ab = 6 (54) = 324 ⇒ ab = 18 Thus, GM = 18. 54. Since, x y, and z are in GP. ∴ y2 = x z Now, taking log10 on both sides, we get 2 log10 y = log10 x + log10 z 1 1 1 ⇒ 2 + ⇒ log x 10 logy 10 logz 10 ⇒ log x 10, logy 10 and logz 10 are in HP. (2 log 2)2 + 2 1 + 2 log 2 + +... 2 ! = 2 (elog 2 ) + 2 (e 2 log 2 ) = 2 × 2 + 2elog 4 = 4 + 2 × 4 = 12 56. Let Tn be the nth term of the given series. n n(+ 1) n + 1 (n − 1) + 2 ∴ Tn = = = n ! (n − 1)! (n − 1)! ∴ S T n = 1 n = 2(n − 2)! n = 1(n − 1)! = e + 2e = 3 e 57. Given, sum of n terms of an AP = 240 n ⇒ n(2 + n − 1) = 240 ⇒ n n(+ 1) = 15 × 16 ⇒ n = 15 58. Given, S and a Let r be the common ratio. a xn + yn ⇒ xn + 1 + yn + 1 = (xy)1 2/ (xn + yn ) ⇒ xn ⋅ x + yn ⋅ y = xn ⋅ x1/ 2 y1/ 2 + yn ⋅ x1/ 2 y1/ 2 ⇒ = 1 y ⇒ n + = 0 ⇒ 60. Let the numbers be a and b. ∴ ab = 10 ⇒ ab = 100 2ab and= 8 a + b 200 ⇒= 8 a + b 55. + + + + + + + + = + + ∴ − = ⇒ = − ⇒ − = ⇒ = ⇒ = 59. x + + + = x =−
  • 29. 3 118 + + ⇒ a + b = 25 On solving Eqs. (i) and (ii), we get a = 5 and b = 20 or a = 20 and b = 5 61. Tn = (2n − 1)3 = 8 n3 − 13 − 3 2⋅n⋅1(2n − 1) = 8 n3 − 1 − 12 n2 + 6 n = 8 n3 − 12 n2 + 6 n − 1 ∴ Sn = ΣTn = 8 Σn3 − 12 Σ n2 + 6 Σ n − Σ1 n n(+ 1) 2 n n(+ 1)(2n + 1) = 8 2 − 12 ...(i) 2 − n = 2n2(n + 1)2 − 2n n( + 1)(2n + 1) + 3 n n( + 1) − n = n n(+ 1)[2n n(+ 1) − 2 (2n + 1) + 3] − n = n n(+ 1)[2n2 + 2n − 4 n − 2 + 3] − n ...(ii) 2 = n n(+ 1)[2n − 2n + 1] − n = n n(+ 1) 2⋅n n(− 1) + n n(+ 1) − n = 2n2(n2 − 1) + n2 = n2(2n2 − 1) 2n 1 1 62. Here, Tn = = − (2n + 1)! 2n ! (2n + 1)! 
  • 30. 4 Complex Numbers The Real Number System Natural Numbers (N ) The numbers which are used for counting are known as natural numbers (also known as set of positive integers), i.e. N = {1, 2, 3, …}. Whole Numbers (W) If ‘0’ is included in the set of natural numbers, then we get the set of whole numbers, i.e. W = {0, 1, 2, … } = {N} + { }0 . Integers (Z or I) If negative natural numbers are included in the set of whole numbers, then we get set of integers, i.e. Z or I = {…, −3, −2, −1, 0, 1, 2, 3, …}. Rational Numbers (Q) p The numbers which are in the form of , (where p, q ∈I, q q ≠ 0), are called as rational numbers. e.g. etc. Irrational Numbers (Q′ ) The numbers which are not rational, i.e. which cannot be p expressed in form or whose decimal part is q non-terminating, non-repeating but which may represent magnitude of physical quantities are called irrational numbers. e.g. 2, 51/3 , π e, etc. Real Numbers (R) The set of rational and irrational numbers is called a set of real numbers, i.e. N ⊂ W ⊂ Z ⊂ Q ⊂ R Ø ● The real number system is totally ordered for any two numbers a, b ∈R. We must say, either a< b or b < a or b = a. Chapter Snapshot ● The Real Number System ● Modulus of a Real Number ● Imaginary Number ● Complex Number ● Algebra of Complex Numbers ● Conjugateof a Complex Number ● Modulus of a Complex Number ● Argument (orAmplitude) of a Complex Number ● Various Forms of a Complex Number ● De-Moivre’sTheorem ● Roots of Unity ● GeometricalApplications of Complex Numbers ● Loci in Complex Plane ● Logarithmof Complex Numbers
  • 31. 4 120 ● All real numbers can be represented by points on a straight line. This line is called as number line. ● Division by zero is meaningless. ● Number zero is neither positive nor negative but it is an even number. ● Square of a real number is always non-negative. ● An integer is said to be even, if it is divisible by 2, otherwise it is an odd number. ● Number ‘0’ is an additive identity. ● The magnitude of a physical quantity may be expressed as a real number times, a standard unit. ● Number ‘1’ is multiplicative identity. ● A positive integer p is called prime, if its only divisors are ± 1 and ± p. ● Between two real numbers, there lie infinite real numbers. ● Infinity (∞) is the concept of the number greater than greatest you can imagine. It is not a number, it is just a concept, so we do not associate equality with it. X Example 1. The value of is (a)1 (b)2 (c)1050 (d)0 Sol. (d) As we know, ∞ is the number greater than greatest we imagine. Also, the value of 1 upon ∞ is tending to zero. Hence, . Intervals Let a, x, b are real numbers, so that x ∈[a b,] ⇒ a ≤ x ≤ b, [a, b] is known as the closed interval a, b. x ∈(a b, )⇒ a < x < b, (a, b) is known as the open interval a, b. x ∈(a b, ]⇒ a < x ≤ b, (a, b] is known as open, closed interval a b, . x ∈[a b,) ⇒ a ≤ x < b, [a, b) is known as closed, open interval a b, . X Example 2. If a = 3 +1, b = 2 2, then the value a of lies in the interval b (a) (−1 0, ) (b) (0,1] (c)[0.5,1.5] (d) (0,1) Sol. (b, c, d) Consider a = 3 + 1 = 0.966 b 2 2 Clearly, 0.966 ∈(0, 1), (0, 1] and [0.5, 1.5]. Modulus of a Real Number The modulus of a real number x is defined as follows: x, when x > 0 | |x 0, when x = 0 x, when x < 0 x − a, when x ≥ a Ø |x − a| (x − a), when x < a X Example 3. If f x( ) = |x − 2| + | |x + |x + 3|, then the value of f x( ) for x ≤ − 3 is (a)3x −1 (b) −3x +1 (c) x − 3 (d) −(3x +1) Sol. (d) |x − 2| = − (x − 2) = 2 − x for x ≤ − 3 | |x = − x for x ≤ − 3 and |x + 3| = − (x + 3) for x ≤ −3 ∴ f x( ) = 2 − x − x − x − 3 = − 3x − 1 = − (3x + 1) Imaginary Number Square root of a negative real number is imaginary number. While solving equation x2 +1 = 0, we get x = ± −1 which is imaginary. So, the quantity −1 is denoted by ‘i’ called ‘iota’. Thus, i = −1. e.g. −2, −3, −4, … may expressed as i 2, i 3, 2i, … . Integral powers of iota As we have seen i = −1, so i2 = −1, i3 = − i and i4 =1. Hence, n ∈ N, in = i, −1, −i, 1 attains four values according to the value of n, so i4n +1 = i i4n + 2 = −1 i4n + 3 = − i i4n or i4n + 4 =1 In other words, n (−1)n/2 , if n is an even integer i n −1 (−1) 2 i, if n is an odd integer
  • 32. 4 121 ( ) i in n + + ∑ 1 13 Ø ● i2 = −1× −1≠ 1 ● −a × −b ≠ ab, so for two real numbers a and b but a ⋅ b = a ⋅ b possible, if both a, b are non- negative. ● i is neither positive, zero nor negative. Due to this reason, order relations are not defined for imaginary numbers. ● The sum of any four consecutive powers of i is zero, i.e. i4n+1 + i4n+ 2 + i4n+3 + i4n+4= 0 X Example 4. The value of i2014 is (a) i (b) − i (c) 1 (d) −1 Sol. (d) Consider i 2014 = (i 2 1007) = −( 1)1007 = − 1 X Example 5. If a < 0, b > 0, then a ⋅ b is equal to (a) − |a b|⋅ (b) |a b i|⋅ ⋅ (c) |a b| (d) None of these Sol. (b) As we can only multiply the positive values in square root. ∴ a b = −| |a b, as a < 0 and b > 0 i.e. −1⋅ | |a b = i | |a b = | |a b⋅⋅ i X Example 6. The value of the sum , where i = −1 is n =1 (a) i (b) i −1 (c) − i (d)0 Sol. (b) Since, the sum of any four consecutive powers of i, is zero. 13 13 13 ∴ ∑(i n + i n + 1) = ∑ i n + ∑ i n + 1 …(i) n = 1 n = 1 n = 1 = (i + i 2 + i 3 + + i13 ) + (i 2 + i 3 + i 4 + + i14 ) = i − 1 [from Eq. (i)] X Example 7. The value of 2n (1 + i)2n + , n ∈I, is equal to 2n n (1 + i) 2 (a)0 (b)2 (c){1 + −( 1) }n ⋅ in (d) None of these Sol. (c) Here, 2n + (1 + i)2n X Example 8. If i = −1, then the number of values of in + i− n for different n ∈I is (a) 3 (b) 2 (c) 4 (d) 1 Sol. (a) As i n + i − n can be written as n 1 i 2n + 1 x = i + = n If n = 4, x = If n = 5, x = If n = 6, x = If n = 7, x = − 1 and so on. Which shows there exist three different solutions for n ∈ I. Complex Number The complex number z = a + ib = (a b, ) can be represented by a point P, whose coordinates are referred to rectangular axes XOX ′ and YOY ′, which are called real and imaginary axis respectively. The plane formed by rectangular axes is called argand plane or argand diagram or complex plane or Gaussian plane. (i) A number of the form z = x + iy = Re z + iIm z is called a complex number. (ii) Two complex numbers are said to be equal, ifand only if their real parts and imaginary parts are separately equal, i.e. a + ib = c + id ⇔ a = c and b = d. (iii) The complex numbers do not possess theproperty of order, i.e. + = + + + + + + = + = + = = − + = − + = ⋅ − + X′ X O Y′ Y Pab ( , ) b M a Realaxis + = + = + − =− + =
  • 33. 4 122 x + iy < (or) > c + id is not defined. (iv) A complex number z is purely real, if Im( )z = 0 and said to be purely imaginary, if Re( )z = 0. The complex number 0 = 0 + i. 0 is both purely real and purely imaginary. 1 X Example 9. If z = −( 5 )i i , then Im ( )z is 8 equal to (a)1 (b)0 (c) −1 (d) None of these Sol. (b) Consider z = −( 5 )i 1 i 8 Let us first express z in the format a + ib, then z = i 2 = − (−1) = + i0 ⇒ Im( )z = 0 X Example 10. If 4x + i(3x − y) = 3 + i(−6), where x and y are real numbers, then the value of x and y respectively are (a)3, 33 (b) (c) (d) None of these Sol. (c) We have, 4x + i(3x − y) = 3 + i(−6) …(i) Equating the real and imaginary parts of Eq. (i), we get 4x = 3 and 3x − y = − 6, which on solving simultaneously, we get x = 3 and y = Algebra of Complex Numbers Addition of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 + z2 = x1 + iy1 + x2 + iy2 = (x1 + x2 ) + i y( 1 + y2 ) ⇒ Re (z1 + z2 ) = Re (z1 ) + Re (z2 ) and Im (z1 + z2 ) = Im (z1 ) + Im (z2 ) Properties of Addition of Complex Numbers (a) z1 + z2 = z2 + z1 [commutative law] (b) z1 + (z2 + z3 ) = (z1 + z2 ) + z3 [associative law] (c) z + 0 = 0 + z (where, 0 = 0 + i0) [additive identity law] X Example 11. The value of 3(7 + i7) + i(7 + i7) is (a)15 + 27 i (b)14 + 28 i (c)14 − 28 i (d)14 + 23 i Sol. (b) We have, 3(7 + i7) + i(7 + i7) = 21 + 21i + 7 i + 7 i 2 = 21 + 28i − 7 [i 2 = − 1] = 14 + 28i Subtraction of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) = (x1 − x2 ) + i y( 1 − y2 ) ⇒ Re(z1 − z2 ) = Re(z1 ) − Re(z2 ) and Im (z1 − z2 ) = Im (z1 ) − Im (z2 ) Multiplication of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers, then z1 ⋅ z2 = (x1 + iy1 )(x2 + iy2 ) = (x x1 2 − y y12 ) + i x y( 12 + x y2 1 ) ⇒ z1 ⋅ z2 = [Re (z1 )Re (z2 ) − Im (z1 )Im (z2 )] + i[Re (z1 )Im (z2 ) + Re (z2 )Im (z1 )] Properties of Multiplication of Complex Numbers (a) z1 ⋅ z2 = z2 ⋅ z1 [commutative law] (b) (z1 ⋅ z2 )z3 = z1 (z2 ⋅ z3 ) [associative law] (c) If z1 ⋅ z2 =1 = z2 ⋅ z1, then z1 and z2 are multiplicative inverse of each other. Thevalueof 1 3 7 3 4 1 3 4 3 + + + + − − i i i is (a) 5 3 17 3 −i (b) 17 3 5 3 −i c) ( 17 3 5 3 +i (d) 17 5 4 3 −i + + + + − = + + + − + + + = − + + = + + + + − + =
  • 34. 4 123 (d) (i) z1 (z2 + z3 ) = z1 ⋅ z2 + z1 ⋅ z3 [left distribution law] (ii) (z2 + z3 ) z1 = z2 ⋅ z1 + z3 ⋅ z1 [right distribution law] X Example 13. The real values of x and y, if (1 + i x)− 2i (2 − 3i y)+ i + = i, are respectively (3 + i) (3 − i) (a) 3, −1 (b) 3, 1 (c) −3, 1 (d) −3, −1 Sol. (a) (1 + i x) − 2i + (2 − 3i y) + i = i (3 + i) (3 − i) ⇒ {(1 + i x) − 2 i}(3 − i ) + {(2 − 3 i y) + i}(3 + i ) = i(3 + i)(3 − i) ⇒ (1 + i )(3 − i x)− 2 (3i− i ) + (2 − 3 )(3i+ i y) + i(3 + i) = 10 i ⇒ (4 + 2 i ) x − 6 i − 2 + (9 − 7 )i y + 3 i − 1 = 10 i ⇒ (4x − 2 + 9y − 1) + i(2 x − 6 − 7 y + 3) = 10 i ⇒ (4x + 9y − 3) + i(2 x − 7 y − 3) = 10 i On equating real and imaginary parts on both sides, we get 4x + 9y = 3 …(i) and x − 7 y = 13 …(ii) On solving Eqs. (i) and (ii), we get x = 3, y = − 1 X Example 14. The multiplicative inverse of 4 − 3i is 4 3i 4 3i (a) − (b) + 25 25 25 25 4 3i (c) + (d) None of these 16 25 Sol. (b) Let z = 4 − 3i Then, its multiplicative inverse is 1 1 1 4 + 3i 4 + 3i = = × = z 4 − 3i 4 − 3i 4 + 3i 16 − 9i 2 [(a − b)(a + b) = a2 − b2] =[i 2 = − 1] = 4 3i 25 25 Division of Complex Numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 (≠ 0) be two complex numbers, then z1 x1 + iy1 = z2 x2 + iy2 1 = 2 2 [(x x1 2 + y y1 2 ) + i x y( 2 1 − x y1 2 )] x2 + y2 z1 X Example 15. The value of , where z1 = 2 + 3i z2 and z2 =1 + 2i, is 8 1 i (b) − i 5 5 i (d) None of these Sol. (b)z1 = 2 + 3i and z2 = 1 + 2i ∴ z2−1 = 1 = 1 − 2 i 1 + 2 i (1 + 2 i)(1 − 2 i) i z1 = z1 ⋅ z2 −1 = (2 + 3 i) 1 − 2 i Then, z2 5 5 2 + 6 + i 4 + 3 = 8 − 1 i 5 5 5 5 5 5 Aliter Here, x1 = 2, y1 = 3, x2 = 1 and y2 = 2. ∴ z1 = 1 {(x x12 + y y12 )+ i x y( 21 − x y12 )} z2 ={(2 × 1 + 3 × 2) + i(3 × 1 − 2 × 2)} = {(2 + 6) + i(3 − 4)} = 8 − 1 i 5 5 Conjugate of a Complex Number Geometrically, the conjugate of z is the reflection of point image of z in the real axis. ∴ z x iy = + and z x iy = − . X Y Realaxis z z O + + + = + x + +
  • 35. 4 124 e.g. If z = 3 + 4i, then z = 3 − 4 i. Properties of Conjugate z is the mirror image of z along real axis. (i) ( )z = z (ii) z = z ⇔ z is purely real (iii) z = − z ⇔ z is purely imaginary. (xii) If z = f z( 1 ), then z = f z( 1 ) X Example 16. If z1 = 9y2 − 4 −10 ix and z2 = 8y2 + 20i, where z1 = z2, then z = x + iy is equal to (a) −2 + 2 i (b) −2 ± 2 i (c) −2 ± i (d) None of the above Sol. (b) Given, z1 = z2 ⇒ 9y2 − 4 − 10 i x = 8y2 + 20 i ⇒ (y2 − 4) − 10i x( + 2) = 0 Since, complex number is zero. ⇒ y2 − 4 = 0 and x + 2 = 0 ∴ y = ± 2 and x = − 2 Thus, z = x + iy = − 2 ± 2 i X Example 17. If (1 + i z) = (1 − i z) , then z is (a) x(1 − i), x ∈R (b) x(1 + i), x ∈R x + (c) , x ∈R 1 + i (d) None of the above Sol. (a) (1 + i)(x + iy) = (1 − i)(x − iy) ⇒ (x − y) + i x( + y) = (x − y) − i x( + y) ⇒ x + y = 0 ∴ z = x − ix = x(1 − i) Modulus of a Complex Number Let z = x + iybe any complex number. Then, | |z = (x2 + y2 ) is called the modulus of the complex number z, where modulus | |z represents distance of z from origin. e.g. If z = 3 + 2i is a complex number, then | |z = 32 + 22 = 9 + 4 = 13 X Example 18. If z is a complex number satisfying the relation | z +1| = z + 2 1( + i), then z is Sol. (c) Let z = x + iy ∴ |x + iy + 1| = x + iy + 2 1( + i) ⇒ (x + 1)2 + y2 = (x + 2) + i y( + 2) ⇒ (x + 1)2 + y2 = x + 2 and y + 2 = 0 ⇒ (x + 1)2 + 4 = (x + 2)2 and y = − 2 ⇒ 2 x + 5 = 4x + 4 and y = − 2 ⇒ x =and y = − 2 ⇒(1 − 4 i) X Example 19. If z =1 + itan α, where π < α < , then | |z is equal to (a)secα (b) −secα (c)cosec α (d) None of the above Sol. (b) As z = 1 + i tan α ∴ | |z = 1 + tan2 α =|secα| ⇒ | |z = − sec α, as π < α < ( ) iv Re() Re() z z z z = = + 2 (v) Im()z z z i = − 2 ) vi ( z z z z 2 1 2 1 + + = (vii) z z z z 2 1 2 1 − − = ) viii ( zz zz 12 12 = ) ix ( z z z z 1 2 1 2 = ) ,( z2 0 ≠ (x) zz zz zz zz 12 12 12 12 2 2 + = = ) ) Re( Re( ) (xi z z z n n n = = () ( ) =
  • 36. 4 125 Argument (or Amplitude) of a Complex Number Argument of z = θ Argument of z is not unique. General value of argument of z is 2nπ + θ. Principal Value of Argument The value of θ of the argument, which satisfies the inequality −π < θ ≤ π is called the principal value of the argument. Principal values of the argument are θ, π − θ, −π + θ, − θ according as the complex number lies on the Ist, IInd, IIIrd or IVth quadrant. Here, θ = tan −1 | y| , where z = x + iy | |x X' X e.g. arg (1 + i) = tan −1 1 = π 1 4 arg (1 − i) = tan −1 −1 = −π 1 4 arg (−1 − i) = tan −1 − 1 = − π + π −1 4 arg (−1 + i) = tan −1 − 1 = π − 1 4 Ø ● Argument of 0 is not defined. ● If z1 = z2, then |z1| = |z2| and arg(z1) = arg(z2) ● Argument of purely imaginary number is or − ● Argument of purely real number is 0 or π . Example 20. The modulus and argument X of the 1 + 2i complex number is 1 − 3i (a) (b) (c) (d) None of the above [(a + b)(a − b) = a2 − b2] =[i 2 = − 1] = 1 + 9 10 2 ⇒ z = − + i ∴ | |z =a2 + b2 ] ∴ Now, tan θ = 2 1 2 θ = tan−1 Im( )z Re ( )z Y X θ O Realaxis z x iy =( + ) θ θ θ θ arg()= z θ arg()= – z θ π arg()=– z θ arg()=– + z π θ Y Y' = + − ∴ = + − × + + = + + + − 1 2 , 3 4 π 1 2 , − 3 4 π 1 2 , 3 4 π + + − − − + = + − = −+ − +  = + + = = =
  • 37. 4 126 π π = 1 ⇒ tan θ = 1 = tan ⇒ θ = Since, the real part of z is negative and imaginary part of z is positive, so the point lies in IInd quadrant. ∴ arg ( )z = π = θ = π − = 3π 4 Hence, modulus = and arg( )z = Work Book Exercise 4.1 1 If x − 3 + y − 3 = i, where x, y ∈R, then 3 + i 3 − i a x = 2 and y = − 8 b x = − 2 and y = 8 cx = − 2 and y = − 6 d x = 2 and y = 8 2 What is the real part of (1 + i )50? a 0b 225 c − 225 d − 250 3 The complex number z satisfies z + | z| = 2 + 8 i. Then, the value of| z| is a 10 b 13 c 17 d 23 4 If z + z3 = 0, then which of the following must be true on the complex plane? a Re ( )z < 0 b Re ( )z = 0 c Im ( )z = 0 d z4 = 1 5 The sequences S = i + 2i 2 + 3i 3 + upto 100 terms simplifies to, where i = −1 c (1 + i n)2 d None of these 7 Let i = −1. The product of the real part of the roots of z2 − z = 5 − 5 i is a − 25 b − 6 c − 5 d 25 8 Number of complex numbers z satisfying z3 = z is a 1 b 2 c 4 d 5 9Number of real solution of the equation, z3 + iz − 1 = 0 is a zero b one c two d three 10 The diagram shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of F, which is a A b B c C d D 11 Identify the incorrect statement. a No non-zero complex number z satisfies the equation z = − 4z b z = z implies that z is purely real c z = − z implies that z is purely imaginary d If z1, z2 are the roots of the quadratic equation az2 + bz + c = 0 such that Im(z z1 2) ≠ 0, then a, b, c must be real numbers 12 If z = (3 + 7 )(i p + iq), where p, q ∈l − {0}, is purely imaginary, then minimum value of| z|2 is a 0 b 58 c d 3364 13 Consider two complex numbers α and β as a + bi 2 a − bi 2 , where a, b ∈R and a − bi a + bi z − 1 β = , where| z| = 1, then z + 1 a bothα andβ are purely real b bothα andβ are purely imaginary c α is purely real andβ is purely imaginary d β is purely real andα is purely imaginary 14 If z is a complex number having the argument θ, 0 < θ < and satisfying the equality| z − 3 i| = 3. 6 − + − 6 − = + + − + − + + − − + + + − + − + + − − + + + − + + − + − − + + + + − + − + + − − + + O D F A C Imaginaryaxis Real axis B
  • 38. 4 127 Then, cot θ − is equal to z a 1 b −1 ci d − i Various Forms of a Complex Number Polar Form Let z = a + ib be any complex number, then by taking a = rcosθ and b = rsin θ We have, z = a + ib = r(cosθ + isin )θ (known as polar form) of z. X Example 21. The polar form of the complex −16 number is 1 + i 3 π 2π 2 (a) 4 cos + isin (b) cos + isin 3 3 3 3 2π 2 (c)8 cos + isin (d) None of these 3 3 By squaring and adding, we get 16 + 48 = r2(cos2 θ + sin2 θ) which gives, r2 = 64, i.e. r = 8 − 1 Hence, cosθ = 2 s inθ = π ⇒ θ = π − 3 3 Thus, the required polar form is 8 cos2 π + i sin 2 3 3 X Example 22. Let z and w be two non-zero complex numbers, such that | |z = |w| and arg( )z + arg( )w = π. Then, z is equal to (a) w (b) − w (c) − w (d) w Sol. (c) Given| |z =| |w = r and arg(w) = θ Also, arg ( )z + arg (w) = π ⇒ arg( )z = π − θ Now, z = r[cos(π − θ) + i sin(π − θ)] = r[−cosθ + i sin θ] = − r(cosθ + i sin )θ = − w where, r z = and θ= Principalvalueofargument Realaxis a θ r ( ) a,b b π =
  • 39. 4 128 − π Eulerian Form of a Complex Number We have, ei θ = cosθ + isin θ and e− i θ = cosθ − isin θ. These two are called Euler’s notations. Let z be any complex number, such that | |z = r and arg ( )z = θ. Then, in polar form, z can be written as z = r(cosθ + isin )θ Using Euler’s notations, we have z = rei θ This form of z is known as the Eulerian form. X Example 23. Express the following complex numbers in Eulerian form. (i)1 + i (ii) −2 + 2i Sol. (i) Given, z = 1 + i. Then, r =| |z = Let θ be the argument of z. Then, tan θ= =1⇒θ = i π So, Eulerian form of z is 2e 4 . (ii) Given, z = − 2 + 2 i Then, r =| |z = ( 2)−2 + 22 = 2 2 Let θ be the argument of z. Then, tan θ = = − 1 ⇒ θ = 3π i So, Eulerian form of z is 2 2e 4 . (iii)Given, z = − −1 i 3 Then, r =| |z = ( 1)− 2 + −( 3)2 = 2 Let θ be the argument of z. Then, tan θ = − 3 = 3 ⇒ θ = − 2 π −1 − i So, Eulerian form of z is 2e 3 . X Example 24. Complex numbers z1, z2, z3 are the vertices A B C, , respectively of an isosceles right angled triangle with right angle at C. Show that (z1 − z2 )2 = 2(z1 − z3 ) (z3 − z2 ). Sol. In an isosceles ∆ ABC, Bz( 2) AC = BC and BC perpendicular to AC. It means that AC is rotated through angleto occupy the position BC. xii. xiii. xiv. Ø ● If z is unimodular, then |z| = 1. Now, iff( )z is a unimodular, Cz( 3) Az( 1) then it is always be expressed as f( )z = cos θ + i sin θ, θ∈R.. We have, ● Square root of z = a + ib is given by z2 − z3 = e iπ /2 = i z | |z + a + i | |z − a , a =Re( )z z1 − z3 2 2 ⇒ z2 − z3 = + i z( 1 − z3 ) To find the square root of a − ib, replace i by − i in the above ⇒ z2 2 + z3 2 − 2 z z2 3 = − (z1 2 + z3 2 − 2 z z1 3 ) result. ⇒ z12 + z22 − 2 z z12 = 2 z z13 + 2 z z23 − 2 z z12 − 2 z32 ● If x, y ∈R = 2(z1 − z3 )(z3 − z2 ) ⇒ (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) | z1 + z2|2 = | z1|2 + | z2|2 z ⇔ 1 is purely imaginary. z2 |z1 + z2|2 + |z1 − z2|2 = 2{|z1|2 + |z2| }2 |az1 − bz2|2 + |bz1 + az2|2 = (a 2 + b2 )(| z1|2 + | z2| )2 , where a, b ∈R. π (iii) − − 3 1 i + = + + − + + = + − = − − + If x iy ib a c id = − − −
  • 40. 4 129 Properties of Modulus of Complex Number X Example 25. , then i.(x2 + y2 )2 is equal to 2 2 2 ii.(a) a− b 2 (b) a + b c2 − d 2 c2 + d 2 iii.a 2 + b2 (c) (d) None of these iv.c2 − d 2 Sol. (b) Given, x − iy = a − ib v.c − id 1/ 2 a − ib ⇒ x + i(− y) c − id vi.On taking modulus both sides, we get vii.|x + i(− y)| c − id a − ib 1 2/ 1 2/ ⇒ x2 + −(y)2 a − ib c − id iy| = x2 + viii. [|x + 1 2/ ⇒ x2 + y2 a − ib ix. c − id On squaring both sides, we get x.2 2 a − ib x + y c − id ⇒ x2 + y2 = |a − ib| |c − id| a2 + b2 y2 and|z n| =| |z n ]  z1 = z1 z2 z2 xi.⇒ (x2 + y2 ) =[|x − iy| = x2 + y2 ] c2 + d 2 On squaring both sides, we get 2 2 2 a2 + b2 (x + y ) = c2 + d 2 | |z ≥ 0 ⇒| |z = 0 iff z = 0 and | |z > 0 iff z ≠ 0 −| |z ≤ Re ( )z ≤ | |z and −| |z ≤ Im ( )z ≤ | |z | |z = | z| = −| z| = −| z| zz = | |z 2 | z z1 2| = | z1|| z2| In general, | z z z1 2 3 zn | = | z1|| z2|| z3|| zn | z1 = | z1 |, (z2 ≠ 0) z2 | z2| | z1 ± z2| ≤ | z1| + | z2| In general, | z1 ± z2 ± z3 ±± zn | ≤ | z1| + | z2| + | z3| ++ | zn | | z1 ± z2| ≥ || z1| − | z2|| | zn | = | zz |n || z1| − | z2|| ≤ | z1 + z2| ≤ | z1| + | z2| Thus, | z1| + | z2| is the greatest possible value of | z1 + z2| and || z1| − | z2|| is the least possible value of | z1 + z2|. | z z | (z z )(z z )
  • 41. 4 X 130 Example 26. For a complex number z, the minimum value of | |z + | z − 2| is (a) 1 (b) 2 (c) 3 (d) None of these Sol. (b) By using | |z1 + |z2|≥|z1 − z2| We have, | |z + |z − 2|≥|z − (z − 2)| ∴ | |z + |z − 2|≥ 2 X Example 27. If | z1 −1| <1, | z2 − 2| < 2 and | z3 − 3| < 3, then | z1 + z2 + z3| (a) is less than 6 (b) is more than 3 (c) is less than 12 (d) lies between 6 and 12 Sol. (c)|z1 + z2 + z3|=|(z1 − 1) + (z2 − 2) + (z3 − 3) + 6| ≤|z1 − 1| + |z2 − 2| + |z3 − 3| + 6 < 1 + 2 + 3 + 6 = 12 X Example 28. If α, β are two complex numbers, then |α |2 + | |β 2 is equal to (a) (|α+ β|2 − α− β|| )2 (b) (|α+ β|2 + α− β| | )2 (c)|α + β|2 + α − β| |2 (d) None of these Sol. (b)|α + β|2 = α + β α + β = α + β( )( ) ( )(α + β ) = αα + ββ + αβ + αβ =| |α 2 + | |β 2 + αβ + αβ …(i) |α − β|2 = α − β α − β = αα + ββ − αβ − αβ( )() =| |α 2 + | |β 2 − αβ − αβ …(ii) Adding Eqs. (i) and (ii), we get | |α 2 + | |β 2 = {|α + β|2 + |α − β| }2 X Example 29. If z1 and z2 are two complex numbers, such that | z1| <1 < | z2|, then prove that 1 − z z1 2 <1. z1 − z2 Sol. Given,| |z1 < 1and|z2|> 1 …(i) Then, to prove  z1 = | |z1 < 1 z1 − z2 z2 |z2| ⇒ |1 − z z12|<|z1 − z2| …(ii) On squaring both sides, we get (1 − z z12 )(1 − z z12 )< (z1 − z2 )(z1 − z2 ) (| |z 2 = zz) ⇒ 1 − z z1 2 − z z1 2 + z z z z1 1 2 2 < z z1 1 − z z1 2 − z z2 1 + z z2 2 ⇒ 1 + | | |z1 2 z2|2 <| |z1 2 + |z2|2 ⇒ 1 −| |z1 2 −|z2|2 + | | |z1 2 z2|2 < 0 ⇒ (1 −| | )(1z1 2 −|z2| )2 < 0 …(iii) which is true by Eq. (i) as| |z1 < 1 and|z2|> 1. ∴ (1 −| | )z1 2 > 0 and (1 −|z2| )2 < 0 ∴ Eq. (iii) is true, whenever Eq. (i) is true. 1 − z z12 < 1 Hence proved. ⇒ z1 − z2 X Example 30. The value of −8 − 6i is equal to (a)1 ± 3i (b) ± (1 − 3 )i (c) ± (1 + 3 )i (d) ± (3 − i) Then, ⇒ ⇒ X (4 + 3 (a) ±6 (d) ± 3 Sol. (a) We may write, (4 + 3 − 20) = (4 + 6i 1 − z z1 2
  • 42. 4 131 Let (4 + 3 − 20) Then, (4 + 6i ⇒ ⇒ ∴ On solving the equations x2 = y2 = 14 and x2 − y2 = 4, we get x2 = 9 and y2 = 5 ∴ x = ± 3 and y = ± 5 Since, xy > 0, it follows that x and y are of the same sign. = Properties of Arguments i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii. Ø Proper value of k must be chosen, so that RHS of (i), (ii), (iii) and (iv) lies in (−π, π). 3 + i X Example 32. If z =, then the 3 − i fundamental amplitude of z is π (a) − (b) (c)(d) None of these Sol. (b) amp( )z = amp 3 + i 3 − i = amp( 3 + i) − amp( 3 − i) = tan−1 1 − tan−1 −1 = π + π = π 3 3 6 6 3 X Example 33. The value of amp(iω) + amp(iω2 ), where i = −1 and ω = 3 1 = non-real, is arg (z z1 2 ) = arg (z1 ) + arg (z2 ) + 2kπ (k = 0 or 1 or −1) In general, arg(z z z1 2 3 zn ) = arg(z1 ) + arg(z2 ) + arg(z3 ) ++ arg(zn ) + 2kπ (k = 0 or 1 or −1) z1 arg arg(z1 )− arg(z2 )+ 2kπ z 2 (k = 0 or 1 or −1) z arg = 2arg( )z + 2kπ (k = 0 or 1 or −1) z arg(zn ) = narg( )z + 2kπ (k = 0 or 1 or −1) z2 z1 If arg , then arg 2kπ − θ, where z1 z2 k ∈I. 1 arg( )z = − arg( )z = arg 2 If arg( )z = 0 ⇒ z is real. arg(z z1 2 ) = arg(z1 ) − arg(z2 ) | z1 + z2| = | z1 − z2| ⇒ arg(z1 ) − arg(z2 ) = | z1 + z2| = | z1| + | z2| ⇒ arg(z1 ) = arg(z2 ) If | z1| ≤1, | z2| ≤1, then (a) | z1 − z2|2 ≤ (| z1| − | z2|)2 + [arg(z1 ) − arg(z2 )]2 (b) | z1 + z2|2 ≥ (| z1| + | z2|)2 − [arg(z1 ) − arg(z2 )]2 | z1 ± z2|2 = | z1|2 + | z2|2 ± 2| z1|| z2| cos (θ1 − θ2 ) z z1 2 + z z1 2 = 2| z1|| z2|cos (θ1 − θ2 ) where, θ1 = arg( z 1 ) and θ2 = arg( z 2 ) 3 π 6
  • 43. 4 X 132 (a) 0 ( b) (c) π (d) None of these Sol. (c) amp(iω +) amp(iω2 ) = amp(i 2 ⋅ω3 )= amp( 1)− = π X Example 34. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then z1 z2 arg arg equals to z4 z3 (a) 0 (b) (c) (d) π Sol. (a) We have, z2 = z1 and z4 = z3 ∴ z z12 =| |z1 2 and z z3 4 =|z3|2 z1 + arg z2 = arg z z1 2 Now, arg z4 z3 z z43 = arg || |zz13|22 = arg zz31 2 = 0 [argument of positive real number is zero] De-Moivre’s Theorem (a) If n ∈I (the set of integers), then (cosθ + isin )θ n = cos nθ + isin nθ. (b) If n ∈Q (the set of rational numbers), then cos nθ + isin nθ is one of the values of (cosθ + isin θ)n . Remark (i) The theorem is also true for (cos θ − isin θ), i.e. (cos θ − isin θ)n = cos nθ − isin nθ, because (cosθ − isin θ)n = [cos(−θ) + isin(−θ)]n = cos( (n −θ) + isin ( (n −θ)) = cos(−nθ) + isin(−nθ) = cos nθ − isin nθ (ii) = (cosθ + isin )θ −1 = cosθ − isin θ (iii) If z = (cos θ1 + isin θ1 ) (cos θ2 + isin θ2 ) (cos θn + isin θn ) Then, z = cos (θ1 + θ2 ++ θn ) + isin(θ1 + θ2 ++ θn ) (iv) If z = r(cosθ + isin )θ and n is a positive integer, then 1/n 1/n 2k 2k z = r cos n + isin n , where k = 0, 1, 2, …, (n −1) Ø ● (sin θ ± i cos θ)n ≠ sin nθ ± i cos nθ n ● (sinθ + i cosθ)n = cos π 2 − + i sin π 2 − nπ nπ cos n i sin − n 2 2 ● (cosθ + i sin φ)n ≠ cos nθ + i sin nφ X Example 35. If z . Then, arg ( )z is (a)2θ(b)2θ − π (c) π + 2θ (d) None of these Sol. (a) z = (cos θ + i sin θ)2 = cos2θ + i sin 2θ where, < 2θ < π Clearly, arg ( )z = 2θ  π < arg( )z < π 2 Example 36.If xr = cos 2r + isin 2r , then the value of x x x1 2 3 ∞ is (a) −1 (b) 1 (c) 0 (d) None of these Sol. (a)xr = cos πr + i sin 2 πr 2 ∴ x i
  • 44. 4 133 x i   (cosθ + isin )θ 4 X Example 37. is equal to (sin θ + icos )θ 5 (a)cosθ − isin θ (b)cos9θ − isin9θ (c)sin θ − icosθ (d)sin9θ − icos9θ (cos θ + i sin θ)4 (cos θ + i sin θ)4 Sol. (d) = 5 5 (sin θ + i cos θ) 5 1 i sin θ + cos i (cos θ + i sin θ)4 (cos θ + i sin θ)4 = = i(cos θ − i sin θ)5 i(cos θ + i sin θ)− 5 = 1 (cosθ + i sin )θ 9 = sin9θ − i cos9θ i 3 i 5 3 i 5 X Example 38. If z + − , 2 2 2 2 then (a) Re ( )z = 0 (b)Im ( )z = 0 (c) Re ( )z > 0, Im ( )z > 0(d) Re ( )z > 0, Im ( )z < 0 Sol. (b) Given, 5 5 3 i 3 i z − 2 2 2 2 1 The maximum and minimum values of| z + 1|, when| z + 3| ≤ 3 are a (5, 0) b (6, 0) c (7, 1) d (5, 1) 5 5 = cos 6 + i sin 6 + cos 6 − i sin 6 π π π π = cos+ i sin+ cos− i sin= 2cos Hence, Im( )z = 0. X Example 39. The product of all the values of π 3 4/ cos + isin is 3 3 (a) −1 (b) 1 (c) 3/2 (d) −1/2 3 4/ Sol. (b) Given, cos π + i sin = [cos π + i sin π]1 4/ 3 3 Since, the expression has only 4 different roots, therefore on putting n = 0, 1, 2, 3 in 2n 2 The region represented by inequalities arg z, Im (z) ≥ 1in the argand diagram is given by ⇒ x x x …= + + π π π π  + + = + + + π π π π   = − + − π π =− + = π π
  • 45. 4 X 134 Work Book Exercise 4.2 3 If 1 + 2i = r(cos θ + i sin θ), then 2 + i a r = 1,θ = tan−1 3 b r = 5,θ = tan−1 c r = 1,θ = tan−1 d None of these 4 If z1 and z2 are two non-zero complex numbers, such that| z1 + z2| =| z1| + | z2|, then arg (z1) − arg (z2 ) is equal to a − π b − c 0 d 5 If z = 1 − sin α + i cos α, where 0, , then 2 the modulus and the principal value of the argument of z are respectively a 2(1 − sin α), π + b 2(1 − sin α), π − 4 2 4 2 c 2(1 + sin α), π + d 2(1 + sin α), π − 4 2 4 2 π 8 6 The expression 1 + sin π 8 + i cos π 8 is equal to 1 + sin − i cos 8 8 a 1 b − 1 c i d − i 7 If zr = cos π + i sin π, r = 0, 1, 2, 3, 4, …, then z z z z z1 2 3 4 5 is equal to a − 1 b 0 c 1 d None of these 8 If zn , then lim (z1 ⋅ z2 ⋅ z3 zn ) is equal to n → ∞ ab cos π + i sin π 6 6 c d None of these 9 If z1, z2 are two complex numbers and a, b are two real numbers, then|az1 − bz2|2 + |bz1 + az2|2 is equal to a (a + b) [2 | z1|2 + | z2| ]2 b (a + b)[| z1|2 + | z2| ]2 c (a2 − b2)[| z1|2 + | z2| ]2 d (a2 + b2)[| z1|2+ | z2| ]2 10 All real numbers x, which satisfy the inequality 1 + 4i − 2− x ≤ 5, where i = −1,x ∈R are a [−2, ∞) b (−∞, 2] c [0, ∞) d[−2 0, ] 11 Square root of x2 + x 1 2 − 4 i x − 1 x − 6, where x ∈R is equal to a x − 1 + 2i b x − 1 − 2i x x c ± x + 1 + 2i d ± x + 1 − 2i x x 12 The minimum value of|1 + z| + |1 − z|, where z is a complex number, is 3 a 2 b 2 c 1 d 0 13 If arg(z + a) = and arg(z a) , then a z is independent of a b | |a =| z + a| c z = a, c is d z = a, c is b a ° 60 2 1 1 2 60° 2 1 1 2 Y X X Y c d 60° 1 2 1 2 60° 1 2 1 2 Y X Y X
  • 46. 4 135 14 Let z be a complex number satisfying the equation (z3 + 3)2 = − 16, then| z| has the value equal to a 51/2 b 51/3 c 52/3 d 5 d | z1|4 + | z2|4 =| z3|8 Roots of Unity Cube Roots of Unity Let x = 3 1 ⇒ x3 −1 = 0 ⇒ (x −1) (x2 + x +1) = 0 −1 + i 3 −1 − i 3 Therefore, x =1, , 2 2 If second root is represented by ω, then third root will be ω2. Therefore, cube roots of unity are 1, ω, ω2 and ω, ω2 are called the imaginary cube roots of unity. Properties i. ii. iii. iv. Important Identities (i) x2 + x +1 = (x − ω)(x − ω2 ) (ii) x2 − x +1 = (x + ω)(x + ω2 ) (iii) x2 + xy + y2 = (x − yω)(x − yω2 ) (iv) x2 − xy + y2 = (x + yω) + (x + yω2 ) (v) x2 + y2 = (x + iy x)( − iy) (vi) x3 + y3 = (x + y x)( + yω)(x + yω2 ) (vii) x3 − y3 = (x − y) = (x − yω)(x − yω2 ) (viii) x2 + y2 + z2 − xy − yz − zx = (x + yω + zω2 )(x + yω2 + zω) or (xω + yω2 + z)(xω2 + yω + z) or (xω + y + zω2 ) (xω2 + y + zω) (ix) x3 + y3 + z3 − 3xyz = (x + y + z)(x + ωy + ω2 y x)( + ωy2 + ωz) (x) Two points P z( 1 ) and Q z( 2 ) lie on the same side or opposite side of the line az + az + b accordingly as az1 + az1 + b and az2 + az2 + b have same sign or opposite sign. Example 40. If x2 − x +1 = 0, then the value of 5 2 n 1 ∑ x + n is n =1 x (a) 8 (b) 10 (c) 12 (d) None of the above Sol. (a) x2 − x + 1 = 0 ⇒ x = 1 ± 3i = −ω, − ω2 2 5 ∴ n∑= 1 x2n + 12n + 2 = x2 + x2 + x4 + 1 + 2 x6 + 1 + 2 0, if r is not a multiple of 3 1 + ωr + ω2 r 3, if r is not a multiple of 3 ω3 =1or ω3 r =1 ω3r +1 = ω, ω3r + 2 = ω2 It always forms an equilateral triangle. 15 = − + = − + = + − Σ = = + − Σ + = − x + x + x +
  • 47. 4 X 136 x8x8 + 2 x10 x110 + 2 = (ω2 + ω4 + ω6 + ω8 + ω10 ) + 1 2 + 1 4 + 1 6 + 1 8 + 1 10 + 10 ω ω ω ω ω = − 1 − 1 + 10 = 8 [x = − ω or − ω2 ] X 49 + 3 + 3 100 ⇒ 2 2 ∴ x + iy = −i ⇒ x = − ∴ k = − X Example 42. If z2 − z +1 = 0, then zn − z− n , where n is a multiple of 3, is (a) 2( 1)− n (b) 0 (c)(−1)n +1 (d) None of the above Sol. (b) As z2 − z + 1 = 0 ⇒ z = − ω, − ω2 ∴ zn − z− n = −( 1)nωn − −(1)−n ω− n = −( 1)n (ωn − ω−n ) , if n is a multiple of 3. = −(1)n (1 − 1) = 0 nth Roots of Unity Let z =11/n . Then, z = (cos0° + isin0° )1/n z = (cos2rπ + isin2rπ)1/n , r ∈Z 2rπ 2rπ ⇒ z = cos + isin , r = 0, 1, 2, …, (n −1) n n [using De-Moivre’s theorem] i r2 π ⇒ z = e n , r = 0, 1, 2, …, (n −1) ⇒ z = {ei2π/n }r , r = 0, 1, 2, …, (n −1) i2π z = α r , α = e n , r = 0, 1, 2, …, (n −1) Thus, nth roots of unity are 1, α, α 2 ,…, α n −1 , where i2π n = cos 2π + isin 2π α = e n n Properties of nth Roots of Unity i. ii. iii. iv. v. vi. Example 41. x = ky, then k is If 3 (x iy) i 2 2 and nth roots of unity form a GP with common i2π ratio e n . Sum of nth roots of unity is always zero. Sum of pth powers of nth roots of unity is zero, if p is not a multiple of n. Sum of pth powers of nth roots of unity is n, if p is a multiple of n. Product of nth roots of unity is (−1)n −1 . nth roots of unity lie on the unit circle | |z =1 and divide its circumference into n equal parts. ) (a − 1 3 (b) 3 (c) − 3 ) (d − 1 3 x + + = = − +
  • 48. 4 137 X Example 43. If z1, z2, z3, …, zn are nth, roots of unity, then for k =1, 2, …, n (a)| zk | = k z| k +1| (b)| zk +1| = k z| k | (c)| zk +1| = | zk | + | zk +1| (d)| zk | = | zk +1| Sol. (d) The nth roots of unity are given by i2π ( k − 1) zk = e, k = 1, 2, …, n ∴ |zk| e n = 1, ∀ k = 1, 2, …, n ⇒ |zk| =|zk + 1|, ∀ k = 1, 2, …, n X Example 44. If n is a positive integer greater than unity and z is a complex number satisfying the equation zn = (z +1)n , then (a) Re ( )z < 0 (b) Re ( )z > 0 (c) Re ( )z = 0 (d) None of these n Sol. (a) We have, zn = (1 + z)n ⇒ z = 1 X Example 45. If α is non- real and α = 5 1, then z + 1 1+ α + α 2 + α − 2 − α −1 z Sol. (a)α5 = 1 z z ⇒ x + = 0 [taking z = x + iy] ⇒ x =Re( )z < 0 ∴ Work Book Exercise 4.3 1 If ω is a non-real cube root of unity, then the 10 expression (1 − ω)(1 − ω2 )(1 + ω4 )(1 + ω8 ) is equal to a 0 b 3 c 1 d 2 2 x3m + x3n − 1 + x3r − 2, where m, n, r ∈N, is divisible by a x2 − x + 1 b x2 + x + 1 c x2 + x − 1 d None of these 11 3 If ω is a non-real cube root of unity, then is equal to 12 a − 1 b 2ω c 0 d − 2ω 4 If ( 3 + i )n = ( 3 − i )n, n ∈N, then least value of n is a 3 b 4 c 6 d None of these 13 5 If x3 − 1 = 0 has the non-real complex roots α β, , then the value of (1 + 2α + β)3 − (3 + 3α + 5β)3 is a − 7 b 6 c − 5 d 0 ⇒ = 11/ n z + 1 z ⇒ is nth root of unity. z + 1 the value of 2 (a) 4 (c) 1 is equal to (b) 2 (d) None of these ⇒ 1 ⇒ | | = 1 z + 1 |z + 1| ⇒ | |z =|z + 1| − π ∴ + + = + + + − − + − − α α α α α α α α + + + + − = α α α α α = − − − = = = × = 4 α α α α α = − ⇒
  • 49. 4 X 138 6 If ( 3 − i )n = 2n, n ∈I, the set of integers, then n is 14 a multiple of a 6b 10 c 9 d 12 7 If z is a complex number satisfying z4 + z3 + 2 z2 + z + 1 = 0, then| z| is equal to 15 a b c 1 None of these 8 If z is a non-real root of − 1, then z86 + z175 + z289 is equal to 16 a 0 b − 1 c 3 d 1 9 Non-real complex number z satisfying the equation z3 + 2 z2 + 3z + 2 = 0 are a c 2 2 If α is the non-real nth root of unity, then 1 + 3α + 5α 2 + + (2n − 1) α n − 1 is equal to n 1 − α d None of these −1∞ is equal to, where ω is the imaginary cube of root of unity and i = − 1. a ω orω2 b − ω or − ω2 c 1 + i or 1 − i d −1 + i or −1 − i If α = e i 2π / n, then (11 − α )(11 − α 2 ) (11 − α n − 1) is equal to 1 bc 11n − 1 − 1 d 11n − 1 − 1 a 11n − 10 11 The complex number w satisfying the equation ω3 = 8 i and lying in the IInd quadrant on the complex plane is 3 1 a − 3 + i b − + i 2 2 c −2 3 + i d − 3 + 2 i Let z be a complex number satisfying the equation z6 + z3 + 1 = 0. If this equation has a root re i θ with 90° < θ < 180°, then the value of θ is a 100° b 110° c 160° d 170° If ω is an imaginary cube root of unity, then the value of (p + q)3 + (pω + qω2 3) + (pω2 + qω)3 is a p3 + q 3 b 3(p3 + q 3) c 3(p3 + q 3) − pq p( + q ) d 3(p3 + q 3) + pq p( + q ) If z2 − z + 1 = 0, then the value of 1 2 2 1 2 3 1 2 z + z z + 2 z+ z3 z +  z24 + 1 24 is equal to z a 24 b 32 c 48 d None of these − ± − + − − − + − − −α −α − − − −
  • 50. 4 139 27 If p = a + bω + cω2 q = b + cω + aω2 and r = c + aω + bω2, where a, b, c ≠ 0 and ω is the complex cube root of unity, then a p + q + r = a + b + c b p2 + q 2 + r2 = a2 + b2 + c2 c p2 + q 2 + r2 = − 2(pq + qr + rp) d None of the above 18 If a and b are imaginary cube of unity, then α n + βn is equal to roots a 2cosb cos c 2i sind i sin 19 If the six solutions of x6 = − 64 are written in the form a + bi, where a and b are real, then the product of those solutions with a > 0 is a 4 b 8 c 16 d 64 20 If cos θ + i sin θ is a root of the equation xn + a x1 n − 1 + a x2 n − 2 + + an − 1x + an = 0, then n the value of ∑ ar cos r θ is r = 1 a 0 b 1 c − 1 d None of the above 21 If ω is a complex nth root of unity, then n ∑ (ar + b)ωr − 1 is equal to r = 1 a n n( + 1)a b nb 22 If α, β respectively are the fifth and fourth non-real roots of unity, then the value of (1 + α )(1 + β)(1 + α 2 )(1 + β2 )(1 + β3 )(1 + α 3 ) is a 0 b (1 + α + α2)(1 − β2) c(1 + α)(1 + β + β2) d 1 2 − n na 23 When the polynomial 5x3 + Mx + N is divided by x2 + x + 1, the remainder is 0. The value of (M + N) is equal to a −3 b 05 c −5 d 15 24 If z and w are two complex numbers simultaneously satisfying the equations, z3 + w5 = 0 and z2 ⋅ w 4 = 1, then a z and w both are purely real b z is purely real and w is purely imaginary c w is purely real and z is purely imaginary dz and w both are imaginary 25 Number of ordered pairs (z, ω) of the complex numbers z and ω satisfying the system of equations, z3 + ω7 = 0 and z5 ⋅ω11 = 1is a 7 b 5 c 3 d 2 26 If 1, z 1, z 2, z 3, …, z n − 1 is the nth roots of unity and w is a non-real complex cube root of unity, n − 1 then the product of ∏ (ω − zr ) is cannot be r = 1 equal to a 0 b 1 c −1 d 1 + ω 27. If Zr , r = 1, 2, 3, …, 50 are the roots of the equation, then the value of cd None of these n − 1 r = 0 is r = 1 r a − 85 b− 25 c 25 d 75 π π π π
  • 51. 4 140 Z − 1 X Example 47. If A, B, C are three points in the argand plane representing the complex numbers z1, z2, z3 such that z1 = , where λ ∈R, then the distance of A from the line BC is (a) λ (b) (c) 1 (d) 0 Sol. (d) As z1 = which shows z1 divides z2, z3 in the ratio of 1: λ. Thus, the points are collinear. ∴Distance of A from line BC is zero. X Example 48. Find the relation, if z1, z2, z3, z4 are the affixes of the vertices of a parallelogram taken in order. Sol. As the diagonals of a parallelogram bisect each other, therefore affix of the mid-point of AC is same as the affix of the mid-point of BD. z1 + z3 = z2 + z4 i.e. 2 2 ⇒ z1 + z3 = z2 + z4 Equation of the Straight Line Equation of the Line Passing through the Points z1 and z2 Let z be any point on the line joining z1 and z2, then z − z1 arg = π or 0 z2 − z1 z − z1 ⇒ must be real. z2 − z1 ⇒ z z( 1 − z2) − z z( 1 − z2 ) + z z1 2 − z z2 1 = 0 General Equation of a Line az + az + b = 0, represents a straight line, where b is a real number and a is a complex number. Geometrical Applications of Complex Numbers Basic Concepts in Geometry Distance formula The distance between two points P z( 1 ) and Q z( 2 ) is given by PQ = | z2 − z1| = | affixof Q − affix of P | Qz( 2) Pz( 1) For any complex number z, | z | = | z − 0 | = | z − (0 + i0)| Thus, modulus of a complex number z represented by a point in the argand plane is its distance from origin. X Example 46. Length of the line segment joining the points −1 − i and 2 + 3i is (a) −5 (b) 15 (c) 5 (d) 25 Sol. (c) Let z1 = − 1 − i and z2 = 2 + 3 i Then, required distance = | z2 − z1 | = | 2 + 3i + 1 + i | = 5 Section formulae If R z( ) divides the line segment joining P z( 1 ) andQ z( 2 ) in the ratio m1 :m2 (m1, m2 > 0), then m z + m z (i) For internal division, z = 1 2 2 1 m1 + m2 m z − m z (ii) For external division, z = 1 2 2 1 m1 − m2 If R z( ) is the mid-point of PQ, then affix of R is z1 + z2 2 ∴ Requiredequationis z z z z z z z z − − = − − 1 2 1 1 2 1 ⇒ z z z z z z 1 1 1 0 1 1 2 2 =
  • 52. 4 141 1 2 2 1 Parametric Equation of a Line z = z1 + t z( 2 − z1 ), where t is real parameter = (1 − t z) 1 + t z2, represents the complete line through z1and z2. X Example 49. Find the general equation of line joining the points z1 = (1 + i) and z2 = (1 − i). Sol. Clearly, the equation of a line is given by z z( 1 − z2 ) − z z( 1 − z2 ) + z z1 2 − z z2 1 = 0 where, z1 = 1 + i and z2 = i − i On substituting the values of z1 and z2, we get z(1 − i − 1 − i) − z(1 + i − 1 + i) + (1 + i)(1 + i) − (1 − i)(1 − i) = 0 ⇒ z( − 2i) − z(2i) + (1 − 1 + 2i) − (1 − 1 − 2i) = 0 ⇒ − 2iz − 2iz + 4i = 0 ⇒ z + z − 2 = 0, which is the required equation. Condition of Collinearity Three points z1, z2 and z3 are collinear, if 1 = 0 X Example 50. If z1, z2, z3 are three complex numbers such that 5z1 −13z2 + 8z3 = 0, then prove that ` 1 = 0. Sol. 5z1 − 13z2 + 8z3 = 0 ⇒ = z2 ⇒ z1, z2 and z3 are collinear. z1 z1 1 ⇒ z2z2 1 = 0 [condition of collinear points] z3 z3 1 Hence proved. Length of Perpendicular The length of perpendicular from a point z1 to az + az + b = 0 is given by az1 + az1 + b 2 a X Example 51. The length of perpendicular from P(2 − 3 )i to the line (3 + 4 )i z + (3 − 4 )i z + 9 = 0 is equal to (a)9 (b) (c) (d) None of these Sol. (c) Let PM be the required length, then PM = |(2 − 3i)(3 + 4i) + (3 − 4i)(2 + 3i) + 9| = 45 9 10 2 Slope of a Line Slope of the Line Segment Joining Two Points If A, B represent complex numbers z1, z2 in the argand plane, then the complex slope of AB is defined by z1 − z2 z − z Re (z − z ) and real slope is defined by . Im (z2 − z1 ) Slope of Line az + az + b = 0 The complex slope of the line −a −Coefficient of z az + az + b = 0 is = a Coefficient of z and real slope of the line az + az + b = 0 is Re (a) −i a( + a) – = Im ( )a (a − a) Ø ● If w1 and w2 are the complex slope of two lines on the argand plane, then the lines are (a)perpendicular, if w1 + w2 = 0 (b)parallel, if w1 = w2 ● The equation of a line parallel to the line az + az + b = 0 is az + az + λ = 0, where λ ∈R . ● The equation of a line perpendicular to the line az + az + b = 0 is az − az + λ =i 0, where λ ∈R . z z z z z z 1 1 2 2 3 3 1 1 z z z z z z 1 1 2 2 3 3 1 1 − =
  • 53. 4 142 X Example 52. If a point z1 is the reflection of a point z2 through the line b z + bz = c b, ≠ 0 in the argand plane, then bz2 + b z1 is equal to (a) 4c (b)2c (c) c (d) None of these Sol. (c) If P (z1)is the reflection of Q z( 2 )through the line bz + bz = c in the argand plane. Then, R z1 + z2 lies on the line. 2 b z1 + z2 + b z1 + z2 = c 2 2 ⇒ b z1 + b z1 + b z2 + b z2 = 2c …(i) Since, PQ is perpendicular to the line bz + bz = c. Therefore, Slope of PQ + Slope of the line = 0 z2 − z1 b = 0 ⇒ z2 − z1 b ⇒ b (z2 − z1) − b ( z2 − z1) = 0 ⇒ b z2 − bz1 − bz2 + bz1 = 0 …(ii) Adding Eqs. (i) and (ii), we get 2(bz1 + bz2 ) = 2c ⇒ bz1 + bz2 = c Concept of Rotation In this section, we shall learn about the effect of multiplication of a complex number by eiα which will also be interpreted geometrically. Complex Number as a Rotating Arrow in the Argand Plane 1. Let z = r (cosθ + isin θ) = reiθ be a complex number, represented by a point P in the argand plane. Then, OP = r and ∠XOP = θ Now, zeiα = reiθ ⋅ eiα = rei (θ + α) This shows that zeiα is the complex number whose modulus is r and argument θ + α. Clearly, zeiα is represented by a point Q in the argand plane such that OQ = r and ∠XOQ = θ + α. In other words, to obtain the point representing zeiα , we rotate OP through angle α in anti-clockwise sense. Thus, multiplication by eiα to z rotates the vector OP in anti-clockwise sense through an angle α and vice-versa. Similarly, multiplication of z with e− αi will rotate the vector OP in clockwise sense. Remark Let z1 and z2 be two complex numbers represented by points P and Q in the argand plane, such that X¢X ∠POQ = θ. Then, z e1 iθ is vector of magnitude | z1| = OP along OQ and z e1 iθ is a unit vector | z1| z eiθ along OQ. Consequently, | z |⋅ 1 2 is a vector of | z1| magnitude | z2| = OQ along OQ. i.e. z= | z2| z ei θ ⇒ z = z2 ⋅ z ei θ 2 1 21 Pz ( ) 1 z+z 1 2 2 R bz+bz=c Qz ( ) 2 ( ) Y O Qze i ( ) α Pz () Y′ X′ X x α θ Y O Q (z ) 2 P(z ) 1 Y¢ q
  • 54. 4 143 | z1| z1 X Example 53. The point represented by the Then, z3 − z1 = OQ (cosα + isin α) complex number 2 − i is origin z2 − z1 OP rotated about CA iα through an anglein the clockwise direction, the = e BA new position of point is = | z 3 − z 1 | iα Sol. (b) Here, z = 2 − i Let z1 be the required complex number. ∴z1 = (2 − i) cos 2 + i sin 2 = (2 − i) cos π – i sin 2 2 = (2 − i)(0 − i) = − (2i − i 2 ) = – 2i − 1 ∆OPQ and ∆ABC are congruent, X Example 54. A particle P starts from the point OQ CA z0 =1 +2i, where i = −1. It moves first horizontally ∴ OP = BA away from origin by 5 units and then vertically z3 − z1 away from origin by 3 units to reach a point z1. or amp From z1, the particle moves 2 units in the z2 − z1 direction of the vector  i +  j and then it moves X Example 55. A man walks a distance of through an angle π/2 in anti-clockwise direction on 3 units from the origin towards the North-East a circle with centre at origin to reach a point z2. (N 45° E) direction. From there, he walks a The point z 2 is given by distance of 4 units towards the North- West (a)6 + 7i (b) −7 +6i (N 45°W) direction to reach a point P. Then, the (c) 7 + 6i (d) −6 + 7i position of P in the argand plane is (a)1 + 2i (c)2 + i (b) −1 − 2i (d) −1 + 2i Y e | z − z | 2 1  = −π z z Q ) – ( 1 3 z P z ( – ) 1 2 Az ) ( 1 Bz ( ) 2 Cz ) ( 3 X O α
  • 55. 4 144 Sol. (d) Imaginary axis (a)3eiπ/4 + 4i (b) (3 − 4i e) iπ/4 (c) (4 + 3i e) iπ/4 (d) (3 + 4i e) iπ/4 associated with A is 3 eiπ / 4. If z is the complex number associated with P, then 2 2 sin45°) = (7, 6) = 7 + 6i By rotation about (0, 0), z2 = eiπ / 2 ⇒ z2 = z2 eiπ2 z2′ z2 = (7 + 6i) cos π + i sin 2 2 = (7 + 6 )( )ii = − 6 + 7i 2. Let z1, z2 and z3 be the vertices of a ∆ABC described in anti- clockwise sense. Draw OP and OQ parallel and equal to AB and AC, respectively. Then, point P is z2 − z1 and Q is z3 − z1. If OP is rotated through ∠α in anti-clockwise sense, it ⇒ coincides with OQ. ⇒ z = (3 + 4i e) iπ / 4 X Example 56. The complex numbers z1, z2 and z1 − z3 1 − i 3 z3 satisfying =are the vertices of a z2 − z3 2 triangle which is (a) of area zero (b) right angled isosceles
  • 56. 4 145 1 2 = z1 − z3 z1 − z3 3 Hence, the triangle is an equilateral. Area of Triangle (i) Area of the triangle with vertices z1, z2 and z3 is (z2 − z3 )| z1 |2 sq unit. 4iz1 (ii) The area of triangle whose vertices are z, izand z + iz is | |z 2 . (iii) The area of triangle whose vertices are z, ωz and 3 2 z + ωz is| z | . 4 X Example 57. If the area of the triangle on the complex plane formed by the points z, iz and z + iz is 50 sq units, then | z | is (a) 5 (b) 10 (c) 15 (d) None of these Sol. (b) We know that, the area of the triangle formed by + iz is 2. the points z, iz and z ∴ | z| = 50 ⇒ | z| = 10 X Example 58. If the area of the triangle on the complex plane formed by complex numbers z, ωz and z + ωz is 4 3 sq units, then | z | is (a) 4 (b) 2 (c) 6 (d) 3 Sol. (a) The area of the triangle formed by z, ω z and z + ωz 3 2 is| |z . 4 ∴ | |z 2 = 4 3 4 ⇒ | |z = 4 Applications of Triangle 1. Centroid The centroid of the triangle (in the argand plane) formed by z1, z2 and z3 is given by 1 (z1 + z2 + z3 ) 3 2. Incentre The incentre of the triangle (in the argand plane), formed by z1, z2 and z3 is az1 + bz2 + cz3 , where a + b + c a = | z2 − z3|, b = | z3 − z1|, c = | z1 − z2| 3. Excentres The excentres of the triangle (in the argand plane), for meet by z1, z2 and z3 are given by −az1 + bz2 + cz3 (i) I1 = −a + b + c az1 − bz2 + cz3 (ii) I2 = a − b + c az1 + bz2 − cz3 (iii) I3 = a + b − c where, a = | z2 − z3|, b = | z3 − z1| and c = | z1 − z2| 4. Circumcentre The circumcentre of the triangle (in the argand plane), formed by z1, z2, z3 is given by 5. Orthocentre The orthocentre of the triangle (in the argand plane), formed by z1, z2, z3 is given by X Example 59. If z1, z2 and z3are affixes of the vertices A, B and C, respectively of a ∆ABC having centroid at G such that z = 0 is the mid-point of AG, then (a) z1 + z2 + z3 = 0 (b) z1 + 4z2 + z3 = 0 (c) z1 + z2 + 4z3 = 0 (d) 4z1 + z2 + z3 = 0 Sol. (d) The affix of G is z1 + z2 + z3 . Since, z = 0 is the 3
  • 57. 4 146 mid-point of AG. Therefore, affix of the mid-point of AG is 0. z1 + z2 + z3 + z1 ⇒ 3 = 0 ⇒ 4z1 + z2 + z3 = 0 1 + 1 X Example 60. The centre of a square ABCD is at the origin and point A is represented by z1. The centroid of ∆BCD is represented by z1 z1 (a) (b) − 3 3 iz1 iz1 (c) (d) − 3 3 Sol. (b) The affixes of the vertices B, C and D are iz1 − z1 and − iz1, respectively. Therefore, the affix of the centroid of ∆BCD is iz1 – z1 − iz1 = − z1 3 3 X Example 61. Let P e( i θ1 ),Q e( i θ2 ) and R e( i θ3 ) be the vertices of ∆PQR in the argand plane. Then, the orthocentre of the ∆PQR is (a) ei (θ1 + θ2 + θ3) (b) ei (θ + θ + θ12 3) (c) ei θ1 + ei θ2 + ei θ3 (d) None of these Sol. (c) We have, |ei θ1| =|ei θ2| =|ei θ3| = 1 ⇒ OP = OQ = OR = 1, where O is the origin. ⇒ Origin O is the circumcentre of ∆PQR. The affix of the centroid is (ei θ1 + ei θ2 + ei θ3 ) Let z be the affix of the orthocentre. Since, centroid divides the segment joining circumcentre and orthocentre in the ratio 1: 2. ⇒ z = ei θ1 + ei θ2 + ei θ3 6. Equilateral Triangle (i) The triangle whose vertices are the pointsz1, z2 and z3 on the argand plane, is an equilateral triangle, if z12 + z22 + z32 = z z1 2 + z z2 3 + z z3 1 1 1 1 or + + = 0 z1 − z2 z2 − z3 z3 − z1 (ii) If the complex numbers z1, z2 and z3 are the vertices of an equilateral triangle and z0 is the circumcentre of the triangle, then z12 + z22 + z32 = 3z02. 7. Isosceles Triangle (i) If z1, z2 and z3 are the vertices of a right angled isosceles triangle, then (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) (ii) If z1, z2 and z3 are the vertices of an isosceles triangle, right angled at z2, then z12 + z22 + z32 = 2z2 (z1 + z3 ) X Example 62. Prove that the complex numbers z1, z2and the origin form an equilateral triangle only, if z12 + z22 − z z1 2 = 0. Sol. If z1, z2 and z3 form an equilateral triangle, then ⇒ z z z z z z z z z ⇒ z z z z z z ⇒ z + z = z z⇒z z z z Hence proved. X Example 63. Let z1and z2be two complex z1 z2 numbers such that + =1, then z2 z1 (a) z1, z2are collinear (b) z1, z2 and the origin form a right angled triangle (c) z1, z2 and the origin form an equilateral triangle (d) None of the above Sol. (c) We have, z1 + z2 = 1⇒ z12 + z22 = z z1 2 z2 z1 ⇒ z12 + z22 + z32 = z z12 + z z1 3 + z z2 3, where z3 = 0 ⇒ z1, z2 and the origin form an equilateral triangle. O D iz (– ) 1 C z (– ) 1 Biz ( ) 1 Az ( ) 1
  • 58. 4 147 Circle Equation of a Circle (i) The equation of a circle whose centre is at pointhaving affix z0 and radius r, is | z − z0 | = r. (ii) If the centre of the circle is at origin and radius r, then its equation is | z | = r. (iii) | z − z0 | < r represents interior of a circle | z − z0| = r and | z − z0 | > r represents exterior of the circle | z − z0| = r. (iv) General equation of a circle The general equation of the circle is zz + az + az + b = 0, where a is complex number and b ∈R. ∴ Centre and radius are −a and | a |2 − b, respectively. (v) Equation of circle in diametric form If end points of diameter represented by A z( 1 )and B z( 2 ) and P z( ) is any point on the circle, then (z − z1 )(z − z2 ) + (z − z2 )(z − z1 ) = 0 which is required equation of circle in diametric form. X Example 64. A circle whose radius is r and centre z0, then the equation of the circle is (a) zz − zz0 − zz0 + z z0 0 = r2 (b) zz + zz0 − zz0 + z z0 0 = r2 (c) zz − zz0 + zz0 − z z0 0 = r2 (d) None of the above Sol. (a) Equation of circle|z − z0|2 = r2 ⇒ (z − z0 )(z − z0 ) = r2 ⇒(z − z0 )(z − z0 ) = r2 zz − zz0 − zz0 + z z0 0 = r2 . X Example 65. The set of values of k for which the equation zz + ( 3− + 4 )i z − (3 + 4 )i z + k = 0 represents a circle is (a) (−∞ , 25) (b) (25, ∞) (c) (5, ∞) (d) (−∞, 5) Sol. (a) We have, zz + −( 3 + 4 )i z − (3 + 4 )i z + k = 0 This equation represents a circle with centre a + (3 − 4 )i and Radius = ( 3−− 4 )i 2 − k = 25 − k For circle to exist, we must have 25 − k > 0 ⇒ k < 25 Hence, the given equation will represent a circle if k < 25. Loci in Complex Plane If z is a variable point and z1, z2 are two fixed points in the argand plane, then (i) | z − z1| = | z − z2| ⇒ Locus of z is the perpendicular bisector of the line segment joining z1and z2. (ii) | z − z1| + | z − z2| = k, if | k | > | z1 − z2| ⇒ Locus of z is an ellipse. (iii) | z − z1| + | z − z2| = | z1 − z2| ⇒Locus of z is the line segment joining z1 and z2 (iv) | z − z1| − | z − z2| = | z1 − z2| ⇒ Locus of z is a straight line joining z1and z2 but z does not lie between z1 and z2. (v) | z − z1| – | z − z2| = k, where k <| z1 − z2| ⇒ Locus of z is a hyperbola. (vi) | z − z1|2 + | z − z2|2 = | z1 − z2|2 ⇒ Locus of z is a circle with z1 and z2 as the extremities of diameter. (vii) | z − z1| = k z| − z2|, (k ≠1) ⇒ Locus of z is a circle. z − z1 (viii) arg = α (fixed) z − z2 ⇒ Locus of z is a segment of circle. z − z1 (ix) arg = ± π / 2 z − z2 ⇒ Locus of z is a circle with z1and z2 as the vertices of diameter. z − z1 (x) arg = 0 or π z − z2 ⇒ Locus of z is a straight line passing through z1 and z2. X Example 66. The complex numbers z = x + iy z − 5i which satisfy the equation=1, lie on z + 5i (a) the X-axis (b) the straight line y = 5 r Cz ( ) 0 Pz ()
  • 59. 4 148 (c) a circle passing through the origin (d) None of the above Sol. (a) Given, z − 5i = 1 ⇒ |z − 5i| =|z + 5i| z + 5i [if|z − z1| =|z – z2|, then it is a perpendicular bisector of z1 and z2] ∴ Perpendicular bisector of (0, 5) and (0, − 5) is X-axis. X Example 67. The equation | z − i| + | z + i| = k k,> 0 can represent an ellipse, if k 2 is (a) <1 (b) < 2 (c) > 4 (d) None of these Sol. (c)| z − z1| + | z – z2 | = k represents ellipse, if | k|>|z1 − z2| Thus,| z − i | + | z − i | = k represents ellipse, if | k|>|i + i| or | k|>|2i| ∴ | k|> 2 or k2 > 4 X Example 68. The equation | z + i| − | z − i| = k represents a hyperbola if (a) −2 < k < 2 (b) k >2 (c)0 < k < 2 (d) None of these Sol. (a)|z − z1| −|z − z2| = k, represents hyperbola, if| k|<|z1 − z2| Thus, |z + i| −|z − i| = k, represents hyperbola, if | k|<| i + i | or| k|< 2 ⇒ −2 < k < 2 X Example 69. If z =1 − t + i t 2 + t + 2, where t is a real parameter. The locus of z in the argand plane is (a) a hyperbola (b) an ellipse (c) a straight line (d) None of these Sol. (a) x + iy = 1 − t + i ⇒ x = 1 − t , y = Eliminating t, y2 = t 2 + t + 2 = (1 − x)2 + 1 − x + 2 2 = x − 3 + or y2 − x − 7 X 2 4 which is a hyperbola. X Example 70. Identify the locus of z, if ⇒ | z − a|2 = r2 ⇒ | z − a| = r X Hence, locus of z is circle having centre a and radius r. X Example 71. If the equation | z − z1|2 +| z − z2|2 = k represents the equation of a circle, where z1 = 2 + 3i, z2 = 4 +3i are the extremities of a diameter, then the value of k is (a) (b) 4 (c) 2 (d) None of these Sol. (b) As z1 and z2 are the extremities of diameter. ⇒ |z − z1|2 + |z − z2|2 =|z1 − z2|2 ⇒ k =|z1 − z2|2 =|2 + 3i − 4 − 3i|2 =| − 2|2 = 4 X X Example 72. If | z +1| = 2 | z – |1, then the locus described by the point z in the argand diagram is a (a) straight line (b) circle (c) parabola (d) None of these Sol. (b)| z + 1| = 2 |z − 1| Putting z = x + iy ⇒ x + iy + 1 = 2|x + iy − 1| ⇒ |(x + 1) + iy| = 2 |(x − 1) + iy| ⇒ (x + 1)2 + y2 = 2[(x − 1)2 + y2 ] ⇒ x2 + y2 − 6x + 1 = 0 which is the equation of a circle. Y (0, 5) (0,– 5) Y¢ X¢ X O z a r a z = + − > 2 0 , . = + − ⇒ − = − ⇒ − − = + + + + =
  • 60. 4 149 π + − π ≤ ≤ z −1 X Example 73. The locus of z given by=1 z − i is (a) a circle (b) an ellipse (c) a straight line (d) a parabola ⇒ 2 x = 2 y or x − y = 0 which is the equation of a straight line. Example 74. If z = x + iy and | z − 2 + i| = | z − 3 − i|, then locus of z is (a)2x + 4y − 5 = 0 (b)2x − 4y − 5 = 0 (c) x + 2y = 0 (d) x − 2y + 5 = 0 Sol. (a) |z − 2 + i| =|z − 3 − i| ⇒ |(x − 2) + i (y + 1)| =|(x − 3) + i (y − 1)| ⇒ (x − 2)2 + (y + 1)2 = (x − 3)2 + (y − 1)2 ⇒ x2 + 4 − 4x + y2 + 1 + 2 y = x2 + 9 − 6x + y2 + 1 − 2 y ⇒ 2 x + 4y − 5 = 0 Example 75. If z = z0 + A z( − z0 ), where A is a constant, then prove that locus of z is a straight line. Sol. z = z0 + A (z − z0 ) ⇒ Az − z − Az0 + z0 = 0 …(i) ⇒ A z− z − Az0 + z0 = 0 …(ii) Adding Eqs. (i) and (ii), we get (A − 1) z + (A − 1) z − (Az0 + Az0 ) + z0 + z0 = 0 This is of the form az + az + b = 0, where a = A − 1 and b = − (Az0 + Az0 ) + z0 + z0 ∈R Hence, locus of z is a straight line. Example 76. Plot the region represented by π z +1 2π ≤ arg in the argand plane. 3 z −1 3 Sol. Let us take z + 1 2 π z arg , clearly z − 1 3 lies on the minor arc of the circle passing through (1, 0) and (−1 0, ). z + 1 π Similarly, arg z − 1 3 means that z is lying on the major arc of the circle passing through (1,0) and (−1 0, ). Now, if we take any point in the region included between the two arcs, say P z1( 1), we get ≤ arg 2 π π z + 1 Thus, represents the shaded region 3 z − 1 3 excluding the points (1, 0) and (−1 0, ). Some Important Results i. Ø Points z1, z2, z3 and z4 (not necessarily in order) will be concyclic, z2 − z4 z1 − z3 if is positive or negative. z1 − z4 z2 − z3 X Example 77. Show that the points 3 + 4i, 3 − 4i, −4 + 3i, −4 − 3i are concyclic. Sol. Let z1 = 3 + 4i, z2 = 3 − 4i, z3 = − 4 + 3i and z4 = − 4 − 3i Then, z2 − z4 z1 − z3 =(7 − i)(7 + i) z1 − z4 z2 − z3 (7 + 7i)(7 − 7i) = 50 = 25 49 + 49 49 = a real number Hence, the given points are concyclic. X Example 78. If z1, z2 and z3 are complex Four points z1, z2, z3 and z4 in anti-clockwise order will be concyclic, if and only if z − z z − z θ = arg 2 4 = arg 2 3 z1 − z4 z1 − z3 z − z z − z ⇒ arg 2 4 −arg 2 3 = 2nπ , n ∈I z1 − z4 z1 − z3 z − z z − z ⇒ arg 2 4 1 3 = 2nπ z1 − z4 z2 − z3 z − z z − z 2 4 1 3 is real and positive. z1 − z4 z2 − z3 − − = ⇒ − − = ⇒ |( x x + = − + − ⇒ x x − + = + − (1, 0) (–1, 0) 2 /3 π π/3
  • 61. 4 150 2 1 1 numbers, such that = + , then show that z1 z2 z3 the points represented by z1, z2 and z3 lie on a circle passing through the origin. ⇒ 2 1 = 1 3 ⇒ 2 1 = − 2 z z1 2 z z3 1 z3 − z1 z3 z2 − z1 = arg z2 ⇒ arg z3 − z1 z3 z2 − z1 = π + arg z2 ⇒ arg z3 − z1 z3 z2 − z1 = π − arg z3 ⇒ arg z3 − z1 z2 ⇒ α = π − β ⇒ α + β = π Hence, the points are concyclic. ii. X Example 79. greatest and least values of | |z . ∴ X Example 80. Let z1 and z2 be two non-real complex cube roots of unity and z − z1 2 + z − z2 2 = λ be the equation of a circle with z1, z2 as ends of a diameter, then the value of λ is = = Logarithm of Complex Numbers Let z = α + iβ = rei(θ + 2nπ) log z = log(rei(θ + 2nπ) ) = log r + i(θ + 2nπ) = log| |z + iarg z + 2n iπ If we put n = 0, we get principal value of log z. ∴ Principal value of log z = log| |z + i arg z. If z + = a, then the greatest and least If 4 z + z = a, then find the z 1 valuesof z arerespectively a a + + 2 4 2 and − + + a a 2 4 2 + = ⇒ − = − − − − Rz ( ) 2 S(0) Qz ) ( 1 Pz ) ( 3 β α = + = + = = + + = + + = =− + +
  • 62. 4 151 π π π Work Book Exercise 4.4 1 z − 1 = 1, represents z + 1 a a circle b an ellipse c a straight line d None of these 2 | z − 4| <| z − 2|, represents the region given by a Re ( )z > 0 b Re ( )z < 0 c Re ( )z > 2 d None of these 3 If 2 z1 − 3z2 + z3 = 0, then z1, z2, z3 are represented by a three vertices of a triangle b three collinear points c three vertices of a rhombus d None of the above 4 If z = x + iy, such that| z + 1| =| z − 1| and z − 1 π amp , then z + 1 4 a x = 2 + 1, y = 0 b x = 0, y = 2 + 1 c x = 0, y = 2 − 1 d x = 2 − 1, y = 0 5 If z8 = (z − 1)8, then the roots of this equation are a collinear b concyclic c the vertices of irregular polygon d None of the above 6 The equation zz + az + az + b = 0, b ∈R represents a circle, if a | |a 2 = b b | |a 2 ≥ b c | |a 2 < b d None of these 7 If z = (λ + 3) + i locus of z is 3 − λ2 , where|λ| < 3, then a circle b parabola c line d None of these 8 If a point P denoting the complex number z moves on the complex plane such that |Re (z)| + |Im (z)| = 1, then the locus of z is a a square b a circle c two intersecting lines d a line 9 The figure formed by four points 1 + 0i, −1 + 0i, 25 3 + 4i and on the argand plane is −3 − 4i a a parallelogram but not a rectangle b a trapezium which is not equilateral c a cyclic quadrilateral d None of the above 10 If i = −1 , then define a sequence of complex number by z1 = 0, zn + 1 = zn 2 + i for n ≥ 1. In the complex plane, how far from the origin is z111 ? a 1 b 2 c 3 d 110 11 The complex numbers whose real and imaginary parts are integers and satisfy the relation zz3 + z z3 = 350, forms a rectangle on the argand plane, the length of whose diagonal is a 5 units b 10 units c 15 units d 25 units 12 The locus of z, for arg z = − is a same as the locus of z for arg z = b same as the locus of z for arg z = c the part of the straight line 3x + y = 0 with y < 0, x > 0 d thepartofthestraightline3x + y = 0 with y > 0, x < 0 13 If z1 and z2 are two complex numbers and z1 + z2 = π but| z1 + z2| ≠| z1 − z2|, then the arg z1 − z2 2 figure formed by the points represented by 0, z1, z2 and z1 + z2 is a a parallelogram but not a rectangle or a rhombus b a rectangle but not a square c a rhombus but not a square d a square 14 z1 and z2 are two distinct points in an argand plane. If a z| 1| = b z| 2|, where a, b ∈R, then the az1 + bz2 is a point on the point bz2 az1 a line segment [−2, 2] of the real axis b line segment [−2, 2] of the imaginary axis c unit circle| z| = 1 d the line with arg z = tan−1 2 15 The roots z1, z2, z3 of the equation z3 + 3αz2 + 3βz + γ = 0 correspond to the points ⇒ z = e−π / 2 Example 82. If z = ilog (2e − value of cos z is 3), then the (a)2 (b) −2 (c)2i (d) −2i Example 81. The value of ii is (a) e−π 2/ (b) eπ / 2 (c) eπ/4 (d) None of these Sol. (a) Let z = i i, loge z = loge i i = i loge i = i loge eiπ / 2 = i 2 loge e = −
  • 63. 4 152 α A, B and C on the complex plane. Then, the triangle is equilateral, if a α2 = β b α = β2 c α2 = 3β2 d 3α2 = β2 16 If z1 2 + z2 2 + 2 z z1 2 cos θ = 0, then the points represents by z1, z2 and the origin form a equilateral triangle b right angled triangle c isosceles triangle d None of these 17 Complex numbers z1, z2, z3 are the vertices A, B and C, respectively of an isosceles right angled triangle, right angled at C. Then, (z1 − z2 )2 is equal to a 2(z1 − z3)(z3 − z2) b (z1 + z3)(z3 − z1) c ( z 1 − z 3)( z 3 − z 2) d None of these 18 A, B and C are points represented by complex numbers z1, z2 and z3. If the circumcentre of the ∆ABC is at the origin and the altitude AD of the triangle meets the circumcircle again at P, then the complex number representing point P is z z1 2 b z = − z z3 2 a z = z3 z1 z z1 2 d z = z z3 2 c z = − z3 z1 19 If z1 and z2 are the roots of the equation z2 + pz + q = 0, where the coefficients p and q may be complex number. Let A and B represent z1 and z2 in the complex plane. If ∠AOB = α ≠ 0 and OA = OB, where O is the origin, then p2 is equal to a 4q cos2 α b 2q cos2α 2 c q cos d None of these 20 If Re z + 4 = 1, then z is represented by a 2 z − i 2 point lying on a a circle b an ellipse c a straight line d None of these 21 Suppose z1, z2 and z3 are the vertices of an equilateral triangle inscribed in the circle| z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the clockwise sense, then a z2 = 1 − 3i, z3 = 2 b z2 = 2, z3 = 1 − 3i c z2 = − 1 + 3i, z3 = − 2 d None of the above 22 Suppose z , z and z are the vertices of an 1 2 3 equilateral triangle circumscribing in the circle | z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the anti-clockwise sense, then z2 is a b c 3 i ) d None of the above 23 The straight line (1 + 2 )i z + (2i − 1)z = 10 i on the complex plane, has intercept on the imaginary axis equal to a 5 b c − d −5 24 If the complex number z satisfies the condition | z| ≥ 3, then the least value of z + is equal to 5 ab cd None of these 25. If z1 = 2 , z3 = a − bi, for a, b ∈R and 1 − i 2 + i z1 − z2 = 1, then the centroid of the triangle formed by the points z1, z2, z3 in the argand plane is given by a b (1 + 7i ) c (1 − 3i ) d (1 − 3i ) 26. Let λ ∈R. If the origin and the non-real roots of 2 z2 + 2 z + λ = 0 form three vertices of an equilateral triangle in the argand plane, then λ is a 1b 2 /3 c 2 d −1 27 If a, b, c and u, v, w are complex numbers representing the vertices of two triangles, such that c = (1 − r a) + rb and w = (1 − r u) + rv . When r is a complex number, then the two triangles a have the same area b are similar c are congruent d None of these 28. On the complex plane, ∆OAP and ∆OQR are = 1. If similar and I OA( ) the points P and Q denote the1) complex numbers z1 and z2, then the complex numbersX z denoted by the point R is given by z1 a z z1 2 b z2 c z2 d z1 + z2 z1 z2 29 Intercept made by the circle zz + αz + αz + r = 0 on the real axis on complex plane, is a (α + α −) r b (α + α)2 − 2r c (α + α)2 + r d (α + α)2 − 4r − − = Rz () Pz ( Qz ( ) 2 θ θ O A (1, 0) Y
  • 64. 4 153 30 The equation of the radical axis of the two circles represented by the equations,| z − 2| = 3 and | z − 2 − 3 i| = 4 on the complex plane is a 3y + 1 = 0 b 3y − 1 = 0 c 2 y − 1 = 0 d None of the above
  • 65. 154 α α 2 2 WorkedOut Examples Type 1. Only One Correct Option Ex 1. If a = cosα + isin α, b = cosβ + isin ,β a b c c = cos γ + isin γ and + + =1, then b c a cos (α − β +) cos (β − γ ) + cos (γ − α) is equal to (a) (b) − (c) 0 (d) 1 + + Sol. cis β cis γ = 1, cis γ cis α where, cis θ represents cos θ + i sin θ. ⇒ cis(α − β) + cis(β − γ) + cis(γ − α) = 1 Equation real parts of both sides cos (α + β) + cos (β − γ) + cos(γ − α) = 1 Hence, (d) is the correct answer. Ex 2. The locus of the centre of a circle which touches the circles | z − z1| = a and | z − z2 | = b externally (z z, 1 and z2 are complex numbers) will be (a) an ellipse (b) a hyperbola (c) a circle (d) None of these Sol. Let A z( 1), B z( 2)be the centres of given circles and Pbe the centre of the variable circle which touches given circles externally, then |AP| = a + r and |BP| = b + r, where r is the radius of the variable circle. On subtraction, we get |AP|− |BP| = a − b ⇒ ||AP | − | BP || = | a − b |, a constant. Hence, locus of P is (i) right bisector AB, if a = b. (ii) a hyperbola, if | a − b | < |AB| = |z2 − z1 |. (iii) an empty set, if |a − b| > |AB | = | z2 − z1 |. (iv) set of all points on line AB except those which lie between A and B, if |a − b| = |AB| ≠ 0. Hence, (d) is the correct answer. Ex 3. If arg (z1 ) = arg (z2 ), then (a) z2 = kz1 −1 , k > 0 (b) z2 = kz1 , k > 0 (c) | z2 | = | z1 | (d) None of these z z1 1 = |z1|2 z1−1 Sol. z1 = z1 ⇒ arg (z1 −1 ) = arg(z1) = arg(z2) ⇒ z2 = kz1 −1 (k > 0) Hence, (a) is the correct answer. tan α − i sin + cos Ex 4. Ifis purely α 1 + 2isin 2 i maginary, then α is given by π (a) nπ + (b) nπ − (c) (2n + 1) π (d) 2nπ + α tanα − i sin + cos 1 − 2isin 2 0 2 α ⇒ sinα(1 − cosα) = cosα(1 − cosα) 4 4 π ⇒ Re tan cos sin sin c sin sin tan α α α α α α α − + − − ⋅ + + 2 2 2 2 2 2 2 i os sin α α 2 1 4 2 0 2 + = ⇒ sin sin tanα α α + = 2 2 2 ⇒ sin cos sin cos α α α α − + = 1 ⇒ sin cos cos cos sin α⋅ α= α+ α− α 2 a b b c c a =1 ⇒ cis cis α β + + Re tan cos sin sin α α α α + − + = i i 2 2 2 1 2 0 ⇒ Re 2 2
  • 66. 155 ⇒ sinα = cosα, cosα = 1 ⇒ or α = 2nπ, n ∈I Hence, (a) is the correct answer. a − ib Ex 5. The expression tan ilog a + ib reduces to 2ab (a)(b) a2 − b2 ab 2ab (c)(d) a2 − b2 a2 + b2 a − ib Sol. Given, tan ilog a + ib Let a + ib = reiθ , r2 = a2 + b2 ⇒ a − ib = re− θi , tan θ = b a a − ib −2i θ ⇒= e a + ib a − ib −2iθ = 2θ ilog ilog(e ) a + ib a − ib ⇒ tan ilog a + ib tan2θ Hence, (b) is the correct answer. Ex 6. If z1 = a + ib and z2 = c + id are complex numbers such that | z1| = | z2| =1 and Re (z z1 2 ) = 0, then the pair of complex numbers a + ic = w1 and b + id = w2 satisfies (a) |w1 | ≠ 1 (b) |w2 | ≠ 1 (c) Re(w w12 ) = 0 (d) None of these Sol. z1 = a + ib z, 2 = c + id |z1| = |z2| = 1 ⇒ a2 + b2 = c2 + d2 = 1 w1 = a + ic w,2 = b + id As Re ( ) ⇒ ac + bd = 0 ⇒ ac = − bd w w12 = (a + ic) (b − id) = (ab + cd) + i bc(− ad) We have, a2 + b2 = c2 + d2 ⇒ a2 − c2 = d2 − b2 ⇒ a2 − c2 + 2i ac = d2 − b2 − 2ibd [as ac = − bd] ⇒ (a + ic)2 = (d − ib)2 ⇒ a + ic = (d − ib) or −d + ib ⇒ a = d and c = − b or a = − d,b = c ⇒ c2 + d2 = b2 + d2 a2 + c2 = a2 + b2 ⇒ a2 + c2 = 1, b2 + d2 = 1 ⇒ |w1| = |w2| = 1 Also, ab + cd = − cd + cd = 0 ⇒ Re (w w12) = 0 Hence, (c) is the correct answer. z1 Ex 7. If =1and arg (z z1 2 ) = 0, then z2 (a) z1 = z2 (b) | z2 |2 = z z1 2 (c) z z1 2 = 1 (d) None of these z1 = 1 Sol. Let z1 = r1 (cos θ1 + isinθ1), then z2 ⇒ |z1| = |z2| ⇒ |z1| = |z2| = r1 Now, arg (z1, z2) = 0 ⇒ arg(z1) + arg (z2) = 0 ⇒ arg (z2) = −θ1 Therefore, z2 = r1[cos(−θ1) + isin (−θ1)] = r1(cos θ1 − isinθ1) = z1 ⇒ z2 = (z1) = z1 ⇒ |z2|2 = z z1 2 Hence, (b) is the correct answer. Ex 8. If z is a complex number, then z2 + z 2 = 2 represents (a) a circle (b) a straight line (c) a hyperbola (d) an ellipse Sol. Let z = x + iy, then z2 + z2 = 2 ⇒ (x + iy)2 + (x − iy)2 = 2 ⇒ x2 − y2 = 1 which represents a hyperbola. Hence, (c) is the correct answer. ab b a 2 2 + = − 2 1 2 tan tan θ θ = − = − 2 1 2 2 2 2 2 ba b a ab a b /
  • 67. 4 156 Ex 9. If = A + iB, then A2 + B 2 equals (a) 1 (b) α2 (c) −1 (d) −α2 Sol. A + iB = ⇒ A − iB = ⇒ (A + iB) (A − iB) = = 1 ⇒ A2 + B2 = 1 Hence, (a) is the correct answer. Ex 10. If| z1| = | z2|and arg (z1 ) + arg (z2 ) = π/2, then (a) z z1 2 is purely real (b) z z1 2 is purely imaginary (c) (z1 + z2 )2 is purely real (d) arg(z1 −1 ) + arg(z2 −1 ) = Sol. Let |z1| = |z2| = r ⇒ z1 = r(cosθ + isin )θ and z2 = r cos 2 − isin 2 − ⇒ z z1 2 = r i2 , which is purely imaginary. z1 + z2 = r [(cosθ + sin )θ + i(cosθ + sin )]θ ⇒ (z1 + z2)2 = 2r2 ⋅ (cosθ + sin )θ 2 ⋅ i which is purely imaginary. Also, arg (z1 −1 ) + arg (z2 −1 ) = − Hence, (b) is the correct answer. Ex 11. The value of the expression 1 1 1 1 2 1 1 + ω2 + 3 2 2 + ω2 1 1 + 4 3 3 + 2 +… + (n +1) ω 1 1 n n + 2 , ω where ω is an imaginary cube root of unity, is n n( 2 + 2) n n( 2 − 2) (a) (b) 3 n n n (d) None of these Sol. tn = (n + 1) n + ω1 n + ω12 n3 + n2 ω 1 2 + ω 1 + 1 1 1 + n 1 + 2 + 1 ω = n3 + n2 (ω + ω2 + 1) + n(ω + ω2 + 1) + 1 = n3 + 1 n n 2 2 3 n n( + 1) ∴ Sn = ∑tr = ∑ (r + 1) = n r = 1 r = 1 Hence, (c) is the correct answer. Ex 12. If z1 and z2 are two complex numbers satisfying the equation z1 + iz2 =1, then z1 z1 − iz2 z2 is (a) purely real (b) of unit modulus (c) purely imaginary (d) None of the above Sol. (z1 + iz2)(z1 − iz2) = (z1 − iz2)(z1 + iz2) ⇒ z z1 2 = z z1 2 z1 = z1 ⇒ z2 z2 ⇒ z1 is purely real. z2 Hence, (a) is the correct answer. Ex 13. If z = − 2 + 2 3i, then z2n + 22n zn + 24n may be equal to (a) 22n (b) 0 (c) 3 4⋅ 2n , n is multiple of 3 (d) None of the above Sol. z = − 2 + 2 3i = 4ω ∴z2n + 22n nz + 24n = 42nω2n + 22n ⋅ 4n ⋅ωn + 24n = 42n[ω2n + ωn + 1]
  • 68. 157 x y y x m n + 1 y x n m = 0, if n is not a multiple of 3. = 3 4⋅ 2n , if nis a multiple of 3. Hence, (c) is the correct answer. Ex 14. The complex number z =1 + i is rotated through an angle 3π/2 in anti-clockwise direction about the origin and stretched by additional 2 units, then the new complex number is (a) − 2 − 2 i (b) 2 − 2 i (c) 2 − 2 i (d) None of the above Sol. If z1 is the new complex number, then |z1| = | |z + 2 = 2 2 z1 = |z1|⋅ ei3π/ 2 Also, z | |z 3π 3 ⇒ z1 = z⋅ 2 cos 2 + isin 2 = 2(1 + i)(0 − i) = − 2i + 2 = 2(1 − i) Hence, (d) is the correct answer. 1 1 Ex 15. If 2cosθ = x + and 2cosφ = y + , then x y (a) + = 2cos (θ + φ) (b) x y = 2cos (mθ + nφ) m n x y xm yn (c) + = 2cos (mθ + nφ) 1 (d) xy + = 2cos (θ − φ) xy 1 1 Sol. 2cosθ = x + , 2cosφ = y + x y ⇒ x2 − 2xcos θ + 1 = 0 2cos θ ± 4cos2 θ − 4 ⇒ = ⇒ + = 2cos(θ − φ) x ym n = 2cos(mθ + nφ) = 2cos(mθ − nφ) 1 xy + = 2cos(θ + φ) xy Hence, (b) is the correct answer. Ex 16. The complex numbers z1 and z2 are such that z1 ≠ z2 and | z1| = | z2|. If z1 has positive real part and z2 has negative imaginary part, then z1 + z2 may be z1 − z2 (a) zero (b) real and positive (c) real and negative (d) purely imaginary Sol. Given, |z1| = |z2| Re (z1) > 0, Im (z2) < 0 z1 + z2 = 1 z1 + z2 + z1 + z2 Re z1 − z2 2 z1 − z2 z1 − z2 1 (z1 + z2)(z1 − z2) + (z1 + z2)(z1 − z2) = 2 (z1 − z2)(z1 − z2) z z1 1 − z z1 2 + z z2 1 − z z2 2 + z z1 1 + z z1 2 1 = z z2 1 − z z2 2 2 |z1 − z2| z1 + z2 is purely imaginary. z1 − z2 Hence, (d) is the correct answer. Ex 17. If z z1 − z2 = k, (z1, z2 ≠ 0), then x 2 ⇒ x = cosθ ± isinθ = e±iθ Similarly, y = e± φi 2 |z1 − z2|2 1 2|z |2 − |z |2 = 0 x y y x xy mn + 1 x y y x m n n m +
  • 69. 4 158 z z1 + z2 (a) for k ≠ 1, locus z is a straight line (b) for k ∉{1, 0}, z lies on a circle (c) for k ≠ 0, z represents a point (d) for k ≠ 1, z lies on the perpendicular bisector z2 − z2 of the line segment joining and z1 z1 z − z2 Sol. Given, z z z z1 +− z z2 2 = k z z1 2 = k 1 z + z1 Clearly, if k ≠ 0, 1 then z would lie on a circle. If k = 1, zwould lie on a perpendicular bisector of the z2 and − z2 . line segment joining z1 z1 If k = 0, z represents a point. Hence, (b) is the correct answer. Ex 18. If A z( 1 ), B z( 2 ) and C z( 3 ) are the vertices of a π AB ∆ABC in which ∠ABC = and = 2, then 4 BC the value of z2 is equal to (a) z3 + i z( 1 + z3 ) (b) z3 − i z( 1 − z3 ) (c) z3 + i z( 1 − z3 ) (d) None of the above AB Sol. Given, = 2 BC Considering the rotation about B, we get z1 − z2 = |z1 − z2|⋅ eiπ/ 4 = AB ⋅ eiπ/ 4 z3 − z2 |z3 − z2| BC 1 i = 2 + = 1 + i 2 2 ⇒ z1 − z2 = (1 + i z)( 3 − z2) ⇒ z1 − (1 + i z) 3 = z2(1 − 1 − i) ⇒ iz2 = − z1 + (1 + i z) 3 ⇒ z2 = iz1 − i(1 + i z) 3 ∴ z2 = z3 + i z( 1 − z3) Hence, (c) is the correct answer. Ex 19. If z z1 2 ∈C, z z R, z1 (z1 2 − 3z2 2 ) = 2 and z z z , then the value of z1 2 + z2 2 is (a) 5 (b) 6 (c) 10 (d) 12 Sol. Here, z z1( 1 2 − 3z2 2 ) = 2 …(i) z2(3z1 2 − z2 2 ) = 11 …(ii) On multiplying Eq. (ii) by i and adding it to Eq. (i), we get z1 3 − 3z z2 2 1 + i(3z z1 2 2 − z2 3 ) = 2 + 11i ⇒ (z1 + iz2) = 2 + 11i …(iii) Again multiplying Eq. (ii) by i and subtracting it from Eq. (i), we get z1 3 − 3z z2 2 1 − i(3z z1 2 2 − z2 3 ) = 2 − 11i ⇒ (z1 − iz2)3 = 2 + 11i On multiplying Eqs. (iii) and (iv), we get (z12 + z22 3) = 4 + 121 ⇒ z12 + z22 = 5 Hence, (a) is the correct answer. …(iv) Ex 20. 1 − c2 = nc −1 and z = eiθ , then c n (1 + nz) 1 + is equal to 2n z (a) 1− ccosθ (b) 1+ 2ccosθ (c) 1+ ccosθ (d) 1− 2ccosθ Sol. Here, 1 − c2 = nc − 1 ⇒ 1 − c2 = n c2 2 − 2nc + 1 c 1 ∴ = …(i) or (1 + c 1 nz) 1 + + n z + z 2n + n⋅ (2cosθ )} [using Eq. (i)] = 1 + ccos θ Hence, (c) is the correct answer. Ex 21. Consider an ellipse having its foci at A z( 1 ) and B z( 2 ) in the argand plane. If the eccentricity of the ellipse is e and it is known that origin is an interior point of the ellipse, then | z1 + z2 | (a) e 0, n n 2 1 2 + n z n 1 1 2 = + 1 2 + n = + + 1 1 1 2 2 n n { = + + + (1 ) 2 cos 1 2 2 n n n θ = + + 1 2 1 2 n n cos θ
  • 70. 159 2 2 | z1 | + |z2 | | z1 − z2 | (b) e 0, | z1 | + | z2 | | z1 + z2 | (c) e 0, | z1 | − |z2 | (d) Can’t be determined Sol. If P z( )is any point on the ellipse. Then, equation of the ellipse is |z1 − z2 | …(i) |z − z1| + |z − z2| = e If we replace z by z1or z2, Eq. (i) becomes |z1 − z2|. For P z( )to lie on ellipse, we have |z1 − z2 | |z − z1| + |z − z2|< e It is given that origin is an interior point of the ellipse. |z1 − z2 | ⇒ |0 − z1| + |0 − z2|< e |z1 − z2 | ⇒ e 0, |z1| + |z2| Hence, (b) is the correct answer. Ex 22. If| z − 2 − i| = | |z sin π − arg ( )z , then 4 locus of z is (a) a pair of straight lines (b) a circle (c) a parabola (d) an ellipse Sol. We have, |(x − 2) + i y(− 1)| = | |z 1 cosθ − 1 sinθ where, θ = arg ( )z |x − y | which is a parabola. Hence, (c) is the correct answer. Ex 23. α1, α 2, α 3, …, α100 are all the 100th roots of unity. The numerical value of ∑∑ (α αi j )5 is 1 ≤ i < j ≤100 (a) 20 (b) 0 (c) (20)1 20/ (d) None of these Sol. − (α101+ α102+ α103+ α104 + α510 + …) = 0 − 0 = 0 r r r 100, if r = 100k Because (α1 + α2 + … + α100) 0 , if r ≠ 100k Hence, (b) is the correct answer. Ex 24. The maximum area of the triangle formed by the complex coordinates z, z1 and z2, which satisfy the relations | z − z1| = | z − z2|, z − z1 + z2 r, where r >| z1 − z2|, is 2 1 By the given conditions, the area of ∆ABC is |z1 − z2| r Hence, (b) is the correct answer. Ex 25. Locus of z, if 3π , when | z | ≤ −| z 2| 4 arg[z − + =(1i)] π , when | z | > −| z2|, is 4 (a) straight line passing through (2, 0) (b) straight line passing through (2, 0), (1, 1) (c) a line segment (d) a set of two rays Sol. The given equation is written as ( ) ( ) x y − + − = 2 1 1 2 2 2 ( ) a 1 2 2 1 2 | | z z − ) b ( 2 2 1 | | z z r − ( ) c 1 2 1 2 2 2 | | z z r − (d) 1 2 1 2 2 | | z z r − Az () r Cz ) ( 2 Bz ) ( 1 zz 12 + 2 Y O X
  • 71. 4 160 1 2 3π − (1 + i)] 4 , when x ≤ 2 arg[z −π , when x > 2 4 The locus is a set of two rays. Hence, (d) is the correct answer. 3 n A(∩ B) is equal to (a) 1 (b) 2 (c) 3 (d) 0 Sol. We can observe that, 3 + 3i ∈ A but ∉ B. ∴ n A(∩ B) = 0 Hence, (d) is the correct answer. Ex 27. Dividing f z( )by z − i, we obtain the remainder i and dividing it by z + i, we get the remainder 1 + i. The remainder upon the division of f z( ) by z2 +1, is 1 1 (a)(z + 1) + i (b)(iz + 1) + i 2 (c)(iz − 1) + i (d)(z + i) + 1 Sol. f z( ) = g z( ) (z − i z)( + i) + az + b; where a, b ∈C f i( ) = i ⇒ ai + b = i …(i) f (−i) = 1 + i ⇒ a(−i) + b = 1 + i …(ii) From Eqs. (i) and (ii), we get i 1 a = , b = + i 2 2 Hence, required remainder = az + b = iz + + i Ex 28. If z1 = a1 + ib1 and z2 = a2 + ib2 are complex numbers such that | z1| =1, | z2| = 2 and Re (z z1 2 ) = 0, then the pair of complex ia2 numbers ω1 = a 1 + and ω2 = 2 b 1 + ib 2, 2 satisfy (a) |ω1 | = 1 (b) |ω 2 | = 2 (c) Re (ω ω12 ) = 0 (d) Im (ω ω12 ) = 0 Sol. a12 + b12 = 1, a22 + b22 = 4 and a a1 2 = b b1 2 a22 + b22 = 4a12 + 4b12 (a2 + 2ia1)2 = (2b1 + ib2)2 ⇒ a2 = ± 2b1 2 |ω1|2 = a1 + a12 + a2 = a12 + b12 = 1 4 ⇒ |ω1| = 1 and |ω2|2 = 4b1 2 + b2 2 = 4 ⇒ |ω2| = 2 a b2 2 = 0 Re (ω ω12) = 2a b1 1 − 2 Im (ω ω1 2) = a b1 2 + a b2 1 = 2a1 2 + 2b1 2 = 2 Hence, (a), (b) and (c) are the correct answers. Ex 29. If from a point P representing the complex number z1 on the curve | |z = 2, pair of tangents are drawn to the curve | |z =1, meeting at point Q z( 2 ) and R z( 3 ), then z1 + z2 + z3 (a) complex number will lie on the Ex 26. If A z |arg ( )z = 4 and 2 B z |arg (z − 3 − 3 )i = , z ∈C. Then, Hence, (b) is the correct answer. Type 2. More than One Correct Option (0, 2) (1, 1) (2, 0) Re()=1 z O (3, 3) A B Y X 2 1 2
  • 72. 161 z3 3 (d) orthocentre and circumcentre of ∆PQR will coincide Sol. Options (a) and (d) are true as PQR is an equilateral triangle, so orthocentre, circumcentre and centroid will coincide. (b) z1 + z2 + z3 = 1 ⇒ |z1 + z2 + z3|2 = 9 3 ⇒ (z1 + z2 + z3) (z1 + z2 + z3) = 9 4 1 1 4 1 1 ⇒ + + + + 9 z1 z2 z3 z1 z2 z3 (c) ∠QOR = 120° Hence, (a), (b), (c) and (d) are the correct answers. Ex 30. One vertex of the triangle of maximum area that can be inscribed in the curve | z − 2i| = 2, is 2 + 2i, remaining vertices is/are (a) − +1i(2 + 3) (b) − −1i(2 + 3) (c) − +1i(2 − 3) (d) − −1i(2 − 3) Sol. Clearly, the inscribed triangle is equilateral. Hence, options (a) and (c) are the correct answers. Ex 31. Let z be a complex number and a be a real parameter, such that z2 + az + a 2 = 0, then (a) locus of z is a pair of straight lines (b) locus of z is a circle (c) arg( )z = ± (d) |z | = |a | Sol. z2 + az + a2 = 0 ⇒ z = aω, aω2 where, ω is non-real root of cube unity. ⇒ Locus of z is a pair of straight lines and arg( )z = arg ( )a + arg ( )ω or arg ( )a + arg (ω2 ) ⇒ arg ( )z = ± Also, | |z = | | |a ω | or | | |a ω2 | ⇒ | |z = | |a Hence, (a), (c) and (d) are the correct answers. Type 3. Assertion and Reason Directions (Ex. Nos. 32-36) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement IIis a correct explanation for Statement I (b) Statement I is true,Statement II is true; Statement II isnot a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true Ex 32. Statement I If A z( 1 ), B z( 2 ), C z( 3 ) are the vertices of an equilateral ∆ABC, then z2 + z3 − 2z1 π arg z3 − z2 4 Statement II If ∠B = α, then z2 + z3 z + − 2z − z1 Sol. arg 2 z3 1 = arg 2 z3 − z2 z3 − z2 3 curve | | z =1 ) b ( 1 1 1 1 4 4 9 1 3 3 2 2 1 z z z z z z + + + + = (c) arg z2 2 = π Y Qz ( ) 2 Pz ( ) 1 Rz ( ) 3 O Y¢ X¢ X ⇒ z z z z e i 2 0 0 1 2 3 − − = π , z z z z e i 0 3 0 1 2 3 − − = − π ⇒ z i 2 3 2 1 + =− + ) ( and z i 3 1 2 3 =− + − ( ) z2 z i 1(2+2) z3 zi 0(2) z z z z AB BC ei 1 2 3 2 − − = α or arg z z z z 2 1 2 3 − − =α α Bz () 2 Cz () 3 Az () 1
  • 73. 4 162 [as AD ⊥ BC ] 2 Hence, (d) is the correct answer. 1 Ex 33. Statement I If x + =1 and x p = x4000 + 1 and q is the digit at unit place 4000 x 2n in the number 2 +1, n ∈ N and n >1, then the value of p + q = 8. Statement II ω, ω2 are the roots of 1 3 1 x + = −1, x + = 2. x x3 1 2 Sol. x + = 1 ⇒ x − x + 1 = 0 x ∴ x = −ω − ω, 2 Now, for x p Similarly, for x = −ω2 , p = −1 For n > 1, 2n = 4k ∴ (2)2n = 24k = (16)k = anumber with last = 6 ⇒ q = 6 + 1 = 7 Hence, p + q = −1 + 7 = 6 Hence, (d) is the correct answer. Ex 34. Statement I If z1, z2, z3are complex numbers represent the points A B C, , such that 2 1 1 = + . Then, A B C, , passes through z1 z2 z3 origin. Statement II If 2z2 = z1 + z3, then z1, z2, z3 are concyclic. 2 1 1 1 1 1 1 Sol. = + ⇒ − = − z1 z2 z3 z1 z2 z3 z1 z2 − z1 = z1 − z3 ⇒ z z1 2 z z1 3 z2 − z1 = − z2 ⇒ z3 − z1 z3 z2 −z1 = arg z2 ⇒ arg z3 − z1 z3 z2 −z1 = π − arg z2 = π − arg z2 ⇒ arg z3 − z1 z3 z3 ⇒ ∠CAB = π − ∠COB ⇒ ∠CAB + ∠COB = π ⇒ Points O A B C, , , are concyclic. Hence, (b) is the correct answer. Ex 35. Statement I 3 + ix y2 and x2 + y + 4i are complex conjugate numbers, then x2 + y2 = 4. Statement II If sum and product of two complex numbers is real, then they are conjugate complex number. Sol. If 3 + ix y2 and x2 + y + 4i are conjugate, then x y2 = − 4 and x2 + y = 3 ⇒ x2 = 4, y = − 1 ⇒ x2 + y2 = 5 Hence, (d) is the correct answer. Ex 36. Statement I If | |z < 2 −1, then | z2 + 2z cos α | <1 Statement II | z1 + z2| ≤ | z1| + | z2|, also |cos α ≤| 1. Sol. |z2 + 2zcos α | < |z2 | + |2zcos α | < | |z 2 + 2| | |coszα| Hence, (a) is the correct answer. Type 4. Linked Comprehension Based Questions = π Az ( ) 1 Cz ( ) 3 Bz ( ) 2 D zz 23 + 2
  • 74. 163 Passage I (Ex. Nos. 37-39) In argand plane, | z | represents the distance of a point z from the origin. In general, | z1 − z2| represents the distance between two points z1and z2. Also, for a general moving point z in argand plane, if arg( )z = θ then z = | |z ei θ , where ei θ = cos θ + i sinθ. Ex 37. The equation | z − z1| + | z − z2| =10, if z1 = 3 + 4i and z2 = − 3 − 4i represents (a) point circle (b) ordered pair (0, 0) (c) ellipse (d) None of these Sol. As z1 − z2 = 6 + 8i =10 ∴ z − z1 + z − z2 = z1 − z2 ⇒ z lies between z1 and z2. i.e. a line segment. Hence, (d) is the correct answer. Ex 38. z − z1| − | z − z2 = t, t >10 where t is a real parameter, always represents (a) ellipse (b) hyperbola (c) circle (d) None of these Sol. Given equation can represent a hyperbola, since |z1 − z2| = 10 ⇒ z lies on hyperbola. Hence, (b) is the correct answer. π Ex 39. If | z − (3 + 2i)| = z cos − arg ( )z ,then 4 locus of z is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola Sol. Given, |z − (3 + 2i)|= |(x − 3) + i y( − 2)| which is a parabola. Hence, (b) is the correct answer. Passage II (Ex. Nos. 40-42) The complex slope of a line passing through two points represented by complex z2 − z1 numbers z1 and z2 is defined by and we shall z2 − z1 denote by ω. If z0 is complex number and c is a real number, then z z0 + z z0 = 0 represents a straight line. Its z0 . Now ,consider two lines complex slope is − z0 αz +αz + iβ = 0…(i) and az + az + b = 0 …(ii) where, α β, and a b, are complex constants and let their complex slopes be denoted byω1 andω2, respectively. Ex 40. If the lines are inclined at an angle of 120° to each other, then (a) ω ω21 = ω ω1 2 (b) ω ω212 = ω ω1 22 (c) ω12 = ω 22 (d) ω1 + 2ω 2 = 0 Sol. ⇒ω31 = ω32 ⇒ω13 ω ω1222 = ω23 ω ω12 22 ⇒ ω1 ω1 2 ω22 = ω2 ω2 2ω12 ⇒ω ω112= ω ω212,as ω1= ω2 ∴ ω2 ω12 = ω1 ω22 Hence, (b) is the correct answer. Ex 41. Which of the following must be true? (a) a must be pure imaginary (b) β must be pure imaginary (c) a must be real (d) β must be imaginary Sol. Since, i β is real. ∴ β is pure imaginary. Hence, (b) is the correct answer. Ex 42. If line (i) makes an angle of 45° with real axis, 2 then (1 + i) is (a) 2 2 (b) 2 2i (c) 2(1− i) (d) − 2(1+ i) Sol. e i 2 ∴ (1 + i) = ± 2 (− 1 + i) Hence, (c) is the correct answer. Passage III (Ex. Nos. 43-45) Consider ∆ABC in argand plane. Let A(0), B(1) and C(1+ i) be its vertices and M be the mid-point of CA. Let z be a variable complex number in the plane. Let us another variable complex number defined as, u = z2 +1. Ex 43. Locus of u, when z is on BM, is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola Ex 44. Axis of locus of u, when z is on BM, is = + || cos sin z 1 2 1 2 θ θ ,where θ= arg()z ⇒ ( ) ( ) | | y x x y = − + − + 2 3 1 2 2 2 − = =± ± α α π i 2
  • 75. 4 164 (a) real axis (b) imaginary axis (c) z + z = 2 (d) z − z = 2i Ex 45. Directrix of locus of z, when z is on BM, is (a) real axis (b) imaginary axis (c)z + z = 2 (d) z − z = 2i Sol. (Ex. Nos. 43-45) BM ≡ y − 0 = − 1(x − 1) x + y = 1 ∴ u − 1 = t + i (1 − t) u = 2t + 2i t (1 − t) x = 2t and y = 2 (1t − t), where u = x + iy (x − 1)2 = − 2 y − 1 , which is parabola. 2 Axis is x = 1, i.e. z + z = 2 Directrix is y = 1, i.e. z − z = 2i 43. (b) 44. (c) 45. (d) Type 5. Match the Columns Ex 46. Match the following : ⇒ 75 × 25 = z1 + 1 ∴ z1 − (− 1) = 1875 ⇒ z1 lies on circle. Locus of the point equationRe (z ) = Re(z + z),isa/an Ex. 47. If z1, z2, z3, z4 are the roots of the equation B. Locus of the point z satisfying the q. straight line z4 + z3 + z2 + z +1 = 0, then equation Sol. A. Put z = x + iy ∴ Re (x + iy)2 = Re (x + iy + x − iy) 2 x2 2 − y2 = 2x Sol. The given equation is z 5 − 1 = 0, which means that or x − y − 2x = 0 z − 1 Rectangular hyperbola, eccentricity = 2 z1, z2, z3, z4 are four out of five roots of unity except 1. B. For ellipse, λ > |z1 − z2| A. z14 + z24 + z34 + z44 + 14 = 0 and for straight line, 4 λ = |z1 − z2| ⇒ ∑zi4 = 1 4 B. ∑ zi 5 is equal to q. i = 1 4 4 r. C.(zi + 2)is equal to i = 1 1 D. Least value of [| z1 + z2|], where s. [ ] represents greatest integer function, is 11 − + − = λ λ∈ + − λ< − + = = − ∈ + = − + ∏ = ∑
  • 76. 165 1 3 1 3 C.  i = 1 B. z z z z ⇒ ⇒ i = 1 C. z4 + z3 + z2 + z + 1 = (z − z1) (z − z2) (z − z3) (z − z4) Putting z = − 2on both the sides, we get ⇒ 4 ∏ (zi + 2) = 11 i = 1 D. |z1 + z2| =2 + 2cos144° for minimum 5 − 1 D. Given, z = 25 = 2cos72° = 2 Let whose greatest integer is 0. A → r; B → q; C → s; D → p Type 6. Single Integer Answer Type Questions Ex 48. Consider an equilateral triangle having vertices Sol. (5) A z( 1) = 2i at points incircle, then AP 2 + BP 2 + CP 2 is equal to _______ . BC D Hence, any point on incircle i.e. P z( )is cos α, sinα i.e. (cosα + isinα) Solving for |AP|2 + |BP|2 + |CP| ,2 we get AP2 + BP2 + CP2 = 5 Ex 49. Let A1, A2,..... An be the vertices of a regular polygon of n sides in a circle of radius unity and a = | A A1 2|2 + | A A1 3|2 + .... + | A A1 n |2 , a b = | A A1 2|| A A1 3|....| A A1 n |, then = ____. b Sol. (2) Let us assume thatO is the centre of the polygon and z0, z1,…, zn − 1 represent the affixes of A1, A2,…, An such that i2π z0 = 1, z1 = α, z2 = α2 ,..., zn−1 = αn−1 , where α = e n Now, |A A1 r |2 = α| r − 1|2 = |1− αr |2 2rπ 2rπ 2 = 1 − cos − isin n n 2r 2 2r 2 2rπ i Bz i ) ( 2 2 3 3 2 1 2 1 3 − = − = Cz i i ) ( 3 2 3 3 2 2 1 3 − = − − =− Radiusofincircleof ∆ABC , i.e. r = 1 3 unit. 1 3 2 1 i z z m − + = z i z m − + = 2 1 2 For m = 2, z z − + = 1 2 1 1 ⇒ z i z − = + 2 1 | |,i.e.straightline z z 1 1 75 =− + ∴ 75 1 1 z z = + or 75 1 1 z z = + 3 O A r B e A e i i 2 3 2 3 6 2 π π − , and C e i 2 3 5 6 − π .If Pz () isanypointonits
  • 77. 4 166 1 − cos sin = 2 − 2 cos n n n n n ∴r∑= 1 |A A12|2 = ∑r = 1 2 − 2 cos rn 2π 4π 2(n − 1) = 2 (n − 1) cos+ cos + ... + cos n n n = 2 (n − 1) − 2⋅ Real part of (α + α2 + ... + αn−1 ) = 2 (n − 1) − 2 (1) [since, {1 + α + α2 + + αn−1 = 0}] ∴ |A A12|2 + |A A13|2 + ... |+ A A1n|2 = 2n Also, let E = |A A12| |A A13||A A1 n| = |1 − α| |1 − α2 | |1 − α3 |... |1 − αn−1 | = (|1 − α)(1 − α2 )(1 − α3 )(1 − αn−1 )| Since, 1 α α, 2 ,…, αn−1 are the roots of zn − 1 = 0. n ⇒ (z − 1) (z − α) (z − α2 )... (z − αn−1 ) = z − 1 z −1 ⇒ (z − α) (z − α2 ) (z − αn−1 ) = zn − 1 = 1 + z + z2 + … + zn−1 Substituting z = 1,we have (1 − α) (1 − α2 )... (1 − αn−1 ) = n |1 − α| |1 − α2 |... |1 − αn−1 | = n a 2n Hence, the value of = = 2 b n
  • 78. 167 1 5 (c) 1 10 Target Exercises Type 1. Only One Correct Option i592 + i590 + i588 + i586 + i584 1. The value of − 1 is i582 + i580 + i578 + i576 + i574 (a) −1 (b) −2 (c) −3 (d) −4 2. i57 + 1 , when simplified has the value 125 i (a) 0 (b) 2i (c) −2i (d) 2 3. in + in + 1 + in + 2 + in + 3 is equal to (a) 1 (b) − 1 (c) 0 (d) None of these 4. 1+ i2 + i4 + i6 + + i2n is (a) positive (b) negative (c) 0 (d) Can’t be determined 5. The value of i19 1 25 2 is i (a) 4 (b) − 4 (c) 2 (d) − 2 6. If n is any positive integer, then the value of i4n + 1 − i4n − 1 equals 2 (a) 1 (b) −1 (c) i (d) −i 7. For a positive integer n, the expression (1− i)n 1− 1 n equals i (a) 0 (b) 2in (c) 2n (d) 4n (1− i)n 8. If the number is real and positive, then nis (1+ i)n − 2 (a) any integer (b) 2λ (c) 4λ + 1 (d) None of these 9. The smallest positive 1+ i n = − 1is 1− i integer n for which (a) 1 (b) 2 (c) 3 (d) 4 10. The smallest positive (1+ i)2n = (1− i)2n is number n for which (a) 4 (b) 8 (c) 2 (d) 12 1+ 2i 11. The complex number lies in the 1− i (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant (1− i)3 12. The value of is equal to 1− i3 (a) i (b) − 1 (c) 1 (d) − 2 1+ i 2 1− i 2 13. is equal to 1− i 1+ i (a) 2i (b) −2i (c) −2 (d) 2 (1+ i)2 14. The value of Re is equal to 3 − i (a) − (b) (d) − 15. If a2 + b2 = 1, then (1 + b + ia) is equal to (1+ b − ia) (a) 1 (b) 2 (c) b + ia (d) a + ib 16. If b + ic = (1+ a z) and a2 + b2 + c2 = 1, then equals a − ib a − ib a + ib a + ib 1 1 + − iz iz
  • 79. 4 168 6 20 i 5 12 π (a)(b)(c) (d) 1 − c 1 + c 1 − c 1 + c −3i 1 17. If 4 3i −1 = x + iy, then 3 i (a) x = 3, y = 1 (b) x = 1, y = 3 (c) x = 0, y = 3 (d) x = 0, y = 0 18. 2i equals (a)1 + i (b)1 − i (c) − 2i (d) None of these 19. If z is a complex number such that | |z ≠ 0 and Re( )z = 0, then (a) Re (z2 ) = 0 (b) Im (z2 ) = 0 (c) Re (z2 ) = Im (z2 ) (d) None of these 20. If (x + iy)1 3/ = a + ib, then x + y is equal to a b (a) 2 (a2 − b2 ) (b) 4 (a2 − b2 ) (c) 8 (a2 − b2 ) (d) None of these 21. If 8iz3 + 12z2 − 18z + 27i = 0,then 32. If x + iy = u + iv , then x2 + y2 is equal to 2u − iv (a) | |z = (b) | |z = (c) | |z = 1(d) | |z = 3(a) 1 (b) −1 (c) 0 (d) None of these 22. If z = 1+ i, then the multiplicative inverse of z2 is 33. If (1+ i) (1+ 2 ) (1i + 3 )... (1i + ni) = α + iβ, then i i 2 (a)1 − i (b) (c) − (d) 2i 2 5 10⋅ ⋅... (1+ n )is equal to 2 2 2 2 (a) α − βi (b) α − β 1+ i 3 1− i 3 (c) α2 + β2 (d) None of these 23. If = x + iy, then (x y, )is equal to 1− i 1+ i 34. The principal argument of the complex number (a) (0, 2) (b) (− 2, 0) (c) (0, −2) (d) None of these 24. The multiplicative inverse of (6 + 5 )i 2 is 11 60 11 60 (a) (c) −(d) (a) − i (b) − i 60 61 61 61 9 60 35. The complex number i + 3 in polar form can be (c) − i (d) None of these 61 61 written as 3 + 4i (a) 1 sin π + icos (b) 2 cos π + isin 25. The multiplicative inverse ofis 2 6 6 6 6 4 − 5i 1 π π 8 31 8 31 (c) 2 sin 6 + icos 6 (d) 4 cos 6 + icos 6 (a) − + i (b) − i 25 25 25 25 (c) 8 31i (d) None of these 36. If z is a complex number, then − − 25 25 (a) |z2 | > | |z 2 (b) |z2 | = | |z 2 (c) |z2 | < | |z 2 (d) |z2 | ≥ | |z 2 ) ( ( ) ) ( 1 3 1 2 3 5 2 + + + − − i i i i is 19 12 π ( ) b − 7 12 π
  • 80. 169 z 26. If z = x + iy lies in III quadrant, thenalso lies in 37. If z = x + iy x, and y are real, then | |x + | y|≤ k z| |, 27. The conjugate complex number of is the least value and greatest value of |z + 1|are (a) 1, 6 (b) 0, 6 (c) 2, 8 (d) None of these 39. If z satisfies |z + 1|< | z − 2|, then w = 3z + 2 + i satisfies (a) |w + 1| < |w − 8| (b) |w + 1| < |w − 7| (c) |w + w| > 7 (d) |w + 5| < |w − 4| 28. 40. For any complex number z, the minimum value of | z | + |z − 1|is (a) 1 (b) 0 (c) (d) 29.41. The greatest positive argument of complex number satisfying | z − 4| = Re ( )z , is 1− ix 2 2 (a) 30. If = a − ib and a + b = 1, where a and b are 1+ ix (c) real, then x is equal to 2a 2b (a)(b)42. The square roots of − +2 2 3 i are (1 + a)2 + b2 (1 + a)2 + b2 (a) ± (1 + 3 )i (b) ± (1 − 3 )i 2a 2b ± ( 1− + (c)(d)(c) 3 )i (d) None of these 31. The modulus ofis (a) (b) (c) (d) (d) None of these III quadrant, if (a) x > y > 0 (c) y < x < 0 z (b) x < y < 0 (d) y > x > 0 where k is equal to (a) 1 (b) 2 (c) 3 (d) None of these 38. If z is any complex number such that | z + 4 |≤ 3, then (1 + b)2 + a2 (1 + b)2 + a2 (3 + 2i)2 43. If z = a + ib, where a > 0, b > 0,then ( ) b ( ) ||z a b ≥ + 1 2 ( ) 4 3 − i 11 5 π 3 π 2 (b) 2π 3 (d) π 4 ) a ( ) ( ||z a b ≥ − 1 2 ( ) c ( || ) z a b < + 1 2
  • 81. 4 170 π 2 π 3 π 6 π 6 2 18 44. If for the complex numbers z1 and z2, |1− z z1 2 |2 − |z1 − z2 |2 = k (1− | z1 |2 ) (1− | z2 |2 ), then k is equal to (a) 1 (b) −1 (c) 2 (d) 4 45. The range of real number α for which the equation z + α| z − 1| + 2i = 0 has a solution, is (a) (b) (c) 0, 2 (d) , − 2 2 , 46. If | |z = 2 and locus of 5z − 1 is the circle having radius aand z1 2 + z2 2 − 2z z1 2 cos θ = 0, then | z1 |:| z2 | is equal to (a) a:1 (b) 2a:1 (c) a:10 (d) None of these 47. The maximum value of | z | when z satisfies the (a) 48. If z1 ≠ z2 and | z1 + z2 | = (a) atleast one of z1, z2 is unimodular (b) z1 ⋅ z2 is unimodular (c) z1 ⋅ z2 is non-unimodular (d) None of the above 49. If 1,ω ω, 2 ,,ω n − 1 are the n, nth roots of unity and z1, z2 are any two complex numbers, then n−1 ∑ | z1 + ω k z2 |2 is equal to k =0 (a) n z[| 1|2 + |z2| ]2 (b) (n − 1)[|z1|2 + |z2| ]2 (c) (n + 1)[|z1|2 + |z2| ]2 (d) None of these 50. If z satisfies the equation | z | − z = 1+ 2i, then z equal to is (a) + 2i (b) − 2i 3 3 (c) 2 − i (d) 2 i 2 2 51. If 3 + i = (a + ib) (c + id ), then tan −1 b + tan −1 d a c is equal to (a) + 2nπ, for some integer n (b) − + nπ, for some integer n (c) + nπ, for some integer n (d) − + 2nπ, for some integer n 52. If z is any complex number satisfying | z − 1| = 1, then which of the following is correct? (a) arg (z − 1) = 2 arg ( )z (b) 2 arg ( )z = 2 3/ arg (z2 − z) (c) arg (z − 1) = arg (z + 1) (d) arg ( )z = 2arg (z + 1) 53. If z = −1, then the principal value of arg (z2 3/ ) is equal to (a) (b) or 2π or (c)(d) π z − z π 54. If z1 = 8 + 4i z, 2 = 6 + 4i and arg 1 = , then z − z2 4 z satisfies (a) |z − 7 − 4i | = 1 (b) |z − 7 − 5i | = (c) |z − 4i | = 8 (d) |z − 7i | = 55. If arg ( )z < 0, then arg (−z)− arg (z) is equal to (a) π (b) − π (c) − (d) 56. If z is a point on the argand plane such that | z − 1| = 1 z − 2 , then is equal to z (a) tan (arg z) (b) cot (arg z) (c) itan (arg z) (d) None of these 57. If C 2 + S 2 = 1, then 1 + C + iS is equal to 1+ C − iS (a) C + iS (b)C − iS (c) S + iC (d) S − iC 58. The imaginary part of (z − 1) (cosα − isin α) + (z − 1)−1 × (cosα + isin α)is zero, if (a) |z − 1| = 2 (b) arg (z − 1) = 2α (c) arg (z − 1) = α (d) | |z = 1 8 sin + icos 8 8 condition z z + = 2 2is 3 1 − (c) 3 3 2 3 2 , ∪ 5 5 5 2 5 2 , 5 ( ) b 3 1 + ) d ( 2 3 + z z 2 1 1 1 + ,then π 3 5 3 π
  • 82. 171 2 2 2 2 59. The value ofis 8 sin − icos 8 8 (a) −1 (b) 0 (c) 1 (d) 2i 4 π 4 π 1 + a 3n 60. If a = cos + isin , then the value of is 3 3 2 (a) (−1)n (b) ( − 1 3n )n n (d) (−1)n + 1 2 1+ i 8 1− i 8 61. The value of is equal to 2 2 (a) 4 (b) 6 (c) 8 (d) 2 62. The value of ∑10 sin 2πk − icos 2πk is k = 1 11 11 (b) −1 (c) i (a) 1 (d) −i 63. The value ofis 10 (a)(1 + i) (c) (1 + i) (d) None of these 64. The value of (cos 2θ − isin 2θ)7 (cos 3θ + isin 3θ)−5 is (cos 4θ + isin 4θ)12 (cos 5θ + isin 5 )θ −6 (a) cos 33θ + i sin 33θ (b) cos 33θ − i sin 33θ (c) cos 47θ + i sin 47θ (d) cos 47θ − i sin 47θ 65. (cos 2θ + isin 2 )θ −5 (cos 3θ − isin 3 )θ 6 (sin θ − icos θ)3 is equal to (a) cos 25θ + i sin 25θ (b) cos 25θ − i sin 25θ (c) sin 25θ + i cos25θ (d) sin 25 θ − i cos 25θ 66. The principal value of the arg ( )z and | z | of the (1+ cosθ + isin θ)5 complex number z =are (cosθ + isin θ)3 (a) − θ , 32 cos5 θ (b) θ , 32 cos5 θ 2 2 (c) − θ ,16 cos4 θ (d) None of these 67. If z = cos θ + isin θ then (a) zn + 1 = 2cos nθ (b) zn + 1 = 2n cos nθ nn zz n 1 n n 1 n (c) z − n = 2 i sin nθ (d) z − n = (2 )i sin nθ z z 2n z − 1 68. If z = cos θ + isin θ then , where n is an z2n + 1 integer, is equal to (a) icotnθ (b) itannθ (c) tannθ (d) cotnθ 69. If a = cos 2α + isin 2 ,α b = cos 2β + isin 2 ,β c = cos 2γ + isin 2γ and d = cos 2δ + isin 2δ, then 1 abcd + is equal to abcd (a) 2 cos (α + β + γ + δ) (b) 2cos (α + β + γ + δ) (c) cos (α + β + γ + δ) (d) None of these m 70. e2mi cot−1 p ⋅ pi + 1 is equal to pi − 1 (a) 0 (b) 1 (c) − 1 (d) None of these 71. If α = cos 8 isin 8 , then 11 11 Re (α + α2 + α3 + α4 + α5 )is equal to (a) (b) − (c) 0 (d) None of these 72. If ω (≠ 1) is cube root of unity satisfying 1 1 1 2 + + = 2ω and a + ω b + ω c + ω 1 1 1 + + = 2 ω, then the value of 2 2 2 a + ω b + ω c + ω 1 1 1 + + is equal to a + 1 b + 1 c + 1 2 5 2 75 4 75 30 30 04 ) (cos sin .( cos sin ) ° °+ °+ ° i i (b) 10 2 (1 ) − i
  • 83. 4 172 (a) 2 (b) −2 (c) − 1 + ω2 (d) None of these 73. Ifα and β are the roots of the equation x2 − 2x + 4 = 0, then the value ofα6 + β6 is (a) 64 (b) 128 (c) 256 (d) None of these 1 74. If z is any complex number such that z + = 1, then z the value of z99 + 1 is 99 z (a) 1 (b) −1 (c) 2 (d) −2 75. The value of (− 1+ −3)62 + −( 1− − 3)62 is (a) 262 (b) 264 (c) −262 (d) 0 76. If ω is a cube root of unity, then (3 + 5ω + 3ω 2 )2 (3 + 3ω + 5ω 2 )2 is equal to (a) 4 (b) 0 (c) − 4 (d) None of these 1+ i 3 6 1− i 3 6 77. The value of is 1− i 3 1+ i 3 (a) 2 (b) −2 (c) 1 (d) 0 78. If z1 = 3 + i 3 and z2 = 3 + i , then the complex z 1 50 number lies in the quadrant number z2 (a) I (b) II (c) III (d) IV 79. If x = a + b, y = aω + bω 2 , z = aω 2 + bω, then xyz is equal to (a) (a + b)3 (b) a3 + b3 (c) a3 − b3 (d) (a + b)3 + 3ab a( + b) 80. If 1, ω and ω 2 are the cube roots of unity, then the value of (1+ ω)3 − (1+ ω 2 )3 is (a) 2ω (b) 2 (c) −2 (d) 0 81. If 1, ω and ω 2 are the three cube roots of unity and α β, and γ are the roots of p p, < 0, then for any x y, xα + yβ + zγ and z, the expression equals xβ + yγ + zα (a) 1 (b) ω (c) ω2 (d) None of the above 82. The value of the expression 1 2( − ω) (2 − ω 2 ) + 2 3( − ω) (3 − ω 2 )+ .... + (n − 1) (n − ω) (2 − ω 2 ), where ω is an imaginary cube root of unity, is n n( + 1) 2 n n( + 1) 2 (a) (b) − n 2 2 n n( + 1) 2 (c) + n (d) None of these 2 83. If α and β are the complex cube roots of unity, then α3 + β3 + α−2 β−2 is equal to (a) 0 (b) 3 (c) − 3 (d) None of these 84. If ω is a complex cube root of unity, then (1− ω + ω 2 ) (1− ω 2 + ω 4 ) (1− ω 4 + ω 8 ) (1− ω 8 + ω16 )is equal to (a) 12 (b) 14 (c) 16 (d) None of these 85. If x = ω − ω 2 − 2, then the value of x4 + 3x3 + 2x2 − 11x − 6 is (a) 1 (b) − 1 (c) 2 (d) None of these 86. If 1, ω and ω 2 are the three cube roots of unity, then (1+ ω) (1+ ω 2 ) (1+ ω 4 ) (1+ ω 8 ) … to 2n factors is equal to (a) 1 (b) − 1 (c) 0 (d) None of these 87. If x = a + b + c y, = aα + bβ + c and z = aβ + bα + c where, α and β are complex cube roots of unity, then xyz is equal to (a) 2 (a3 + b3 + c3 ) (b) 2 (a3 − b3 − c3 ) (c) a3 + b3 + c3 − 3abc (d) a3 − b3 − c3
  • 84. 173 88. The common roots of the equations z3 + 2z2 + 2z + 1= 0and z1985 + z100 + 1= 0are (a) −1, ω (b) −1, ω2 (c) ω ω, 2 (d) None of the above 89. The values of (16)1 4/ are (a) ± 2, ± 2i (b) ± 4, ± 4i (c) ±1, ± i (d) None of the above 90. If p q, and r are positive integers and ω is an imaginary cube root of unity and f x( ) = x3p + x3q + 1 + x3r + 2 , then f (ω)is equal to (a) ω (b) −ω2 (c) 0 (d) None of the above 91. If1,α α, 2 ,,αn − 1 are the n, nth roots of unity, then n − 1 1 ∑ is equal to i = 1 2 − αi (b) (n − 2)⋅ 2n (n − 2)⋅ 2n − 1 (c) (d) None of these 2n − 1 1+ i 92. The triangle formed by the points 1,and i as vertices in the argand diagram is a/an (a) scalene (b) equilateral (c) isosceles (d) right angled 93. If the points represented by complex numbers z1 = a + ib z, 2 = a′ + ib′ and z1 − z2 are collinear, then (a) ab′ + a b′ = 0 (b) ab′ − a b′ = 0 (c) ab + a b′ ′ = 0 (d) ab − a b′ ′ = 0 94. If α + β =i tan −1 z z, = x + iy and α is constant, then the locus of z is (a) x2 + y2 + 2x cot 2α = 1 (b) (x2 + y2 ) cot 2α = 1 + x (c) x2 + y2 + 2y tan 2α = 1 (d) x2 + y2 + 2x sin 2α = 1 95. The equation | z + 1− i| = |z + i − 1|represents (a) a straight line (b) a circle (c) a parabola (d) a hyperbola 96. If the roots of (z − 1)n = i z( + 1)n are plotted in the argand plane, they as (a) on a parabola (b) concyclic (c) collinear (d) the vertices of a triangle 97. Locus of the point z satisfying the equation |iz − 1| + | z − i| = 2is (a) a straight line (b) a circle (c) an ellipse (d) a pair of straight lines 98. For x1 ,x2 , y1 , y2 ,∈R, if 0< x1 ,< x2 , y1 = y2 and z1 = x1 + iy1, z2 = x2 + iy2 and z z z , then z1 , z2 , z3 satisfy (a) |z1| = |z2| = |z3 | (b) |z1| < |z2| < |z3| (c) |z1| > |z2| > |z3| (d) |z1| < |z3| < |z2| 99. The equation not representing a circle is given by 1 + z (a) Re 0 1 − z (b) zz + iz − iz + 1 = 0 z − 1 π (c) arg = z + 1 2 z − 1 (d) = 1 z + 1 100. If z1 , z2 and z3 are three complex numbers in AP, then they lie on (a) a circle (b) a straight line (c) a parabola (d) an ellipse 101. If the area of the triangle on the argand plane formed by the complex numbers −z iz z, , − iz is 600 sq units, then | |z is equal to (a) 10 (b) 20 (c) 30 (d) None of these 102. The complex numbers given by1− 3 4i, + 3i and 3 + i represent the vertices of (a) a right angled triangle (b) an isosceles triangle (c) an equilateral triangle (d) None of the above z + 2i 103. If Re = 0,then z lies on a circle with centre 2
  • 85. 4 174 z + 4 (a) (− 2, −1) (b) (− 2 1, ) (c) (2, −1) (d) (2,1) z + 2i 104. If Im = 0, then z lies on the curve z + 2 (a) x2 + y2 + 2x + 2y = 0 (b) x2 + y2 − 2x = 0 (c) x + y + 2 = 0 (d) None of the above 105. In the argand plane, all the complex numbers satisfying |z − 4i| + |z + 4i| = 10lie on (a) a straight line (b) a circle (c) an ellipse (d) a parabola 106. If z = x + iy and a is a real number such that | z − ai| =|z + ai|, then locus of z is (a) X-axis (b) Y-axis (c) x = y (d) x2 + y2 = 1 107. If P represents z = x + iy in the argand plane and | z − 1|2 + |z + 1|2 = 4,then the locus of P is (a) x2 + y2 = 2 (b) x2 + y2 = 1 (c) x2 + y2 = 4 (d) x + y = 2 108. All the roots of the equation a z1 3 + a z2 2 + a z3 + a4 = 3, where |ai |≤ 1, i = 1, 2, 3, 4, lie outside the circle with centre origin and radius (a) 1 (b) (c) (d) None of the above 109. The region of argand diagram defined by | z − 1| + |z + 1|≤ 4 is (a) interior of an ellipse (b) exterior of a circle (c) interior and boundary of an ellipse (d) None of the above 110. The locus of the complex number z in an argand plane satisfying the inequality | z − 1| + 4 2 log1/2 3|z − 1| − 2 > 1 where,|z − 1|≠ 3 is (a) a circle (b) interior of a circle (c) exterior of a circle (d) None of the above 111. The closest distance of the origin from a curve given as az + az + aa = 0(a is a complex number), is |a | (a) 1 (b) 2 Re ( )a Im ( )a (c) (d) |a | |a | 112. Let a be a complex number such that | |a < 1 and z1 , z2 ,…, zn be the vertices of a polygon such that zk = +1 a + a2 +…ak , then the vertices of the polygon lie within the circle 113. If z1 , z2 , z3 and z4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such that z 1 − z 4 = z 2 − z 3 and z4 − z1 = π, then the quadrilateral is a amp z2 − z1 2 (a) square (b) rhombus (c) cyclic quadrilateral (d) None of the above 114. The locus of z satisfying Im (z2 ) = 4 is (a) a circle (b) a rectangular hyperbola (c) a pair of straight lines (d) None of the above 115. The curve represented by Re (z2 ) = 4 is (a) a parabola (b) an ellipse (c) a circle (d) a rectangular hyperbola ) a ( | | | | a z a = − − 1 1 ) (b | | | | z a − = − 1 1 1 c ( ) z a a − − = − 1 1 1 | |1 (d) None of these 1 3 2 3
  • 86. 175 116. A point z is equidistant from three distinct points z1 , z2 , z3 in argand plane. If z z, 1 and z2 are z3 − z1 collinear, then arg will be (z1 , z2 , z3 in z3 − z2 anti-clockwise sense) (a)(b) (c)(d) Type 2. More than One Correct Option 117. If z1 , z2 , z3 and z4 are roots of the equation a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 where, a0 , a1 , a2 , a3 and a4 are real, then (a) z 1, z 2, z 3, z 4 are also roots of equation (b) z1 is equal to atleast one of z1, z2, z3, z4 (c) −z1, − z2, − z3, − z4 are also roots of equation (d) None of the above 118. If z3 + (3 + 2i z) + −( 1+ ia) = 0 has one real root, then the value of a lies in the interval (a ∈R) (a) (−2 1, ) (b) (−1 0, ) (c) (0,1) (d) (−2 3, ) 119. The real value of θ for which the expression i + icosθ is a real number, is n n I n n I n n I n n I 120. Ifα is a complex constant such thatαz + z + a = 0has a real root, then (a) α + α = 1 (b) α + α = 0 (c) α + α = −1 (d) the absolute value of the real root is 1 121. If |z1 | = | z2 | = 1and arg z1 + arg z2 = 0,then (a) z z1 2 = 1 (b) z1 + z2 = 0 (c) z1 = z2 (d) None of the above 122. If |z1 | = 15and | z2 − 3 − 4i| = 5,then (a) |z1 − z2|min = 5 (b) |z1 − z2|min = 10 (c) |z1 − z2|max = 20 123. If |z − (1/ )|z = 1,then (d) |z1 − z2|max = 25 satisfying | z1 2 − z2 2 | = | z1 2 + z2 2 − 2z z1 2 |, then z1 is purely imaginary (b) z1 is purely real (a) z2 z2 (c) |arg z1 − arg z2| = π (d) |arg z1 − arg z2| = 125. If |z − 1| = 1,then (a) arg {(z − 1 − i z)/ }can be equal to −π /4 (b) (z − 2)/z is purely imaginary number (c) (z − 2)/z is purely real number (d) If arg (z) = θ, where z ≠ 0and θ is acute, then 1 − 2/ z = i tanθ 126. Let complex number z of the form x + iy satisfy 3z − 6 − 3i π arg and |z − 3 + i| = 3. Then, the 2z − 8 − 6i 4 ordered pairs (x y, )are 4 2 4 2 (a) 4 − ,1 + (b) 4 + ,1 − (c) (6, −1) (d) (0, −1) 127. If z = ω ω, 2 , whereω is a non-real complex cube root of unity, are two vertices of an equilateral triangle in the argand plane, then the third vertex may be represented by (a) z = 1 (b) z = 0 (c) z = − 2 (d) z = − 1 128. If amp (z z1 2 ) = 0and | z1 | = | z2 | = 1,then (a) z1 + z2 = 0 (b) z z1 2 = 1 (c) z1 = z2 (d) None of these π 2 π 6 π 4 2 3 π ) a ( || max z = + 5 1 2 (b) || min z = −1 5 2 (c) || max z = − 2 5 2 (d) || min z = − 5 1 2 If z z 1 2 and aretwocomplexnumbers ( ) z z 1 2 ≠ 5 5 5 5
  • 87. 4 176 129. If z is complex number satisfying z + z−1 = 1, then zn + z−n ,n ∈N has the value (a) 2( 1)− n , when nis a multiple of 3 (b) (−1)n − 1 , when nis not a multiple of 3 (c) (−1)n + 1 ,when nis a multiple of 3 (d) 0, when nis not a multiple of 3 130. Let z1 , z2 be two complex numbers represented by points on the circle | |z = 1and | z | = 2 respectively, then (a) max |2z1 + z2| = 4 (b) min |z1 − z2| = 1 (c) z≤ 3 (d) None of these 131. ABCD is a square, vertices being taken in the anti-clockwise sense. If A represents the complex number z and the intersection of the diagonals is the origin, then (a) B represents the complex number iz (b) D represents the complex number iz (c) B represents the complex number iz (d) D represents the complex number − iz 132. If z0 , z1 represent point P Q, on the locus | z − 1| = 1 and the line segment PQ subtend an angle at the point z = 1, then z1is equal to i (a) 1 + i z( 0 −1) (b) z0 − 1 (c)1 − i z( 0 −1) (d) i z( 0 −1) 133. If | z1 | = |z2 | = |z3 | = 1 and z1 , z2 and z3 are represented by the vertices of an equilateral triangle, then (a) z1 + z2 + z3 = 0 (b) z z z1 2 3 = 1 (c) z z1 2 + z z2 3 + z z3 1 = 0 (d) None of these Type 3. Assertion and Reason Direction (Q. Nos. 134-144) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II isa correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II isnot a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true. 134. Consider z1 = 1, z2 = 2and z Statement I If z 1 + 2 z 2 of z z2 3 + 8z z3 1 + 27z z1 2 Statement II z1 + z2 + z 135. Statement I 7+ >i 5 + i Statement II Cancellation laws does not hold true in complex numbers. 136. Statement I If z1and z2 are two complex numbers z1 = 0. such that |z1 | = | z2 | + | z1 − z2 |,then Im z2 Statement II arg ( )z = 0⇒ z is purely real. 1 137. Statement I If z + = 1(z ≠ 0 is a complex z 5 +1 number), then the maximum value of | |z is. 2 1 Statement II On the locus z + = 1, the farthest z 5 +1 distance from origin is. 2 138. Statement I The number of complex numbers z satisfying | |z 2 + a z| | + b = 0 (a b, ∈R)is atmost 2. Statement II A quadratic equation in which all the coefficients are non-zero can have atmost two roots. 139. Statement I Ifcos (1− i) = a + ib, where a b, ∈R and 1 1 1 1 i = −1, then a = e + cos1, b = e − sin .1 2 e 2 2 Statement II eiθ = cos θ + isin θ 140. Statement I The product of all values of (cos α + isin α)3 5/ is cos 3a + isin 3 .a StatementII The product of fifth roots of unity is 1. 141. Statement I The locus of the centre of a circle which touches the circles | z − z1 | = aand | z − z2 | = b externally (z z, 1 and z2 are complex numbers) will be hyperbola. Statement II | z − z1 | − |z − z2 |< | z2 − z1 | ⇒ z lies on hyperbola. 142. Statement I Consider an ellipse having its foci at A z( 1) and B z( 2) in the argand plane. If the eccentricity of the z 2 1 1 + 3 3 = . z3 3 6 + = , thenthevalue is36. z z z3 3 1 2 ≤ + +
  • 88. 177 ellipse is e and it is known that origin is an interior point of the ellipse, then | z1 + z2 | |z1 + z2 | e , | z1 | + |z2 | |z1 | + |z2 | Statement II If z0 is the point interior to curve |z − z1 | + |z − z2 | = λ | z0 − z1 | + |z0 − z2 |< λ 143. Statement I The equation | z − i| + | z + i| = k k, > 0, can represent an ellipse, if k > 2i. StatementII | z − z1 | + | z − z2 | = k, represents an ellipse, if | |k > | z1 − z2 |. 144. Statement I The equation zz + az + az + λ = 0, where a is a complex number represents a circle in argand plane, if λ is real. Statement II The radius of the circle zz + az + az + λ = 0is aa − λ. Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 145-147) Consider the complex 147. One of the possible argument of complex number numbers z1 and z2 satisfying the relation i z( 1 / z2 ) is z1 + z 1 (a) (c) 0 (d) None of these 145. Complex number z z1 2 is Passage II (Q. Nos. 148-150) If z satisfy the relation (a) purely real (b) purely imaginary 2 | z − {(α − 7α +11) + i}| =1,α ∈R. (c) zero (d) None of these π Also, arg ( )z ≥ is satisfied by atleast one z. 146. Complex number z1 / z2 is 2 (a) purely real 148. The values of α lies in the interval (b) purely imaginary (c) zero (a)[2, 6] (b) 1, 72 (c)[2, 5] (d)[− 4, −2] (d) None of the above 149. Maximum value of | z − i|is AB and points A andE lies in the opposite side of lineBC. If A B, and C are represent by the complex numbers 1, (a) (b) (c) (d) ω and ω2 respectively. 150. Maximum value of arg ( )z , for which | z − i| is151. Angle between AC and DE is equal to maximum, is (a) (b) (c) (d) (a) π − tan−1 4 (b) π − tan−1 9 9 4 (c) π − tan−1 1 (d) π − tan−1 4 9 5 PassageIII (Q. Nos. 151-153) On the sides AB and BC of a ∆ABC, squares are drawn with centres D and E such that points C and D lies on the same side of line AB Type 5. Match the Columns 152. The length of DE is (a) (b) 3 2 153. The length of AE is (a) (b) 2 2 (c) 2 (c) 3 − 3 (d) 6 (d) 3 + 3 3 3 − 3 3 + 3 z z 2 1 2 2 2 = + .
  • 89. 4 178 154. Match the following: 157. 155. For the equation z6 − 6z + 20 = 0, match the items of Column I with that of Column II. A. The number of the roots in the first p. 1 quadrant can be B.The number of the roots in the second q. 2 quadrant can be C.The number of the roots in the third r. 3 quadrant can be Type 6. Single Integer Answer Type Questions 158. If x = a + bi is a complex number such that x2 = 3 + 4i and x3 = 2 + 11i, where i = −1, then (a + b)equals _______. 159. If the complex numbers z is simultaneously satisfy 161. If z is a complex number satisfying z4 + z3 + 2z2 + z + 1= 0, then | |z is equal to _____ . 162. ABCD is a rhombus. Its diagonals AC and BD z − 12 5 z − 4 intersect at the point M and satisfy BD = 2AC. Its equations = , =1, then Re ( )z is ___. z − 8i 3 z − 8 points D and M represent the complex numbers 1+ i and 2 − i respectively. Then, the complex number 1 λ 160. If | |z ≥ 3,then the least value of z + is ,where λ represented by A is z,has Re ( )z = λ{ 1 , λ 2},then the z 3 value of λ1 + λ 2 is __________ . is __________ . C. Ifωk, k ∈ I; 0 ≤ k ≤ n − 1are the nth r. roots of unity, then the maximum value of ( !)n 1/n ω(n−1 2)/ is ω D. If z1, z2, z3 are the affixes points and A, s. B and C are lying on circle centred at origin. If the altitude is drawn from vertex A to base BC, such that it meets the circumcircle at P z( ), then zz1 + z z2 3 is ω2 − + ± ≠± = − − − − − = − π + + = ≠ ∈ − − − + + − = − − π Matchthefollowing: = − + = π π = + − = π π = + = + π π − = + = Match the following: ∞ −− −− = = = ∑ ω−
  • 90. 4 179 Entrances Gallery JEE Advanced/IIT JEE 1. Match the following : [2014] 2k 2k Let zk = cos 10 + i sin 10 ; k = 1 2, ,..., 9. Column I Column II A. For each zk, there exists a zj such zk ⋅ zj = 1 p. True B. There exists k ∈{1, 2, ..., 9}, such that z1 ⋅ z = zk has no solution z in the set of complex numbers q. False C.equals r. 1 10 Codes A B C D A B C D (a) p q s r (b) q p r s (c) p q r s (d) q p s r 2. Let complex numbers α and lie on the circles z0 = x0 + iy0 satisfies the equation 2 z = r2 + 2 , [2013] (a) (c) (d) 3. Let w = and P = {wn :n = 1, 2, 3,...}. Further 2 H1 = z ∈C : Re ( )z > 1 and 2 z ∈C :Re ( )z − 1 , where C is the set of all H2 2 complex numbers. Ifz1 ∈P∩ H1, z2 ∈P ∩ H2 and O represents the origin, then ∠z Oz1 2 equals [2013] (a) Passage (Q. Nos. 4-5) Let S = S1 ∩ S2 ∩ S3 where, S1 = {z ∈C : z < 4}, z −1+ 3 i S2 z ∈C :Im 0 1− 3 i and S3 {z ∈C :Re z > 0}. 4. The area of S equals [2013] z ∈S 6. Let z be a complex number such that the imaginary part of z is non-zero and a = z2 + z + 1is real. Then, a cannot take the value [2012] (a) − 1 (b) (c) (d) 7. If z is any complex number satisfying z − 3 − 2i ≤ 2, then the minimum value of 2z − 6 + 5i is________ .[2011] π 2i 8. Let ω = e 3 and a b c x y z, , , , , be non-zero complex numbers, such that a + b + c = x a, + bω + cω 2 = y and a + bω2 + cω = z. Then, the value of is ________ . [2011] 9. Let z1 andz2 be two distinct complex numbers and z = (1− t z) 1 + tz2 for some real number t with 0< t <1. If arg (w) denotes the principal argument of a non- zero complex number w, then [2010] a z − z1 + z − z2 = z1 − z2 b arg (z − z1) = arg (z − z2) z − z1 z − z1 c = 0 z2 − z1 z2 − z1 1 − z1 1 − z2 ... 1 − z9 (x − x0 )2 + ( y − y0 )2 = r2 and (x − x0 )2 + ( y − y0 )2 = 4r2 , respectively. If − = ∑ π 1 7 a) ( 10 3 π ) b ( 20 3 π ) c ( 16 3 π d ( ) 32 3 π min z i − −3 1 equals [2013] (a) 2 3 2 − ) b ( 2 3 2 + ) c ( 3 3 2 − d) ( 3 3 2 + x y z b a c 2 2 2 2 2 2 + + + + 0 2 1 3 then α equals 1 2 ( ) b 1 2 i + 3 1 3 3 4
  • 91. 180 2. 1 2 5 2 (d) 2 2 + 1 d arg (z − z1) = arg (z2 − z1) 10. Match the statement in Column I with those in Column II. Ø Here, z takes the values in the complex plane and Im z and Re z denote, respectively the imaginary part and the real part of z. [2010] JEE Main/AIEEE 11. A complex number z is said to be unimodular, if | z | = 1. Suppose z1 and z2 are complex numbers such z1 − 2z2 thatis unimodular and z2 is not 2 − z z1 2 unimodular. Then, the point z1 lies on a [2015] (a) straight line parallel to X-axis (b) straight line parallel to Y-axis (c) circle of radius 2 (d) circle of radius 12. If z is a complex number such that z ≥ 2, then the minimum value of z + [2014] (a) is equal to (b) lies in the interval (1, 2) (c) is strictly greater than than but less than (d) is strictly greater 13. If z is a complex number of unit modulus and 1+ z argumentθ, then arg equals [2013] 1+ z (a) − θ (b) − θ (c) θ (d) π − θ 2 z 14. If z ≠ 1and is real, then the point represented by z − 1 the complex number z lies [2012] (a) either on the real axis or on a circle passing through theorigin (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing throughthe origin (d) on the imaginary axis 15. Let α and β be real and z be a complex number. If z2 + α z + β = 0 has two distinct roots on the line Re ( )z = 1, then it is necessary that (a) β ∈ −( 1, 0) (b) β = 1 (c) β ∈(1, ∞) (d) β ∈(0,1) 16. If ω (≠ 1) is a cube root of unity and [2012] (1+ ω)7 = A + Bω. Then, (A B,) equals (a) (1,1) (b) (1, 0) (c) (−1 1, ) (d) (0,1) [2011] 17. The number of complex numbers z such that z − 1 =| z + 1|= z − i equals (a) 0 (b)1 (c) 2 (d) ∞ [2010] 18. If α and β are the roots of the equation x2 − x + 1= 0, thenα2009 + β2009 is equal to [2010] (a) − 2 (b) − 1 (c)1 (d) 2 4 19. If z − = 2, then the maximum value of z is z equal to [2009] (a) 3 + 1 (b) 5 + 1 (c) 2 20. The conjugate of a complex number is . Then, i − 1 that complex number is [2008] 1 1 1 1 (a) − (b) (c) − (d) i − 1 i + 1 i + 1 i − 1 21. If z + 4 ≤ 3, then the maximum value of z + 1 is [2007] (a) 4 (b) 10 (c) 6 (d) 0 22. The value of ∑10 sin 2kπ + icos 2k is [2006] k = 1 11 11 (a) 1 (b) − 1 (c) − i (d) i 23. If z2 + z + 1= 0, where z is complex number, then + − = + = − + = ω= =ω− ω ≤ ω= ω =ω+ ≤ ≤ 5 2 3 2
  • 92. 4 181 2 2 2 the value of z + 1z z2 + z12 z3 + z13 2 + ... z6 + z 1 6 is [2006] (a) 54 (b) 6 (c) 12 (d) 18 24. If the cube roots of unity are 1, ω ω 2 , then the roots of the equation (x − 1)3 + 8 = 0, are [2005] (a) −1,1 + 2ω 1 + 2ω 2 (b) −1,1 − 2ω 1 − 2ω 2 (c) −1, −1, −1 (d) −1, −1 + 2ω, −1 − 2ω2 25. If z1 and z2 are two non-zero complex numbers such that z + z , then arg (z1 ) − arg (z2 ) is equal to [2005] (a) − (c) −π (d) w = and w = 1, then 26. If z lies on [2005] (a) a parabola (b) a straight line (c) a circle (d) an ellipse 27. Let z and w be two complex numbers such that z + iw = 0and arg (zw)= π. Then, arg ( )z equals [2004] (a)(b) (c)(d) 28. If z = x − iy and z1/3 = p + iq, then x y 2 2 / ( p + q )is equal to [2004] p q (a) 1 (b) − 1 (c) 2 (d) − 2 29. If |z2 − 1| = | |z 2 + 1, then z lies on [2004] (a) the real axis (b) the imaginary axis (c) a circle (d) an ellipse 30. Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle. Then, [2003] (a) a2 = b (b) a2 = 2b (c) a2 = 3b (d) a2 = 4b 31. If z and w are two non-zero complex numbers such that zw = 1, and arg ( )z − arg (w) = , then z w( ) is equal to [2003] (a) 1(b) − 1 (c) i (d) − i Other Engineering Entrances 34. If α and β are two different complex numbers with β − α β = 1, thenis equal to [Karnataka CET 2014] 1−αβ (a) 1/2 (b) 0 (c) −1 (d)1 35. If z, then z is equal to [Kerala CEE 2014] (a) 8 (b) 2 (c) 5 (d) 4 (e) 10 36. Let w ≠ ± 1 be complex number. If w = 1 and w − 1 = , then Re ( z )z is equal to[Kerala CEE 2014] w + 1 1 (a) 1 (b)(d) 0 (e) w + w 2 37. The value of z is minimum, when z equals [WB JEE 2014] −i (b) 45 + 3i (a) 2 ii (c)1 +(d)1 − 3 3 38. Convert (i + 1)/ cos π − isin π in polar form. 4 4 [J&K CET 2014] 4 z z 1 2 1 2 + = π 2 (b) 0 z z i − 3 π 4 3 4 π π 2 5 4 π 1 w + (c) Re()w z z i 2 2 3 + − + − 2 3
  • 93. 182 1 2 39. Let z1 ≠ z2 and z1 = z2 . If z1 has positive real part z2 + z1 and z2 has negative imaginary part. Then, z1 − z2 may be [Manipal 2014] (a) 0 (b) real and positive (c) real and negative (d) None of these 40. If z = e2π /3 , then 1+ z + 3z2 + 2z3 + 2z4 + 3z5 is equal to [Kerala CEE 2014] (a) − 3eπ i / 3 (b) 3eπ i/ 3(c) 3e2π i/ 3 (d) − 3e2π i/ 3 (e) 0 x 1+ i 32. If = 1, then [2003] 1− i (a) x = 4n, where nis any positive integer (b) x = 2n, where nis any positive integer (c) x = 4n + 1, where nis any positive integer (d) x = 2n + 1, where nis any positive integer 33. If ω is an imaginary cube root of unity, then (1+ ω − ω 2 )7 equals [2002] (a) 128 ω (b) −128 ω (c) 128 ω2 (d) −128 ω2 41. If ω, ω 2 are the cube roots of unity, then roots of equation (x − 1)3 + 5 = 0are [ RPET 2014] (a) −5, − 5 ω − 5ω 2 (b) −4 1, − 5 ω 1 − 5ω 2 (c) 6 1, − 5 ω 1 + 5ω 2 (d) None of the above 42. The complex number z = x + iy, which satisfies the z − 3i equation= 1, lie on [BITSAT 2014] z + 3i (a) the X-axis (b) the straight line y = 3 (c) a circle passing through the origin (d) None of the above 43. If the complex numbers z1, z2 and z3 denote the vertices of an isosceles triangle, right angled at z1, then (z1 − z2 )2 + (z1 − z3 )2 is equal to [Kerala CEE 2014] (a) 0 (b) (z2 + z3)2 (c) 2 (d) 3 (e) (z2 − z3)2 44. Let z1 and z2 be two fixed complex numbers in the argand plane and z be an arbitrary point satisfying z − z1 + z − z2 = 2 z1 − z2 . Then, the locus of z will be [WB JEE 2014] (a) an ellipse (b) a straight line joining z1 and z2 (c) a parabola (d) a bisector of the line segment joining z1 and z2 45. Let z1 be a fixed point on the circle of radius 1 centred at the origin in the argand plane and z1 ≠ ± 1. Consider an equilateral triangle inscribed in the circle with z1, z2 and z3 as the vertices taken in the counter clockwise direction. Then, z1 z2 z3 is equal to [WB JEE 2014] (a) z1 2 (b) z1 3 (c) z1 4 (d) z1 46. Suppose that z1, z2 and z3 are three vertices of an equilateral triangle in the argand plane. Let α = ( 3 − i) and β be a non- zero complex numbers. The pointsαz1 + β α z2 +β α, z3 +β will be [WB JEE 2014] (a) the vertices of an equilateral triangle (b) the vertices of an isosceles triangle (c) collinear (d) the vertices of a scalene triangle 47. In the argand plane, the distinct roots of 1+ z + z3 + z4 = 0 (z is a complex number) represent vertices of (a) a square (b) an equilateral triangle (c) a rhombus (d) a rectangle [WB JEE 2014] 48. If z1 = 2 2 1( + i) and z2 = +1 i 3, then z z1 2 2 3 is equal to [Kerala CEE 2013] (a) 128i (b) 64i (d) −128i (e) 256 (c) −64i 49.The value of (1+ i)3 + (1− i)3 is equal to[J&K CET 2013] (a) 1 (b) −2 (c) 0 (d) −4
  • 94. 4 183 50. If z(2− i2 3)2 = i( 3 + i)4 , then amplitude of z is [UP SEE 2013] (a) (b) (c)(d) None of these 51. If z1, z2 and z3 are complex numbers such that + = 1, then [AMU 2013] (a) 3 (b) 1 (c) greater than 3 (d) less than 1 52. If ( 3 i + 1)100 = 299 (a + ib), then a2 + b2 is equal to [RPET 2013] (a) 4 (b) 3 (c) 2 (d) 0 53. The value of (1+ 3 )i 4 + (1− 3 )i 4 is [RPET 2013] (a) − 16(b) 16 (c) 14 (d) − 14 54. If the fourth roots of unity are z1 , z2 , z3 and z4 ,then z1 2 + z2 2 + z3 2 + z4 2 is equal to [Karnataka CET 2013] (a) 0 (b) 2 (c) 3 (d) None of these 55. Among the complex number z satisfying condition z + 1− i ≤ 1, the number having the least positive argument is [OJEE 2013] (a) 1 − i (b) 1 + i (c) −i (d) None of the above 56. If z1 and z2 are two complex numbers such that z1 = z 2 + z 1 − z 2 , then [Manipal 2012] (a) Im z1 = 0 (b) Re z1 = 0 z2 z2 z z (c) Re 1 = Im 1 (d) None of these z2 z2 57. If the conjugate of (x + iy)(1− 2i)is 1+ i, then [Karnataka CET 2012] (a) x − iy = (b) x + iy = (c) x = = − (d) x 58. If = 325 (x + iy), where xand yare real, 2 2 then the ordered pair (x y, )is [WB JEE 2012] (a) (−3 0, ) (b) (0, 3) 1 3 (c) (0, −3) (d) , 2 2 59. If z − z + z + z = 2, then z lies on [AMU 2012] (a) a circle (b) a square (c) an ellipse (d) a line 60. If 2x = 3 + 5i, then the value of 2x3 + 2x2 − 7x + 72is [MP PET 2011] (a) 4 (b) −4 (c) 8 (d) −8 2 2 2 z1 61. Ifz1 + z2 = z1 + z2 , then is [MP PET 2011] z2 (a) purely real (b) purely imaginary (c) zero of purely imaginary (d) neither real nor imaginary 62. The value of[Karnataka CET 2011] (a) 20 (c) 63. If z1 and z2 are two non-zero complex numbers such z1 thatz1 + z2 = z1 + z2 , then arg is z2 [Kerala CEE 2011] (a) 0 (b) −π (c) − (e) π 64. If ω ≠ 1is a cube root of unity, then the sum of the series S = 1+ 2ω + 3ω 2 + ... + 3nω 3n − 1 is z z 2 3 1 1 + −π 6 π 6 z z z z 1 2 3 1 1 = = = z z z 1 2 3 + + is i i − − 1 1 2 1 5 3 3 50 +i i i − − 1 1 2 1 5 1 3 1 1 1 2 + + + i i is (b) 9 (d) 4 5
  • 95. 184 [WB JEE 2011] 3n (a)(b) 3n (ω − 1) (c)(d) 0 3n [Kerala CEE 2010] (a) − 20(b) −60 (c) −120 (d) 60 (e)156 68. The modulus of the complex number z such that z + 3 − i = 1 and arg ( )z = π, is equal to [Kerala CEE 2010] 73. (a) 1 (b) 2 (c) 9 (d) 4 (e) 3 69. If z , z ,...., z are complex numbers such that 1 2 n = 1,then z2 + z2 + ... + zn is [Kerala CEE 2010] (b) z1 + z2 + ... + zn 74. (d) n 1− i of z) [WB JEE 2010] (a) 2(1 + i) (b) (1 + i) 2 4 (c) (d) 1 − i If − π < arg ( )z < − , then arg ( )z − arg(−z)is [WB JEE 2010] (a) π (b) −π (c) π /2 (d) −π /2 The value of is equal to [VITEEE 2010] (a) i 1 + 3i (c)(d) 2 2 4 1+ cos θ2 − isin θ2 4n 70. 65. is equal to [UP SEE 2011] 1+ cos θ 2 + isin θ 2 (a) cos nθ − i sin nθ (b) cos nθ + i sin θ (c) cos 2nθ − i sin 2nθ (d) cos 2nθ + i sin 2nθ − 3 + i 3 66. If x =is a complex number, then the value z z 2 71. If z = r (cos θ + isin θ),then the value of+ is of (x2 + 3x)2 (x2 + 3x + 1)is [UP SEE 2011] z z [Kerala CEE 2010] (a) − (b) 6 (a) cos 2 θ (b) 2 cos 2 θ (c) 2 cos θ (d) 2 sin θ (c) − 18 (d) 36 (e) 2 sin 2 θ 67. If (x + iy)1 3/ = 2 + 3i, then 3x + 2yis equal to 72. If z = 4 , then z is (where, z is complex conjugate 1 ω− ω−1 1 − i If i z1 2 4 4 + = sin cos π π and i z2 3 3 = + cos , sin π π 3 then zz 2 1 is 2010] KeralaCEE [ (b) 6 a) ( 2 c) ( 6 d ( ) 3 (e) 2 3 + z z zn 2 1 = = =... equalto ) (a z zzz n 123 ... ) c ( 1 1 1 2 1 z z zn + + + ... (e) n cos sin sin cos 30 30 60 60 ° °+ °− ° i i (b) −i 1 3 − i
  • 96. 185 Work Book Exercise 4.1 Answers 1. (b) 2. (a) 3. (c) 4. (b) 5. (a) 6. (d) 7. (b) 8. (d) 11. (d) 12. (d) 13. (c) Work Book Exercise 4.2 14. (c) 1. (a) 2. (b) 3. (a) 4. (c) 5. (a) 6. (b) 7. (c) 8. (b) 11. (a) 12. (a) 13. (d) Work Book Exercise 4.3 14. (b) 15. (b) 1. (c) 2. (b) 3. (b) 4. (c) 5. (a) 6. (d) 7. (c) 8. (b) 11. (a) 12. (b) 13. (b) 14. (c) 15. (b) 16. (c) 17. (c) 18. (a 21. (c) 22. (a) 23. (c) Work Book Exercise 4.4 24. (a) 25. (d) 26. (c) 27. (b) 1. (c) 2. (d) 3. (b) 4. (b) 5. (a) 6. (b) 7. (a) 8. (a) 11. (b) 12. (c) 13. (c) 14. (a) 15. (a) 16. (c) 17. (a) 18. (b 21. (a) 22. (d) Target Exercises 23. (a) 24. (b) 25. (a) 26. (b) 27. (b) 28. (a 1. (b) 2. (a) 3. (c) 4. (d) 5. (b) 6. (c) 7. (c) 8. (c) 11. (b) 12. (d) 13. (c) 14. (a) 15. (c) 16. (d) 17. (d) 18. (a) 21. (a) 22. (c) 23. (c) 24. (d) 25. (c) 26. (c) 27. (d) 28. (c) 31. (a) 32. (a) 33. (c) 34. (c) 35. (b) 36. (b) 37. (b) 38. (b) 41. (d) 42. (a) 43. (b) 44. (a) 45. (a) 46. (c) 47. (b) 48. (b) 51. (c) 52. (a) 53. (b) 54. (b) 55. (a) 56. (c) 57. (a) 58. (c) 61. (d) 62. (c) 63. (a) 64. (d) 65. (c) 66. (a) 67. (a) 68. (b) 71. (b) 72. (a) 73. (b) 74. (d) 75. (c) 76. (c) 77. (a) 78. (a) 81. (c) 82. (b) 83. (b) 84. (c) 85. (a) 86. (a) 87. (c) 88. (c) 91. (a) 92. (c) 93. (b) 94. (a) 95. (a) 96. (c) 97. (a) 98. (d) 101. (b) 102. (d) 103. (a) 104. (c) 105. (c) 106. (a) 107. (b) 108. (c) 111. (b) 112. (c) 113. (c) 114. (b) 115. (d) 116. (a) 117. (a,b) 118. (a,b,d) 121. (a,c) 122. (a,d) 123. (a,b) 124. (a,d) 125. (a,b,d) 126. (a,b) 127. (a,c) 128. (b,c) 131. (a,d) 132. (a,c) 133. (a,b) 134. (b) 135. (d) 136. (a) 137. (a) 138. (d) 141. (d) 142. (d) 143. (d) 144. (a) 145. (b) 146. (b) 147. (c) 148. (c) 151. (c) 152. (a) 153. (b) 154. (*) 155. (**) 156. (***) 157. (****) 158. (3) 161. (1) 162. (4) * A → r; B → p; C → q; D → s ** A → p, q; B → p, q; C → p, q; D → p, q *** A → s; B → r; C → p; D → q **** A → r,s; B → p; C → q; D → p Entrances Gallery 1. (c) 2. (c) 3. (c,d) 4. (b) 5. (c) 6. (d) 7. (5) 8. (3) 11. (c) 12. (b) 13. (c) 14. (a) 15. (c) 16. (a) 17. (b) 18. (c) 21. (c) 22. (c) 23. (c) 24. (b) 25. (b) 26. (b) 27. (c) 28. (d) 31. (d) 32. (a) 33. (d) 34. (d) 35. (b) 36. (d) 37. (c) 38. (d) 41. (b) 42. (a) 43. (a) 44. (a) 45. (b) 46. (a) 47. (b) 48. (d) 51. (b) 52. (a) 53. (a) 54. (a) 55. (d) 56. (a) 57. (b) 58. (d)
  • 97. 186 61. (b) 62. (d) 63. (a) 64. (a) 71. (b) 72. (d) 73. (a) 74. (a) * A → q; B → p, t; C → p; D → s,q 65. (c) 66. (c) 67. (c) 68. (e)
  • 98. 4 187 Explanations Target Exercises i 584 (i 8 + i 6 + i 4 + i 2 + 1) i 584 1. 574 8 6 4 2 − 1 = 574 − 1 i (i + i + i + i + 1) i = i10 − 1 = − 1− 1 = − 2 2. i 57 + i1251 = (i 4 14) i + (i 4 311) i = i + 1i = i − i = 0 3. We have, i n + i n + 1 + i n + 2 + i n + 3 = i n (1 + i + i 2 + i 3 ) = i n (1+ i − 1 − i ) [i 3 = i 2 ⋅ i = −(1)⋅ i = − i ] = i n (0) = 0 4. Given expression = 1 + i 2 + i 4 + i 6 + ... + i 2n = 1 − 1 + 1 − 1 + ... + (− 1)n which cannot be determined unless n is known. 5. We have, i 1i 25 2 i16 ⋅ i 3 + i 241⋅ i 2 19 2 1 2 2 i⋅ i + i = (− i − i ) = −( 2 )i 2 = 4i 2 = − 4 i 4n ⋅ i − i 4n 4 n n = [i= (i ) = ( )1 = 1] 2 i 7. (1 − i ) n 1 − 1 n = (1 − i ) n (1 + i ) n (1 + i ) n − 2 (1 + i ) n − 2 (1 − i ) n − 2 n − 1 = 2 (− i )n − 1 = 2 (− 1)2 n − 1 This is positive and real, if is even. 2 n − 1 Let = 2λ 2 ∴ n = 4λ + 1 n n n 9. 1 + i 1 + i × 1 + i 1 − 1 + 2i 1 − i 1 − i 1 + i 1 + 1 = i n = − 1, if n = 2 1 − i 1 − i 2 13. We have, 1 + i 2 1 − i 2 1 − i 1 + i 2 2 (1+ i ) (1 − i ) 2i − 2i = (1 − i )2 + (1 + i )2 = − 2i + 2i  (1 + i )2 = 1 + i 2 + 2i = 2i 2 2 and (1 − i ) = 1 + i − 2i = − 2i = − 1 − 1 = − 2 14. Re ( 13+−ii)2 = Re 2i 3 + i = Re 15. Given that, a2 + b2 = 1 1 + b + ia (1 + b + ia)(1 + b + ia) ∴ = (1+ b − ia (1 + b − ia) (1 + b + ia) (1 + b)2 − a2 + 2ia (1 + b) = 2 2 1 + b + 2b + a i = (1 + 1) n= 2n (1 − i ) n 8. = (1 − i ) n (1 − i ) n − 2 × 10. − = + ⇒ =− ∴ 11. + = + × + + = − + =− + ∴ + − 12. − − = − + − + = − − + =− 6. − + − ⋅ = − − = − = − − = − = − = − + − + + = − =−
  • 99. 188 = b + ia 16. We have, b + ic = (1 + a z) b + ic ib − c a + ib 1 + c 17. x + iy = 6 i (3 i 2 + 3) + 3 i (4 i + 20) + 1 12( − 60i ) = 0 − 12 + 60i + 12 − 60i = 0 + 0i ∴ x = 0, y = 0 18. 2i = 1 + i 2 + 2i = (1 + i )2 = 1 + i 19. Let z = x + iy, where x y, ∈R ∴ x2 + y2 ≠ 0 and x = 0 ∴ z = 0 + iy = iy, y ≠ 0 ⇒ z2 = − y2 ⇒ Im (z2 ) = 0 20. We have, (x + iy)1 3/ = a + ib On cubing both sides, we get x + iy = (a + ib)3 = a3 + 3 a2 (ib) + 3 a ib()2 + (ib)3 = a3 + 3 a2 bi + 3 ab i2 2 + i b3 3 = a3 + 3 a bi2 − 3 ab2 − ib3 [i 3 = i 2 ⋅ i = − i] = a a(2 − 3 b2 ) + ib (3 a2 − b2 ) ∴ x = a a( 2 − 3 b2 ) and y = b(3 a2 − b2 ) ⇒ x = a2 − 3 b2 and y = 3 a2 − b2 a b x y 2 2 ∴ + = 4(a − b ) a b 21. We have, 8 iz3 + 12 z2 − 18z + 27i = 0 ⇒ 4z2 (2iz + 3) + 9 i (2iz + 3) = 0 ⇒ (2iz + 3) (4z2 + 9 i ) = 0 ⇒ 2iz + 3 = 0 or 4z2 + 9 i = 0 ∴ z = 22. Multiplicative inverse of z2 1 − i 1 + i Hence, (x y, ) = (0, − 2). 24. Let z = (6 + 5 )i 2 = 36 + 2(6)(5 i ) + 25 i 2 = 36 + 60i − 25 = 11 + 60i 1 − i 1 + i ⇒ i 3 − −( i )3 = x + iy ⇒ − i + i 3 = x + iy ⇒ − i − i = x + iy x + iy = 0 − 2i ∴ = + ⇒ = + ⇒ + − = − + + − + + = + − + + − + × + + + + + + = + + + + + + = + + + + = = = + =− = 23. + − = =− ∴ + − − = + x
  • 100. 189 = − − = and ∴ Multiplicative inverse of − − i z = i 26. Let z = x + iy, where x < 0, y < 0 z x − iy (x − iy)2 x2 − y2 2 xy ∴ z = x + iy = (x2 + y2 ) = x2 + y2 − x2 + y2 i z 2 2 lies in III quadrant, if x − y < 0 and − 2 xy < 0 z x < 0, y < 0 ⇒ − 2 xy < 0 Also, x2 − y2 < 0, if (x + y) (x − y) < 0 If x − y > 0, if x > y [x y, < 0 ⇒ x + y < 0] ∴ lies in III quadrant, if y < x < 0. 27. Let z 2 − i − 2 + i 3 − 4 i ∴= × − 3 − 4 i 3 + 4 i 3 − 4 i i 29. (1 + ix) (1 − ix) 1 − x2 − 2ix a − ib ⇒ 1 + x2 = 1 1 − x2 2 x ⇒ 2 = a and 2 = b 1 + x 1 + x Now, x can be written as = = − −
  • 101. 4 190 32. We have, x + iy = u + iv …(i) u − iv We know that, when two complex numbers are equal their conjugates are also equal. u − iv ∴ x − iy = …(ii) u + iv On multiplying Eqs. (i) and (ii), we get u + iv u − iv (x + iy) (x − iy) = × u − iv u + iv 2 2 2 ⇒ x2 + y2 = u 2 − i v2 2 u − i v 2 2 u 2 + v2 ∴ x + y = 2 2 = 1 u + v 33. Taking modulus and squaring on both sides, we get 1 + i 2 ⋅ 1 + 2i 2 ⋅ 1 + 3 i 2 .... 1 + ni 2 = α + i β 2 ⇒ (1 + 1)⋅ (1 + 4)⋅ (1 + 9) ... (1 + n2 ) = α 2 + β2 ⇒ 2 ⋅ 5⋅ 10 ... (1 + n2 ) = α 2 + β2 12 36. z ⇒ x2 − 8x + 16 + y2 = x2 ⇒ y2 = 8 (x − 2) The given relation represents the part of parabola with focus (4, 0) lying above X-axis and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence, the greatest positive argument of z is . 42. Let a + bi be a square root of − 2 + 2 3 i. ∴ (a + bi )2 ⇒ a2 + 2abi + i b2 2 = + + − ⇒ = π π π π π − + = + − π 41. − = ⇒ x x − + = 0) (4, Y X X′ Y′ O
  • 102. 4 191 + =− =− + = − − − − −  − − = ⇒ (a2 − b2 ) + 2abi = − 2 + 2 ∴ a2 − b2 = − 2 …(i) and 2ab = 2 …(ii) Now, (a2 + b2 2) = (a2 − b2 2) + 4 a b22 = −(2)2 + (2 3)2 = 4 + 12 = 16 ∴ a2 + b2 = 16 = 4 …(iii) On solving Eqs. (i) and (iii), we get 2a2 = 2 or a2 = 1 ∴ a = ± 1 ⇒ 2b2 = 6 or b2 = 3 ∴ b = ± 3. From Eq. (ii), ab = 3, which is positive. Eithera = − 1, b = − i.e. ± (1 + Aliter From Eq. (i), we get| z2| ≥ 2ab ∴ | z|2 + a2 2 ≥ a2 + b2 + 2ab ⇒2 ⇒2 ⇒ 2 z ≥ a + b, as z is positive ∴ (a + b) 44. We have, 1 − z z − z − z =(1 − z z12 ) (1 − z z12 ) − (z1 − z2 ) (z1 − z2 ) [z z = z 2 ] and 1 = 1] =(1 − z z12 ) (1 − z z12 ) − (z1 − z2 ) (z1 − z2 ) [(z1) = z1] = 1 − z z1 2 − z z1 2 + z z z z1 1 2 2 − z z1 1 + z z1 2 + z z1 2 − z z2 2 = 1 + z1 2 z2 2 − z1 2 − z2 2 = (1 − z1 2 ) (1 − z2 2 ) ∴ k = 1 45. Let z = x + iy We have, z + α z − 1 + 2i = 0 ⇒ x + i (y + 2) + α (x − 1)2 + y2 = 0 On equating real and imaginary parts y + 2 = 0 ⇒ y = − 2 and x + α (x − 1)2 + 4 = 0 ∴ x2 = α 2 (x2 − 2 x + 5) or (1 −α 2 )x2 + 2α 2 x − 5α 2 = 0 Since, x is real. ∴ D = B2 − 4 AC ≥ 0 ⇒ 4α 4 + 20 α 2 (1 + α 2 ) ≥ 0 ⇒ − 4 α 4 + 5α 2 ≥ 0 ⇒ 4α 2 α 2 − 5 ≤ 0 4 − 5 5 ∴ ≤ α ≤ 2 2 46. Given, z = 2 = 10 Let) ⇒ ∴Locus of α, i.e. 5z − 1is the circle having centre at − 1 and radius 10. ∴ a = 10 Again, z1 2 + z2 2 − 2 z z12 cos θ = 0 z 2 z + + ≥ + + ≥ ≥ α= − ⇒ α+ = = α+ =
  • 103. 4 192 1 − 2 1 cos θ + 1 = 0 ⇒ ⇒ ∴ ⇒ ⇒ ⇒ ⇒ ⇒ = 0 ⇒ = 1 [z1 ≠ − z2 ] 49. Since, 1, ω ω 2, ...,ω n − 1 are the nth roots of unity. n − 1 n − 1 n − 1 = ∑ k = 0 n − 1 = ∑ k = 0 = n z ∴ ⇒ ⇒ ( ⇒ x2 + 4 = (1 + x) ⇒ 2 x = 3 or 3 ∴ z = x + iy = − 2i 2 5 1. ( 3 + i ) = (a + ib) (c + id ) On taking argument both sides, we get tan− 1 1 = tan− 1 b + tan− 1 d 3 a c
  • 104. 4 193 ⇒ tan− 1 b + tan− 1 d = nπ + π , n ∈Z a c 6 52. Let z − 1 = e iθ ⇒ z = 1 + cos θ + i sin θ [z − 1 = 1] θ θ θ θ = 2 cos cos + i sin 2 2 ∴ arg (z) = = arg (z − 1) or arg (z − 1) = 2 arg (z) 53. arg (z )arg (− 1) = = z − z2 (x + iy) − (6 + 4i ) (x − 6) + i (y − 4) z − z1 = π We have, arg z − z2 4 x + y − 14x − 8y + 64 ⇒ x y x y ⇒ ∴ z − (7 + 5 i ) = 2 55. arg (− z) = π − θ = π + (−θ) = π + arg (z) ∴ arg (− z) − arg (z) = π 56. Since, z − 1 = 1 Let z − 1 = cos θ + i sin θ Then, z − 2 = cos θ + i sin θ − 1 = − 2 sin2 θ + 2 i sin θ cos θ 2 2 2 = 2i sin cos + i sin …(i) 2 2 2 and z = 1 + cos θ + i sin θ = 2 cos2 θ + 2i sin θ cos θ 2 2 2 = 2 cos cos + i sin …(ii) 2 2 2 From Eqs. (i) and (ii), we get z − 2 z 2 arg z = 2 , from Eq. (ii) = i tan = i tan (arg z) 57. We have, C2 + S 2 = 1 Let C ∴ = 1 + C − iS 1 + cos θ − i sin θ cos cos + i sin 2 2 2 = cos cos − i sin 2 2 2 θ 2 cos + i sin 2 2 = cos θ + i sin θ 1 =C + iS 58. Let z − 1 = r (cos θ + i sin θ) = re iθ ∴ Given expression = re iθ ⋅ e −iα + 1 iθ e iα re = re i( θ − α) + 1 e −i( θ − α) r ⇒ = − − − π ⇒ − − − − − − − = x x π ⇒ − =
  • 105. 4 194 + + + − − 16 59. 8 8 cos π + i sin π 8 8 8 sin − i cos 8 8 61. We have, 1 + i 8 1 − i 8 π π 8 = cos 4 4 4 4 = cos 2π + i sin 2π + cos 2π − i sin 2π = 2 cos 2π = 2 (1) = 2 [using De-Moivre’s theorem] 10 = − i ∑ e 11 − 1 k = 1 = − i (Sum of 11th roots of unity − 1) = − i (0 − 1) = i 63. We have, 4 (cos 75° + i sin 75° ) 0 4. (cos 30° + i sin 30° ) 10 {cos (75° − 30° ) + i sin (75° − 30° )} = 2 2 cos 30° + sin 30° 10 = 10 (cos 45° + i sin 45° ) = (1 + i ) 2 64. Using De-Moivre’s theorem, the given expression (cos θ + i sin θ)− 14 (cos θ + i sin θ)− 15 = θ + i sin θ)48 (cos θ + i sin θ)− 30 (cos = (cos θ + i sin θ)− 47 = cos 47θ − i sin 47θ 65. We have, sin θ − i cos θ = − i 2 sin θ − i cos θ = − i (cos θ + i sin θ) ∴The given expression, using De-Moivre’s theorem = −(i )3 [cos (− 25 θ) + i sin (− 25 θ)] = i [cos 25 θ − i sin 25 θ] = sin 25 θ + i cos 25 θ 66. We have, z = (1 + cos θ + i sin θ)3 5 (cos θ + i sin θ) 2 θ θ 5 1 + 2 cos − 1 + 2 i sin cos = 2 2 2 cos 3 θ + i sin 3 θ 5 θ 5 32 cos cos + i sin = 2 2 2 cos 3 θ + i sin 3 θ = 32 cos5 cos 5 θ + i sin 5 {cos 3 θ − i sin 3 θ} 2 2 2 = 32 cos5 2 cos 5 2 θ − 3 θ + i sin 5 2 θ − 3 = 32 cos5 θ 2 cos θ 2 + i sin − θ 2 Hence, the modulus and argument of z are 32 cos5 θ 2 Since, imaginary part of given expression is zero, we have 1 r sin (θ − α ) − sin (θ − α ) = 0 r r 2 − 1 = 0 ⇒r 2 = 1 ⇒ r = 1 ⇒ z − 1 = 1 or sin (θ − α ) = 0 ⇒(θ − α ) = 0 ⇒ θ = α ⇒ π 8 sin + i cos arg (z − 1) = α = π− π= 60. + + + = π π + = π π π + =− π π ∴ + = − + = − π π 62. π π − = ∑ = − − = ∑ π π + =− =− = = ∑ ∑ π π π π
  • 106. 4 195 and , respectively. 2 67. We have, 1 1 = = cos θ − i sin θ z cos θ + i sin θ ∴ zn = (cos θ + i sin θ)n = cos nθ + i sin nθ , 1 n and n = (cos θ − i sin θ) = cos nθ − i sin nθ z Hence, zn + 1 n = 2 cos nθ and zn − 1 n = 2i sin nθ z z 68. We have, [using De-Moivre’s theorem] 69. We have, abcd = cos (2α + 2β + 2γ + 2δ) + i sin (2α + 2β + 2γ + 2δ) ∴ abcd = [cos (2α + 2β + 2γ + 2δ) +i sin (2α + 2β + 2γ + 2δ)]1 2/ or abcd = cos (α + β + γ + δ) + i sin (α + β + γ + δ)…(i) [using De-Moivre’s theorem] 1 ∴ = cos (α + β + γ + δ) abcd − i sin (α + β + γ + δ)…(ii) On adding Eqs. (i) and (ii), we get 1 abcd + = 2 cos (α + β + γ + δ) abcd 70. Let cot− 1 p = θ then p = cot θ ∴ e 2mi cot −1 p ⋅ pi + 1 m = e 2 miθ ⋅ i cot θ + 1 m pi − 1 i cot θ − 1 =e 2miθ. i (cot θ − i ) m = e 2miθ ⋅ cot θ − i m i (cot θ + i ) cot θ + i = e 2miθ ⋅ cos θ − i sin m = e 2miθ ⋅ e − iθ m 8 8 i 71. We have, α = cos i sin e 11 11 Re (α + α 2 + α 3 + α 4 + α 5 ) 1 + (1 + α + α 2 + α 3 + α 4 + α 5 2 3 4 5 = + α + α + α + α + α ) 2 − 1 + 0 =[sum of 11 and 11th roots of unity] 1 = − 2 72. We have, 1 + 1 + = 2ω2 = 2 1 ω 2 and a + ω2 b + ω2 c + ω2 = 2ω = ω ∴ ω and ω2 are roots of the equation 1 1 1 2 cos θ + i sin = e 2miθ (e − 2iθ )m = e 0 = 1 e iθ − + = + − + + θ θ) θ) θ = + − + + θ θ θ θ = − + − + − θ θ) θ θ θ θ+ = + + θ θ θ θ θ θ  =− = θ θ θ θ θ θ θ + + = + + + ω ω ω + +
  • 107. 4 196 + + = . a + x b + x c + x x When x = 1, 1 1 1 2 + + = = 2 a + 1 b + 1 c + 1 1 2 1 3 73. x − 2 x + 4 = 0 ⇒ x = 1 + 3 i = 2 ± i 2 2 = − 2ω − 2ω2 ⇒ α 6 + β6 = 128 1 74. z + = 1 z ⇒ z = − ω or − ω2 ∴ z99 + 1 = −( 1) + 1 = − 2 z99 (− 1) 75. (− 1 + − 3)62 + (− 1 − − 3)62 = 262 ω62 + 262 (ω2 62) = 262 [(ω3 20)ω2 + (ω3 41)ω] = 262 [(ω2 + ω] = − 262 76. Given expression = {3(1 + ω2 ) + 5ω}2 + {3 (1 + ω +) 5ω2}2 = (− 3ω + 5ω)2 + (− 3ω2 + 5ω2 2) = 4ω2 + 4ω = − 4 6 6 ω2 − 1 + i 3 2 2 z2 z2 3 25 2 2 2 z1 50 lies in I quadrant. z2 79.xyz = (a + b) (aω + bω2 ) (aω2 + bω) = (a + b) (a2 − ab + b2 ) = a3 + b3 80. (1 + ω)3 − (1 + ω2 3) = − ω( 2 3) − − ω)( 3 = − ω6 + ω3 = − 1 + 1 = 0 81. p1 3/ = −( p)1 3/ (− 1)1 3/ = − −(p)1 3/, − −(p)1 3/ω − −(p)1 3/ ω 2 = − q, − qω, − qω2, where q = −(p)1/ 3 Let α = − q, β = − qω and γ = − qω2 xα + yβ + z γ − q (x + yω + zω2 ) ∴ x y z = 2 = ω xω + yω + z 82. Given expression n = ∑ (k − 1)(k − ω) (k − ω2 ) k = 2 nn 2 2 2 n n(+ 1) n 2 83. Let α = ω and β = ω2 Now, 84. We have, z2 2 z 50 ⇒ 1 z1 2 25 = − 3 i 25 = 3 25 = − iω 77. ω ω 2 + = ω= ω = − − 78. = − ω +
  • 108. 4 197 (1 − ω + ω2 ) (1 − ω2 + ω4 )(1 − ω4 + ω8 )(1 − ω8 + ω16 ) = (1 + ω2 − ω) (1 − ω2 + ω) (1 − ω + ω2 ) (1 − ω2 + ω) [ω4 = ω3 ⋅ ω = ω;ω8 = (ω3 2) ⋅ ω2; ω16 = (ω3 5) ⋅ ω =ω andω3 = 1] = (− ω − ω) (− ω2 − ω2 ) (− ω − ω) (− ω2 − ω2 ) = (− 2ω) (− 2ω2 )(− 2ω) (− 2ω2 ) = 16⋅ω6 = 16 (ω3 2) = 16 1( )2 = 16 85. We have, x = ω − ω2 − 2 or x + 2 = ω − ω2 On squaring, x2 + 4x + 4 = ω2 + ω4 − 2ω3 = ω2 + ω3 ⋅ ω − 2ω3 = ω2 + ω − 2 [ω2 = 1] = − 1 − 2 = − 3 ⇒ x2 + 4x + 7 = 0 On dividing x4 + 3x3 + 2 x2 − 11x − 6 by x2 + 4x + 7, we get x4 + 3x3 + 2 x2 − 11x − 6 = (x2 + 4x + 7) (x2 − x − 1) + 1 = (0) (x2 − x − 1) + 1 = 0 + 1 = 1 86. We have, (1 + ω) (1 + ω)2 (1 + ω4 ) (1 + ω8 ) ... to 2n factors = (1 + ω) (1 + ω2 ) (1 + ω3 ⋅ ω) (1 + ω6 ⋅ω2 ) ... to 2n factors = (1 + ω) (1 + ω2 ) (1 + ω) (1 + ω2 ) ... to 2n factors [ω3 = ω6 = ... = 1] = [(1 + ω) (1 + ω ...)to n factors] [(1 + ω2 ) (1 + ω2 ) ... to n factors] = (1 + ω) n (1 + ω2 ) n = [(1 + ω) (1 + ω2 )]n = (1 + ω + ω2 + ω3 ) n = (0 + 1) n = 1 [1 + ω + ω2 = 0, ω3 = 1] 87. We have, α = ω and β = ω2 Then, xyz = (a + b + c ) (aω + bω2 + c ) (aω2 + bω + c ) = (a + b + c ) (a2 + b2 + c 2 − ab − bc − ca) = a3 + b3 + c 3 − 3 abc 88. We have, z3 + 2 z2 + 2 z + 1 = 0 ⇒ (z + 1) (z2 + z + 1) = 0 Its roots are − 1, ω and ω2. The root z = − 1 does not satisfy the equation, z1985 + z100 + 1 = 0 but z = ω and z = ω2 satisfy it. Hence, ω and ω2 are the common roots. 89. We have, (16)1 4/ = (24 1 4) / = 2 1( )1 4/ 1 1 0, 1, 2, 3 4 4 = 2 × 1, 2 × i, 2 × − 1, 2 × − i, = ± 2, ± 2i 90. We have, f(ω) = ω3p + ω3q + 1 + ω3r + 2 = ω3p + ω3q ⋅ ω + ω3r ⋅ ω2 = (ω3 ) p + (ω3 )q ⋅ ω + (ω3 ) r ⋅ ω2 = 1 + ω + ω2 = 0 [ω3 = 1] 91. Since 1, α, α 2, ..., α n − 1 are n and nth roots of unity. ∴ xn − 1 = (x − 1) (x − α) (x − α 2 ) ...(x − α n − 1) ⇒ log (xn − 1) = log (x − 1) + log (x − α ) + log (x − α 2 ) + ... + log (x − α n − 1) On differentiating both sides w.r.t. x, we get nxn − 1 1 1 1 1 = + + + ... + xn − 1 x − 1 x − α x − α 2 x − α n − 1 Putting x = 2, we get n2n − 1 1 1 1 1 = + + + ... + n n − 1 − 1 ∴ 1 2 − 1 i = 1 2 − α n − 1 n⋅2n − 1− 2n + 1 Hence, = n i i = 1 2 − 1 92. The vertices of the triangle are A (1, 0), B (1/ 2, 1/ 2 ) and C (0, 1). ∴ AB2 = 2 − 2 BC2 = 2 − 2
  • 109. 4 198 AC2 = 1 + 1 = 2 ∴ AB = BC 93. By definition, z1 = x1 + iy1, z2 = x2 + iy2 and collinear, if z3 = x3 + iy3 are x1 y1 1a x2 y2 1= 0 ⇒a′ x3 y3 1a − a′ ⇒ ab′ = a b′ 94. Let z = x + iy, where x y, ∈R b 1 b′ 1= 0 b − b′ 1 ∴ α + β =i tan− 1 (x + iy) ⇒ α − β =i tan− 1 (x − iy) ⇒ 2α = tan− 1 (x + iy) + tan− 1 (x − iy) = tan− 1 22x 2 1 − x − y ∴ x2 + y2 + 2 x cot 2α = 1 n z − 1 = z + 1 (x − 1)2 + y2 = (x + 1)2 + y2 ∴The roots lies on the Y-axis. 97. Let z = x + iy, where x y, ∈R We have, iz − 1 + z − i = 2 ⇒ ix − y − 1 + x + iy − i = 2 ⇒ (− y − 1)2 + x2 + x2 + (y − 1)2 = 2 ⇒ x2 = 0 or x = 0 ∴The locus of z is a straight line. 98. Clearly, z1 < z3 < z2 , as z3 is mid-point of z1 and z2. 99. (a) Re 1 + z = 0 ⇒ arg 1 + z = π 1 − z 1 − z 2 This represents a circle. (b) z z + iz − iz + 1 = 0 represent a circle. z − 1 π (c) arg z + 1 2 This represents a circle. (d) This represents a straight line. 100. We have, z2 = z + z The point representing z2 divided the line segment joining points representing z1 and z3 in the ratio 1: 1. ∴The points lies on a line. 101. Area of the triangle on the argand plane formed by 3 2 complex numbers − z iz z, , − iz is z . 2 3 2 ∴ z = 600 ⇒ z = 20 2 102. Let z1 = 1 − 3 i, z2 = 4 + 3 i and z3 = 3 + i. Then, z3 divides the line segment joining z1 and z2 in the ratio 2 : 1 internally. So, the points z1, z2 and z3 are collinear. 103. Let z = x + iy z + 2i z + 4 which represents a circle with centre (− 2, − 1). 104. Let z = x + iy z + 2i x + iy + 2i x + (y + 2)i Then, = = − + = ⇒ − = + 95. = − − 96. = − + ∴ = + − ⇒ + = − ⇒ = − + ⇒ ⇒ ⇒ x = ⇒ x = x x x x = + + + + = + + + + = x x x + + + − + + = + + + + + + + + x x x x + + = ⇒ x x = + + +
  • 110. 4 199 z + 2 x + iy + 2 (x + 2) + iy z + 2 which represents a straight line. 105. We have, z − 4 i + z + 4 i = 10 ⇒ z − (0 + 4 i ) + z − (0 − 4 i ) = 10 This represents an ellipse. 4 i + 4 i < 10 i.e. 10 > 8 106. We have, z − a i = z + a i ⇒ x + i (y − a) 2 = x + i (y + a) 2 ⇒ x2 + (y − a)2 = x2 + (y + a)2 ⇒ 4ay = 0; y = 0, which is X-axis. 107. We have, z − 1 2 + z + 1 2 = 4 ⇒ (x − 1) + iy 2 + (x + 1) + iy 2 = 4 ⇒ (x − 1)2 + y2 + (x + 1)2 + y2 = 4 ⇒ 2 (x2 + 1) + 2 y2 = 4 ∴The locus of P is x2 + y2 = 1. ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ∴ ⇒ ⇒ and i.e. A2 − B2 = − 4x …(ii) On dividing Eq. (ii) by Eq. (i), we get ≤ − x ⇒ ≤ 4 − x ⇒ 3x 2 ≤ 12 [squaring and simplifying] x2 y2 or + ≤ 1 4 3 which represents the interior and boundary of an ellipse. 110. We have, log1 2/ 3zz−−11+−42 > 1 = log1 2/ 21 z − 1 + 4 1 ⇒ < < 1 3 z − 1 − 2 2 [loga x is a decreasing function, if a < 1] ⇒ z − 1 + 4 < 3 z − 1 − 2 ⇒ 2 z − 1 > 6 ⇒ z − 1 > 3 which is an exterior of a circle. 111. The closest distance = length of the perpendicular from the origin on the line az + az + aa = 0 a (0) + a = = + + + − + + x x x = + + + + + + + + x x x x + = ⇒ x = + + x − + − x + + x − + + + = = + = + − +
  • 111. 4 200 112. We have, zk = 1 + a + a 1 − a 113. We have, z1 − z4 = z2 − z3 or z1 + z3 = z2 + z4 i.e. the 2 2 diagonals bisect each other. ∴It is a parallelogram. z4 − z1 = π Also, amp z2 − z1 2 ⇒ Angle at z1 is a right angle. ∴ It is a rectangle and hence, a cyclic quadrilateral. 114. We have, Im (z2 ) = 4 ⇒ Im [(x2 − y2 ) + 2 ixy] = 4 [putting z = x + iy] ⇒ 2 xy = 4 or xy = 2, which is a rectangular hyperbola. 115. We have, Re (z2 ) = 4 ⇒ Re [(x2 − y2 ) + 2 ixy] = 4 [putting z = x + iy] ⇒ x2 − y2 = 4 which is a rectangular hyperbola. 116. z − z1 = z − z2 = z − z3 ∴ z must be the mid-point of z1 and z2 and since z − z1 = z − z2 and z z, 1 and z2 are collinear. ⇒ z is circumcentre of ∆ formed by z1, z2 and z3. z3 − z1 = ± π [neglecting − ve value] ∴ arg z3 − z2 2 117. a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 ⇒ a z0 4 + a z1 3 + a z2 2 + a z3 + a4 = 0 [taking conjugate on both sides] ⇒ a0(z4 ) + a z1( )3 + a2(z)2 + a3(z)2 + a z3 + a4 = 0 ∴z is a root of the equation if z is a root. So, option (a) is correct. Also, if z1 is real, z1 = z1. If z1 is non-real complex, then z1 is also a root because imaginary root occurs in conjugate pairs. So, option (b) is also correct. 118. Let z =α be a real root. Then, α 3 + (3 + 2i )α + (− 1 + ia) = 0 ⇒ (α 3 + 3α − 1) − i (a + 2 α ) = 0 ⇒ α 3 + 3α − 1 = 0 and α = − a / 2 ⇒ ⇒ a3 + 12a + 8 = 0 Let f a() = a3 + 12a + 8 ∴ f(− 1) < 0, f(0) > 0, f(− 2) < 0, f(1) > 0 and f(3) = 0 119. We have, ⇒ θ = 2nπ ± where, n is an integer. 120. Let z = c be a real root. Then, αc 2 + c + α = 0 Let α = p + iq Then, (p + iq c) 2 + c + p − iq = 0 ⇒ pc 2 + c + p = 0 and qc 2 − q = 0 …(i) ⇒ c = ± 1 [q ≠ 0] From Eq. (i), we get α ± 1 + α = 0 Also, c = 1 121. Let z2 = r2 (cos θ2 + i sin θ2 ) ⇒ z2 = r2 Also, arg (z1) + arg (z2 ) = 0 ⇒ arg (z1) = − arg (z2 ) = − θ2 ∴ z1 = r2 [cos (− θ2 ) + i sin (− θ2 )] = r2 [cos θ2 − i sin θ2 ] ⇒ z1 = z2 1 ⇒ z1 = z2 ∴ z z12 = 1 + − = + + + − θ θ θ θ θ θ = − + + θ θ θ + − θ θ θ= ⇒ − − − − + ⇒ − − = − < − +  < ∴ lie within the circle − − = −
  • 112. 4 201 z = 15 = 25 ⇒ ⇒ From z ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ 1 lies on ⊥ bisector of 1 and − 1. z2 z1 lies on imaginary axis. ⇒ z2 z1 is purely is imaginary. ⇒ z2 122. = = − − − = = − = − − = + = + O C A B i= z2 –3–4 5 1
  • 113. 4 202 z1 = ± ⇒ arg z2 ∴ arg (z1) − arg(z2 125. Since, arg ((z − 1 − i ) / z) is the angle subtended by the chord joining the pointsO and 1 + i at the circumference Y z 2 z − 0 We have, 2 − z π z − 2 AP ⇒ arg ⇒ = i 0 − z 2 2 OP AP Now, in ∆OAP, tan θ = Thus, z − 2 = 0 OP 126. Given, z = x + iy 6x2 − 36x + 48 + 6y2 − 24y + 18 + i [6xy 4 a ⇒ 6x2 + 6y2 − 36x − 24y + 66 = 12 x − 12 y − 12 ⇒ 6x2 + 6y2 − 48x − 12 y + 78 = 8 ⇒ x2 + y2 − 8x − 2 y + 13 = 0 Again, z − 3 − i = 3 ⇒ x + iy − 3 + i = 3 ⇒ (x − 3)2 + (y + 1)2 = 9 …(i) ⇒ x2 + y2 − 6x + 2 y + 1 = 0 On subtracting Eq. (ii) from Eq. (i), W get − 2 x − 4y + 12 = 0 …(ii) ⇒ x = − 2 y + 6 Putting the value of x in Eq. (ii), we get (− 2 y + 6)2 + y2 − 6 (− 2 y + 6) + 2 y + 1= 0 ⇒ 5y2 − 10 y + 1 = 0 …(iii) ∴ y = ∴ x = − 2 y + 6 = 4 Then, 2 ⇒ x = 1, − 2 128. amp (z z1 2 ) = 0 ⇒ amp (z1) + amp (z2 ) = 0 ∴ amp (z1) = −amp (z2 ) =amp (z2 )  z1 = z2 , we get z1 = z2 . So, z1 = z2 Also, z z12 = z z2 2 = z2 2 = 1, because z2 = 1 129. z + z− 1 = 1⇒ z2 − z + 1 = 0 1 ± 3 i 2 ∴ z == − ω −ω = − 4 π/ − = + = ∴ − =± π ⇒ − Pz () O θ 2 A X 1 x x − − − − = + − − + − − = − − − − + − + − − x x x x − − − 8) = + + − − + − x x − + − − x = + + − − − + − x x x + x x − − − + − = + + = π ∴ π = ⇒ = ± = ±  ∴ x = = + ± +  x x = + + − π = π − + 127. . Let x = x x − = − = − ⇒ x + = + − + − − − = ⇒ x + =±
  • 114. 4 203 2 ∴ zn + z− n = −( ω) n + (− ω)− n or (− ω2 )n + (− ω2 )− n = −(1)nωn + (− 1)− n ⋅ 1 or = −(1) n ωn + = (− 1) n ⋅ (ωn + ω2n ) = −(1) n ⋅ (1 + 1) or (− 1) n ⋅ −(1) Dependening on whether n is a multiple of 3 or not. = 3 131. OB = OD = OA = z amp (z2 ) = θ + and z = z {cos θ + i sin θ} ∴ z1 = z cos θ − 2 + i sin θ − 2 = z {sin θ − i cos θ} = z (− i ) (cos θ + i sin θ) = − iz z2 = z cos θ + 2 + i sin θ + 2 iz ∴ 132.  = − − = − − = z –1=1 Y Pz ( ) 0 Qz ( ) 1 C 1 X ω + − =− − ω ω ω + =− ω ω = − + ω ω ω = =θ =θ− π π Dz ) ( 1 C Az () B ) (z2
  • 115. 4 204 z − 1 ⇒ The vertices are at equal distances from the origin z = 0. ∴The origin is at the centroid of the equilateral triangle. z1 + z2 + z3 = 0 ∴ 3 ∴ z1 + z2 + z3 = 0 with OA as real axis, z1 = 1, z2 = 1(cos 120° + i sin 120° ), z3 = 1(cos 120° − i sin 120° ) 135. Statement I is false, since there is no order relation in the set of complex numbers. Cancellation laws, a + c > b + c ⇒ a > b does not hold true in complex numbers, therefore Statement II is true. 136. Given that, arg (z) = 0 ⇒ z is purely real. ∴ Statement II is true. ⇒ arg(z1) − arg (z2 ) = 0 z1 = 0 ⇒ z1 is purely real. ⇒ arg z2 z2 z1 = 0 ⇒ Im z2 Hence, Statement I and Statement II are true and Statement II is correct explanation of Statement I. 137. We will show that Statement I is true and follows from Statement II. Indeed ⇒ ⇒ But as z > 0, we have ⇒ Maximum value of 138. Statement I is false, since z = 2 > 4b and z is a positive z = c ⇒ z = c (cos θ + i sin θ) ⇒ Infinite complex numbers satisfy the given equation. Statement II is true. A quadratic can have more than two roots, if all coefficients are zero. = + − ∴ − − = ⇒ + = − −θ θ = − + ⇒ θ θ = − ⇒θ θ = − = + − ≤ − ⇒ − = − π π ∴ − − = ⋅ ± + ± = − π π ∴ − − = − =− − 133. = = O 1 1 Bz ( ) 2 Cz ( ) 3 Az ( ) 1
  • 116. 4 205 139. e iθ = cos θ + i sin θ ⇒ e − iθ = cos θ − i sin θ e iθ + e − iθ e i(1 −i) + e − i(1−i) e(1 + i) + e − (1 + i) 2 e 2 e 140. We have, (cos θ + i sin θ)3 5/ = (cos 3θ + i sin 3θ)1 5/ = [cos (2rπ + 3θ) + i sin (2rπ + 3θ)]1 5/ where, r = 0, 1, 2, 3, 4 2πr + 3 i = e 5 , r = 0, 1, 2, 3, 4 Hence, product of all values of (cos θ + i sin θ)3 5/ 3θ i 2π + 3 i 6π + 3 i 8π + 3 i = e 4 e 5 e 5 e 5 = e i3θ + 4πi =e 4iπ e i3θ = cos 3θ + i sin 3θ Also, product of roots of the equation x5 − 1 = 0 is 1. Hence, Statement II is true, but it is not a correct explanation of Statement I. 141. Let A z( 1)and B z( 1)be the centres of the given circles and P be the centre of the variable circle, which touches given circles externally, then AP = a + r and BP = b + r where, r is the radius of the variable circle. On subtraction, we get AP − BP = a − b ⇒ AP − BP =|a − b is a constant. Hence, locus of P is (i) a right bisector of AB, if a = b (ii) a hyperbola if a − b < AB = z2 − z1 (iii) an empty set, if a − b > AB = z2 − z1 (iv) set of all points on line AB except those which lie between A and B, if a − b = AB ≠ 0 Thus, Statement I is false and Statement II is true. 142. If P z( ) is any point on the ellipse, then equation of the ellipse is z − z1 + z − z2 =…(i) e For P z( )to lie in ellipse, we have z1 − z2 z − z1 + z − z2 < e It is given that origin is an interior point of the ellipse. ⇒ 0 − z1 + 0 − z2 < e z1 − z2 ∴ e 0, z1 + z2 Hence, Statement I is false and Statement II is true. 143. As, we know z − z1 + z − z2 = k represents an ellipse, if k > z1 − z2 . Thus, z − i + z + i = k represents an ellipse, if k > i + i or k > 2. ∴ Statement I is false and Statement II is true. 144. The equation can be rewritten as zz + az + az + aa = aa − λ ⇒ (z + a) (z + a) = aa − λ ⇒ z + a = aa − λ Since, aa is real, 1 should be real. aa − λ represents radius of the circle. Hence, Statement I and Statement II both are true and Statement II is the correct explanation for Statement I. Solutions (Q. Nos. 145-147) Given that, | z1 + z2|2 =| z1|2 + | z2|2 ⇒| z1|2 + | z2|2 + z z1 2 + z z1 2 =| z1|2 + | z2|2 ⇒ z z1 2 + z z1 2 = 0 …(i) z1 + z1 = 0 [dividing by z z2 2 ] ⇒ z2 z2 z1 z1 = 0 …(ii) ⇒ z2 z2 z1 − z2 z1 − z2 = − = = + − − = + + − − = + + − ∴ + = = −
  • 117. 4 206 145. From Eq. (i),z z2 2 is purely imaginary. 146. From Eq. (ii), z1/ z2 is purely imaginary 147. Also, i z( 1/ z2 ) is purely real. Hence, its possible arguments are 0 and π. 148. Clearly, according to the least possibility α 2 − 7α + 11 ≤ 1 ⇒ α ∈[2, 5] X 151. zD = B z iA , zE = zB − z iC 1 − i 1 − i ∴Angle between AC and DE = arg C zA − arg zC − zA z − zE − zD zA − zC 1 − i π = i 4 A D B C E
  • 118. 4 207 z + 1 ⇒ z − z = 0, z + z = 0, y = 0, x = 0 Locus of z is portion of pair of lines xy = 0 z2 − 1  2 > 0 z + 1 B. Given,z − cos− 1 cos 12 − z − sin− 1 sin 12| = 8 (π − 3) Since, cos− 1 cos 12 − sin− 1 sin 12= 8(π − 3) ∴ Locus of z is portion of a line joining z1 and z2 except the segment between z1 and z2. C. z2 − i z1 2 = k2 − k1 ∴ x2 − y2 + 2ixy − λi 1 = λ 2 ⇒ x2 − y2 = λ2 and xy = ∴Locus of zis point of intersection of hyperbola. ⇒ Locus of z is the segment joining z1 and z2. 155. There are no real roots of the equation z6 − 6z + 20 = 0. If x + iy is a root, then x − iy is also a root. Let the roots be x1 ± iy1, x2 ± iy2, x3 ± iy3 Sum of the roots = 2 (x1 + x2 + x3 ) = 0 ⇒ x1 + x2 + x3 = 0 ⇒ One of x1, x2, x3 is negative and other two are positive or vice-versa. ⇒ The number of roots in each of the quadrant is either 1 or 2. 156. A. z4 − 1 = 0 ⇒ z4 = 1= cos 0 + i sin 0 ⇒ z = (cos 0 + i sin 0)1 4/ = cos 0 + i sin 0 B. z4 + 1 = 0 ⇒ z4 = − 1 = cos π + i sin π ⇒ z = (cos π + i sin π)1 4/ 8 8 D. iz4 − 1 = 0 ⇒ z = ω ω,2 B. Here, z1 + z2 + z3 + .... + z6 = 0 ... (i) z2 = z e1( i2π )/ n, if e( 2i π )/ n= α ⇒ z2 = z1 α Similarly, z3 = z1α 2, z4 = z2 α 3, z5 = z2 α 4, z6 = z1α 5 On squaring and adding and then using Eq. (i), we get z12 + z22 + z32 + z42 + z52 + z62 = 0 C. 1 + 2 ω + 3ω2 + 4ω3 + ... + nωn − 1 n !ω n n( 2+ 1) 1/ n n [AM ≥ GM] Now, E = 1 + 2 ω + 3ω2 + 4ω3 + ... + nωn − 1 +.... ω + 2 ω2 + 3ω3 + ... + (n − 1)ωn − 1 + nωn D. Let x2 + y2 = 1 where, z1 − i, z2 = 1, z3 = − 1 ∴ z = − i ⇒ zz1 + z z23 = 1 − 1 = 0 158. Consider, x3 2 + 11 i 3 − 4 i 50 + 25 i x = x2 = 3 + 4 i × 3 − 4 i = 25 = − + = − + = π π + + = ⇒ + = = π π ⇒ + = π π + = π π ⇒ − = =− π π/ ⇒ − = π π/ = π π − 157. − = ⇒ = + + ω ω) ω ω ω = + + − = + + − − ω ⇒ = − −ω ∴ = ω− − − − + + = − − π + 1 − = + − − π ∴ + + = − −
  • 119. 4 208 = 2 + i ∴ a + b = 2 + 1= 3 x − 4 + iy x − 8 + iy ⇒ y2 − 25y + 136 = 0 ⇒ 9 (36 + y2 ) = 25 [36 + (y − 8)2 ] ⇒ y = 17, 8 Thus, the required numbers are z = 6 + 17i, 6 + 8 i. Hence, the value of Re (z) is 6. Hence, λ is equal to 8. 161. Consider z4 + z3 + 2 z2 + z + 1 = 0 ⇒ z4 + z3 + z2 + z2 + z + 1 = 0 ⇒ z2 (z2 + z + 1) + (z2 + z + 1) = 0 ⇒ (z2 + z + 1) (z2 + 1) = 0 It is given that, BD = 2 AC ⇒ MD = 2 AM Also, DM is perpendicular to AM. ⇒ (1 − 2)2 + (1 + 1)2 = 4[(x − 2)2 + (y + 1)2 ] …(i) y + 1 1 + 1 and ⋅ = − 1 x − 2 1 − 2 ⇒ 2 (y + 1) = x − 2 With x − 2 = 2(y + 1) , Eq. (i) can be written as ⇒ y ⇒ x = 3, 1 ∴ λ1 + λ 2 = 4 Entrances Gallery 1. A. zk is 10th root of unity ⇒ zk will also be 10th root of unity. Take zj as zk. zk , we can always find z. B. z1 ≠ 0, take z = z1 C. z10 − 1 = (z − 1)(z − z1) (z − z9 ) ⇒ (z − z1)(z − z2 ) (z − sz9 ) = 1 + z + z2 + + z9, ∀ z ∈complex number Put z = 1 (1 − z1)(1 − z2) (1 − z9 ) = 10 D. 1 + z1 + z2 + + z9 = 0 ⇒ Re (1) + Re (z1) + + Re (z9 ) = 0 ⇒ Re (z1) + Re (z2)|+ + Re (z9) = − 1 | z0|2 + |α|2 − r 2 = |α | ⇒ 160. − ≥ + − = − = − ≥ x x − = x ≥ x x > + = ⇒ x = − + = = = ∴ = − ω ω = 162. M C A ( ) x,y B D 1) , (1 –1) (2, 159. − − = ⇒ = ⇒ x x = + − − + ⇒ x = x = − − = ∴ = − = ∑ π = α = = |α |α ∆ θ α α| = + − ∆ θ α |α = − + + − 2 D O A B()α 2r r C ) / (1 α θ z0 Y X
  • 120. 4 209 3 6 2 Possible position of z1 are A1, A2, A3, whereas of z2 are B1, B2, B3 (as shown in the figure). So, possible value of ∠z Oz1 2 according to the given options is . 4. Area of = 42 π 5. Distance of (1, − 3) from y + > 6. Given, is z2 + z + 1 − a = 0 Clearly, this equation do not have real roots, if D < 0 ⇒ 1 − 4 1( − a) < 0 ⇒ 4a < 3 ∴ a < ∴ Minimum value = 5 8. The expression may not attain integral value for all a, b,c. If we consider a = b = c, then x = 3 a y = a(1 + ω + ω2 ) = 0 z = a(1 + ω2 + ω) = 0 ∴ | x|2 + | y|2 + | z|2 = 9|a|2 | x|2 + | y|2 + | z|2 9 ∴ |a|2 + |b|2 + |c|2 = 3 = 3 As, x 2 + y 2 + z 2 = 3(|a|2 +|b|2 +|c|2 ) [using 1+ ω + ω2 = 0] | x|2 +| y|2 +| z|2 ∴ 2 2 2 = 3 |a| +|b| +|c| 9. Given, z = (1 − t z) 1 + tz2 z − z1 = t ⇒ z2 − z1 z − z1 = 0 ...(i) ⇒ arg z2 − z1 = ∩ ∩ = × + × π π + = π 60° x+y 2 2 <16 y+x= 0 3 √ Y X X′ Y′ = 0) (0, B (3 0) , C 2) , (3 A(3,–5/ 2) O Y Y¢ X¢ X α| α| ⇒ α|= = + = π = π π > π <− A A2 A1 B1 B2 B3 x 2 –1/ = x 2 / =1 π/6 O x = − + × > −
  • 121. 4 210 ⇒ ⇒ AP + PB = AB ⇒ | z − z1| + | z − z2| =| z1 − z2| z z 10. A. − i = + i , z ≠ 0 | z| | z| z is unimodular complex number | z| and lies on perpendicular bisector of i and − i z ⇒ = ± 1⇒ z = ± 1| z| ⇒ a is real number | z| ⇒ Im(z) = 0 B. | z + 4| + | z − 4| = 10 z lies on an ellipse, whose focus are (4, 0) and (− 4, 0) and length of major axis is 10. ⇒ 2ae = 8 and 2a = 10 ⇒ e = 4 / 5 |Re(z)| ≤ 5. C. ∴|ω| = 2 ⇒ w = 2(cos θ + i sin θ) x iy i i 2 x ⇒ 2 + (3 / 2) y = 1 2 (5 / 2) e ⇒ e D. |ω| = 1 ⇒ x + iy = cos + i sin θ + cos θ − i sin θ x + iy = 2 cos θ |Re(z)| ≤ 1,|Im(z)| = 0 11. Given, z2 is not unimodular i.e.| z2| ≠ 1 z1 − 2 z2 is unimodular. and − z z1 2 ⇒⇒| z1 − 2 z2|2 =|2 − z z1 2|2 ⇒ (z1 − 2 z 2 ) = (2 − z z [zz =| |z 2 ] ⇒ | z1|2 + 4| z2|2 − 2 z z12 − 2 z z1 2 = 4 + | z1| |2z2|2 − 2 z z12 − 2 z z1 2 ⇒ (| z2|2 − 1)(| z1|2 − 4) = 0  | z2| ≠ 1 ∴ | z1| = 2 Let z1 = x + iy ⇒ x2 + y2 = (2)2 ∴ Point z1 lies on a circle of radius 2. 12.| |z ≥ 2 2 2 13. Given,| |z = 1, arg z = θ ∴ z = e ie 1 But z = z 1 + z ∴ arg = arg(z) = θ 1 + 1 z z2 14. Given, a complex number (z ≠ 1) is purely real. z − 1 To find the locus of the complex number z. z2 Since, (z ≠ 1) is purely real. z − 1 z2 z2 ∴ = z − 1 z − 1 ⇒ z2(z − 1) = z2(z − 1) ⇒ z z2 − z2 = z z2 − z2 ⇒ zzz − z2 = zzz − z2 ⇒ z z| |2 − z2 = z z| |2 − z2 ∴ ≥ + − ≥ − ≥ − ⇒ − = − − − = − − − − − − = Pz () Az ( ) 1 Bz ( ) 2 − − = − −
  • 122. 4 211 On rearranging the terms, we get z z| |2 − z| |z 2 = z2 − z2 ⇒ | z|2 (z − z) = (z − z)(z + z) ⇒ | |z 2 (z − z) − (z − z)(z + z) = 0 ⇒ (z − z)(| z|2 − (z+ z)) = 0 Either (z − z) = 0 or[| z|2 − (z+ z)] = 0 Now, z = z ⇒ Locus of z is real axis and (| z|2 − (z + z)) = 0 ⇒ zz − (z + z) = 0 Locus of z is a circle passing through the origin. Aliter Put z = x + iy, then z2 (x + iy)2 (x2 − y2 ) + i(2 xy) = = z − 1 (x + iy) − 1 (x − 1) + iy 2 2 (x − y ) + i(2 xy) (x − 1) − iy = × (x − 1) + iy (x − 1) − iy z2 Since, (z ≠ 1) is purely real, hence its imaginary z − 1 part should be equal to zero. ⇒ (x2 − y2 )(− y) + (2 xy)(x − 1) = 0 ⇒ y (x2 − y2 + 2 x − 2 x2 ) = 0 ⇒ y x( 2 + y2 − 2 x) = 0 Either y = 0 or x2 + y2 − 2 x = 0 Now, y = 0, locus of z is real axis, and x2 + y2 − 2 x = 0, locus of z is a circle passing through the origin. Locus of z is either real axis or a circle passing through the origin. 15. Let z = x + iy, given Re (z) = 1 ∴ x = 1⇒ z = 1 + iy Since, complex roots are conjugate of each other. ∴z = 1 + iy and 1 − iy are two roots of z2 + α z + β = 0 Product of roots = β ⇒ (1 + iy)(1 − iy) = β ∴ β = 1 + y2 ≥ 1 ⇒ β ∈(1, ∞) 16. (1 + ω)7 = A + Bω, we know 1 + ω + ω2 = 0 ∴ 1 + ω = − ω2 ⇒ (− ω2 7)= A + Bω ⇒ − ω14 = A + Bω [ω14 = ω12 ⋅ω2 = ω2 ] ⇒ − ω2 = A + Bω ⇒ 1 + ω = A + Bω On comparing A = 1, B = 1 17. We have,| z − 1| =| z + 1| =| z − i| Clearly, z is the circumcentre of the triangle formed by the vertices (1, 0), (0, 1) and (− 1, 0), which is unique. Hence, the number of complex number z is one. 18. Since, α and β are roots of the equation x2 − x + 1 = 0. ⇒ α + β = 1, αβ = 1 ⇒ x = 1 + 3i 1 − ⇒ x =or 2 ⇒ x = − ω or Thus, α = − ω2, then β = − ω or α = − ω , then β = − ω2 [where,ω3 = 1] Hence, α 2009 + β2009 = −(ω)2009 + (− ω2 2009) = − [(ω3 669)⋅ω2 + (ω3 1337)⋅ω] = − [ω2 + ω] = − −(1) = 1 19. | z| z − 4 + z ⇒ ⇒ ⇒ | z|2 ⇒ ∴Maximum value of| z| is Ø For minimum value of |z|, we take 4 4 4 4 z − − z − z z z z from which we get | |z ≥5 − 1 O (0, 1) (–1, 0) (1, 0) z i + ± ≤ + − ≤ + ≤ − − ≤ + +
  • 123. 4 212 1 20. Let z = 1 1 1 ∴ z = = − i − 1 − i − 1 i + 1 21. From the argand diagram, maximum value of| z + 1| is 6. Aliter | z + 1| =| z + 4 − 3| ≤| z + 4| + −|3| ≤ 6 Thus, maximum value Y¢ of| z + 1| is 6. 10 22. ∑ sin 2 kπ + i cos 2k k = 1 11 11 10 = i ∑ k = 1 = i ∑ e 11 k = 1 = i ∑10 e − 211kπ − 1 = − i k = 0 23. Given, equation is z2 + z + 1 = 0 ⇒ z = ω ω,2 1 2 Now, z + z2 + 12 2 z3 + z13 2 z z 2 2 2 z4 + 14 z5 + z15 z6 + z16 z = (ω + ω2 2)+ (ω2 + ω)2 + (ω3 + ω−3 2) + (ω + ω2 2)+ (ω2 + ω)2 + (ω6 + ω−6 2) = −(1)2 + (− 1)2 + (1 + 1)2 + (− 1)2 + (− 1)2 + (1 + 1)2 = 1 + 1 + 4 + 1 + 1 + 4 = 12 24. Given that, (x − 1)3 + 8 = 0 −2 x − 1 1 3/ ⇒ ( )1 −2 x − 1 2 Cube roots of are 1, ω, ω . −2 Cube roots of (x − 1) are − 2, − 2ω and − 2ω2. ∴Cube roots of x are − 1, 1 − 2ω and 1 − 2ω2. 25. Let z1 = x1 + iy1 and z2 = x2 + iy2 Given,| z1 + z2| =| z1| + | z2| ∴ On squaring both sides, = x12 + x22 + 2 x x12 + y12 + y22 + 2 y y1 2 = x12 + y12 + x22 + y22 + 2 (x12 + y12 )(x22 + y22 ) ⇒ x x12 + y y12 = (x12 + y12 )(x22 + y22 ) Again, squaring both sides, we get ⇒ = x x1222 + y y1222 + 2 x x y y1 2 1 2 = x x1222 + y y1222 + x y1222 + y x12 22 ⇒ (x y12 − y x12 )2 = 0 y1 = y2 ⇒ x1 x2 ⇒ tan−1 y1 = tan−1 y2 x1 x2 ⇒ arg (z1) = arg(z2 ) ⇒ (x − 1)3 = −( 2)3 ⇒ x − 1 3 = 1 x x + + + + = + + x x − X X¢ Y (–7, 0) (–4,0)(–1, 0) O − π π π
  • 124. 4 213 ∴ arg(z1) − arg(z2 ) = 0 Aliter Given, | z1 + z2| =| z1| + | z2| =| z1|2 + | z2|2 + 2 Re(z z12 ) [squaring both sides] =| z1|2 + | z2|2 + 2| z1|| z2| ⇒ Re (z z12 ) =| z1|| z2| ⇒ | z1|| z2|cos(θ1 − θ2 ) =| z1|| z2| ⇒ θ1 − θ2 = 0 ∴ arg (z1) − arg(z2 ) = 0 z 26. Given, w = and|w| = 1 ⇒ = 1 ⇒ | |z = z − 1 ⇒ z lies on perpendicular bisector of (0, 0) and 0, . ⇒ − + 2 arg(z) = π arg(− i ) = − 2 ∴ arg(z) = 28. Since, z1/ 3 = p + iq ∴ z = (p + iq)3 = p3 − iq3 + 3ip q2 − 3pq2 Given that, z = x − iy ∴ x − iy = p3 − 3pq2 + i(3p q2 − q3 ) ⇒ x = p3 − 3pq2 and − y = 3p q2 − q3 ⇒ x 2 2 = p − 3q p and= − 3p2 + q2 ∴= − 2 p2 − 2q2 ⇒ (p2 + q2 ) p q = − 2 29. Using the relation, if | z1 + z2| =| z1| + | z2| Then, arg(z1) = arg(z2 ) Since, | z2 + (− 1)| =| z2| + −|1| Then, arg(z2 ) = arg(− 1) ⇒ 2 arg(z) = π [arg(− 1) = π] ⇒ arg(z) = ⇒ z lies on Y-axis (imaginary axis). 30. Since, origin, z1 and z2 are the vertices of an equilateral triangle, then z12 + z22 = z z1 2 ⇒ (z1 + z2 )2 = 3z z1 2 ...(i) Again, z1, z2 are the roots of the equation z2 + az + b = 0 Then, z1 + z2 = − a and z z12 = b On putting these values in Eq. (i), we get (− a)2 = 3b ⇒ a2 = 3b 31. Let z = r e1 iθ ⇒ w = r e2 iφ ⇒ z = r e1 iθ Given, | zw| = 1 ⇒ |r e1iθ ⋅ r e2iφ| = 1 ⇒ r r1 2 = 1 ...(i) and arg(z) − arg(w) = ⇒ ...(ii) Now, zw = r e1−i θ ⋅ r e2i φ Hence, z lies on a straight line. 27. Given, z + i w = 0 ⇒ z = − iw ⇒ z = iw ⇒ w = − iz and arg(zw) = π ⇒ arg(− iz2 ) = π ⇒ arg(− i ) + 2 arg(z) = π 3 − − x + x +
  • 125. 4 214 = r r e1 2−i( θ − φ ) = 1⋅e iπ / 2 [from Eqs. (i) and (ii)] ∴ zw = − i 32. Now, 11+− ii x ((11+− ii)()(11++ ii)) x (1 + i )2 x 1 − 1 + 2i x = = 34. Given,|β| = 1 β − α β − α ∴ = [1 = ββ] 1 − αβ ββ − αβ 1 β − α 1 = = × 1 = 1 [| z| =| z|] |β| β − α ( )1 35. Given, z ∴ | |z = | z1| = | z1| | z2| | z2| =[| zn| =| |z n ] (2) (3 5)2 102 ⋅2 =2 = 2 = 2 (10) (10) 36. Given,|w | = 1and z = w + 1 ⇒ z w( + 1) = w − 1 ⇒ w z(− 1) = − 1 − z 1 + z 1 + z ⇒ w =⇒|w | = 1 − z 1 − z |1 + z| ⇒ 1 =⇒ |1 − z| =|1 + z| |1 − z| On squaring both sides, we get 12 + | |z 2 − 2 Re (z) = 12 + | |z 2 − 2 Re (z) ⇒ 4Re (z) = 0 ∴ Re (z) = 0 37. Let z = x + iy ∴ | |z 2 + | z − 3|2 + | z − i|2 =| x + iy|2 + |(x − 3) + iy|2 + | x + i y(− 1)|2 = x2 + y2 + (x − 3)2 + y2 + x2 + (y − 1)2 = x2 + y2 + x2 − 6x + 9 + y2 + x2 + y2 + 1 − 2 y = 3x2 + 3y2 − 6x − 2 y + 10 = 3(x2 − 2 x + 1) + 3 y2 −y + 1 + 10 − 3 − 9 2 = 3(x − 1)2 + 3 y − 1 + 3 It is minimum, when x − 1 = 0 and y − = 0 − i 2 2 1 1 + i x ⇒ = ( )i x = 1 1 − i ⇒ ( )i x = ( )i 4n where, n is any positive integer. ∴ x = 4n [given] 33. Now, (1 + ω − ω2 7)= − ω(2 − ω2 7) = −(2ω2 7) = − 27 ⋅ω14 = − 128(ω3 4) ω2 [1 + ω + ω2 = 0] = − 128ω2 [ω3 = 1] + + + + + + + + + + + + = −
  • 126. 4 215 π = −iπ / 4 e = 2e iπ / 4 + iπ / 4 = 2e 2iπ / 4 = 2e iπ / 2 π = 2 cos 2 + i sin 2 39. Let z1 = x + iy and z2 = p − iq, where x q, > 0 Given,| z1| =| z2| ⇒ x2 + y2 = p2 + q2 ...(i) z1 + z2 = ∴ z1 − z2 =...(ii) If xq + yp ≠ 0, then it is purely imaginary and if x y xq + yp = 0 or= − = λ [say] p q ⇒ x = pλ, y = − qλ From Eq. (i), we get p2 + q2 = λ2(p2 + q2 ) ⇒ λ2 = 1 ⇒ λ = ± 1 For λ = − 1 and z1 ≠ z2, but| z1| =| z2| In this case, Eq. (ii) is zero. 40. Given, z = e 2π / 3 ∴ 1 + z + 3z2 + 2 z3 + 2 z4 + 2 z4 + 3z5 = 1 + e 2π / 3 + 3(e 2π / 3 2)+ 2(e 2π / 3 3) + 2(e 2π / 3 4)+ 3(e 2π / 3 5) 2 4 = 1 cos 3 + i sin 3 4 4 + 3 cos 3 + i sin 3 8π 8 + 2[cos 2π + i sin 2π ] + 2 cos 3 + i sin 3 10 10 + 3 cos 3 + i sin 3 = 1 + [cos 120° + i sin 120° ] + 3[cos 240° + i sin 240° ] + 2[cos 360° + i sin 360° ] + 2[cos 480° + i sin 480° ]+ 3[cos 600° + i sin 600°] 1 3 1 3 = 1 2 +2i + 3 2 − 2 i + 2[1 + 0] + 2[cos 120° + i sin 120° ] + 3[cos 120° − i sin 120° ] 41. Given, (x − 1)3 = − 5 (x − 1) = − 5, − 5ω, − 5ω2 ⇒ x = 1 − 5 1,− 5ω, 1 − 5ω2 ⇒ x = − 4 1,− 5ω, 1 − 5ω2 42. Given, z − 3 i = 1 z + 3 i ⇒ | z − 3 i| =| z + 3 i| [if| z − z1| =| z − z2|, then it is perpendicular bisector of z1 and z2] Hence, perpendicular bisector of (0, 3) and (0, − 3) is X-axis. 43. (z3 − z1) = (z2 − z1)(cos 90° + i sin 90° ) ⇒ (z3 − z1) = i z( 2 − z1) ⇒ (z3 − z1)2 = − (z2 − z1)2 ⇒ (z3 − z1)2 + (z2 − z1)2 = 0 Aliter Let z1 = 0, z2 = 1 − i and z3 = 1 + i Now, (z1 − z2 )2 + (z1 − z3 )2 = [0 − (1 − i )]2 + [0 − (1 + i )]2 = (1 − i )2 + (1 + i )2 = 12 + i 2 − 2i + 12 + i 2 + 2i = 1 − 1 + 1− 1 = 0 44. We know that, ∴ x = = ∴ = + + − = + −π + π π x x + + − − + + − + + − + x
  • 127. 4 216 | z − z1| + | z − z2| = k will represent an ellipse, if | z1 − z2| < k. Hence, the equation | z − z1| + | z − z2| = 2| z1 − z2|, represents an ellipse. 45. Given, z1 ≠ ± 1 Since, z2 and z3 can be obtained by rotating vector representing through and , respectively. ∴ z2 = z1ω and z3 = z1ω2 ∴ z z z1 2 3 = z1 × z1ω × z1ω2 = z13ω3 = z13 [ω3 = 1] 46. Since, z1, z2 and z3 are the vertices of an equilateral triangle, therefore | z1 − z2| =| z2 − z3| =| z3 − z1| = k [say] Also, α = 3 + i ) × 2 = 1 ⇒ |α =| Let A = αz1 + β, B = αz2 + β and C = αz3 + β Now, | AB| = α|z2 + β − α( z1 + β)| =|α(z2 − z1)| =|α|| z2 − z1| =|1|| z2 − z1| = 1| z2 − z1| =| z2 − z1| = k Similarly, BC = CA = k Hence, the points αz1 + β α, z2 + β and αz3 + β are the vertices of an equilateral triangle. 47. Given, equation is 1 + z + z3 + z4 = 0. ⇒ (1 + z) + z3(1 + z) = 0 ⇒ (1 + z)(1 + z3 ) = 1 ⇒ z = − 1, z3 = − 1 ⇒ z = − 1 ∴ z = − 1, − ω, − ω2 Hence, roots are in cubic roots of unity. Hence, these roots are the vertices of an equilateral triangle. 48. Given, z1 = 2 2 1( + i ) and z2 = 1 + i 3 (− 1 − 3 i ) (− 1 + i 3) i(− 2 − 2i 3) 3 1 = − i 4 2 2 ∴ Amplitude 51.| z1| =| z2| =| z3| = 1 ⇒ | z1|2 =| z2|2 =| z3|2 = 1 ∴ z z1 1 = z z22 = z z33 = 1 1 1 ⇒ z1 = , z2 = z1 z2 1 and z3 = z3 ∴ | z1 + z2 + z3| = = 1 = 1 52. Given, ( 3 i + 1)100 = 299(a + ib) 1001 100 99 2 ∴ 2 i + 2 = 2 (a + ib) ω= 2 ⇒ 2(− ω2 100) = a + ib ⇒ 2 × ω200 = a + ib ⇒ 2 × (ω3 66) × ω2 = a + ib 66 − 1 − 3 i ⇒ 2 × ( )1 a + ib 2 ⇒ − 1 − 3i = a + ib ∴ a2 + b2 = 4 53. (1 + 3 i )4 + (1 − 3 i )4 = −(2ω2 4) + (− 2ω)4 1 1 1 + + z1 z2 z3 = =− − + + − + + − − = + + − + = + − = =− = + − = + − − = + − − − = − + × − + + = − −
  • 128. 4 217 − 1 + 3 i 2 − 1 − 3 i ω =and ω 2 2 = 16ω8 + 16ω4 = 16(ω3 2) ω2 + 16 (ω3 )ω = 16(ω2 + ω) = 16(− 1) = − 16 54.11/ 4 = (cos θ + i sin θ)1/ 4 [1 + ω + ω2 = 0] = (cos 2πr + i sin 2πr )1 4/ πr πr = cos + i sin 2 2 where, r = 0, 1, 2, 3 ∴ 11/ 4 = 1, i, − 1, − i ∴Required value = 12 + i 2 + (− 1)2 + (− i )2 = 1 − 1 + 1 − 1 = 0 55. From figure, it is clear that argument ≥ 90°. The given equation | z − −( 1 + i ) ≤ 1|represents the points inside and on the boundary of the circle, centred at (− 1 1, ) and whose radius is 1. It lies in 2nd quadrant. The point (0, 1), i.e. i lies on it such that, it has least argument. ∴The required complex number is i. 56. (| z1| −| z2|)2 =| z1 − z2|2 ⇒ | z1|2 + | z2|2 − 2| z1|| z2| =| z1|2 + | z2|2 − 2| z1|| z2|cos(θ1 − θ2 ) ⇒ cos(θ1 − θ2 ) = 1 ⇒ θ1 = 2nπ + θ2 Thus, if z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ1 + i sin θ2 ) ⇒ z1 = r1(cos θ2 + i sin θ2 ) and z2 = r2(cos θ2 + i sin θ2 ) z r ⇒ 1 = 1 z2 r2 z ⇒ Im 1 = 0 z2 57. Gievn, (x + iy)(1 − 2i ) = 1 + i ⇒ (x + iy)(1 − 2i ) = 1 + i ⇒ (x − iy)(1 + 2i ) = 1 + i 1 + i ⇒ x − iy = ...(i) 1 + 2i 1 + i 1 − i ⇒ x − iy ...(ii) i + 2i 1 − 2i 1 − i ⇒ x + iy = 1 − 2i 50 ⇒ ⇒ ∴ 59. Locus of z is a circle passing through origin. 60. Given, 2 x = 3 + 5 i ⇒ 8x3 = 27 − 125 i + 27 + 5i − 25 × 9 ⇒ 2 x3 = − 198 + 10i 4 ...(i) Also, 4x2 = 9 − 25 + 30i P1 P(0, 1) X Y X' Y' Q G (–1,1) P1
  • 129. 4 218 i ...(ii) [form Eqs. (i) and (ii)] 61. We have,| z1 + z2|2 =| z1|2 + | z2|2 ⇒| z1|2 + | z2|2 + 2Re (z z12 ) =| z1|2 + | z2|2 ⇒ Re (z z12 ) = 0 ⇒ z z1 2 + z z2 1 = 0 z1 = − z1 ⇒ z2 z2 z1 z2 ⇒ z1 is purely imaginary. z2 62. Let z 1i++i2 32 = (1 +3i+ 34)(i 2i ) i + 1 63. We have, | z1 + z2|2 = (| z1|2 + | z2|2 + 2| z1|| z2|) ⇒ =| z1|2 + | z2|2 + 2| z1|| z2|cos(θ1 − θ2 ) =| z1|2 + | z2|2 + 2| z1|| z2| ⇒ cos(θ1 − θ2 ) = 1 ⇒ θ1 = θ2 ⇒ arg (z1) = arg (z2 ) ∴ z arg 1 = 0 z2 64. Given, S = 1 + 2 ω + 3ω2 ++ 3 nω3n − 1 ∴ Sω = ω + 2ω2 + 2ω2 ++ (3 n − 1)ω3n − 1 + 3 nω2n ⇒ S(1 − ω =)1 + ω + ω2 ++ ω3n − 1 − 3 nω3n = 0 − 3 n 3 n 3 n ⇒ S = − = 1 − ω ω − 1 2 4n 65. Let z 2 cos 4 − i2 sin 4 cos 4 2 cos2 2i sin cos 4 4 4 4n cos 4 − i sin 4 cos + i sin θ 4 4 cos nθ − i sin nθ (cos θ + i sin θ)−n = cos nθ + i sin nθ = (cos θ + i sin θ)n = (cos θ + i sin θ)−2n = cos 2nθ − i sin 2nθ 66. Given, x = − 3 + i 3 2 − 1 + i 3 ⇒ x =− 1 = ω − 1 2 ∴ (x2 + 3x) (2 x2 + 3x + 1) = [(ω − 1)2 + 3(ω − 1)] [(2ω − 1)2 + 3(ω − 1) + 1] = (ω2 + ω − 2) (2 ω2 + ω − 1) = −(1 − 2) (2 − 1 − 1) = − 18 [1 + ω + ω2 = 0] 67. Given, (x + iy)1 3/ = 2 + 3 i ⇒ x + iy = (2 + 3 i )3 = 8 + 36 i + 54 i 2 + 27i 3 = − 46 + 9 i On equating real and imaginary parts from both sides, we get x = − 46, y = 9 ∴ 3x + 2 y = − 138 + 18 = − 120 68. Let z = x + iy ∴ | z + 3 − i| =|(x + 3) + i y(− 1)| = 1 ⇒ (x + 3)2 + (y − 1)2 = 1 ...(i) = + − x = + − + = = = − + ⇒ = − + = × × =
  • 130. 4 219  arg z = π ⇒ tan−1 y = π x y ⇒ = tan π = 0 ⇒ y = 0 ...(ii) x From Eqs. (i) and (ii), we get x = − 3, y = 0 ∴ z = − 3 ⇒ | |z = −|3| = 3 69. Given,| z1| =| z2| = =| zn| = 1 n ⇒ arg (z) = π − θ arg (−z) = − θ ∴ arg (z) − arg (− z) = π − θ − −(θ) = π − θ + θ = π 74. cos 30° + i sin 30° cos 60° − i sin 60° = (cos 30° + i sin 30° )(cos 60° + i sin 60° ) = cos 90° + i sin 90° = i ⇒ = = =  ⇒ = = = =  ⇒ = = = + = + +  + + + = + + + =   = + + +  from Eq. (i  = = ∴ = = + = + = = + − − − θ θ θ θ θ θ θ = + + = = + ∴ = − =−π+θ
  • 131. 5 Inequalities and Quadratic Equation Inequality An equation having signs >, <, ≥ or ≤ at the place of sign of equality ( )= is known as inequality or inequation. or Let a and b be two real numbers. If a − b is negative, we say that a is less than b a( < b) and if a − b is positive, then a is greater than b a( > b). Properties of Inequalities We shall learn about some elementary properties of inequalities which will be used in the subsequent discussion. Chapter Snapshot ● Inequality ● Generalised Method of Intervals for Solving Inequalities by Wavy Curve Method (Line Rule) ● Absolute Value of a Real Number ● Logarithms ● Arithmetico-Geometric Mean Inequality ● Quadratic Equation with Real Coefficients ● Formation of a Polynomial Equation from Given Roots ● Symmetric Function of the Roots ● Transformation of Equations ● Common Roots ● Quadratic Expression and its Graph ● Maximum and Minimum Values of Rational Expression ● Location of the Roots of a Quadratic Equation ● Algebraic Interpretation of Rolle’s Theorem ● Condition for Resolution into Linear Factors ● Some Application of Graphs to Find the Roots of Equations
  • 132. 5 221 i. ii. iii. iv. v. vi. vii. If a > b and b > c, then a > c. Generally, if a1 > a2, a2 > a3,...,an −1 > an , then a1 > an If a > b, then a ± c > b ± c, ∀ ∈c R a b b a If a > b, then for m > 0, am > bm, > and for m < 0, bm > am, > m m m m If a > b > 0, then (a) a 2 > b2 (b) | |a > | |b (c) 1 < 1 a b and if a < b < 0 , then (a) a 2 > b2 (b) | |a > | |b (c) 1 > 1 a b If a < 0 < b, then (a) a 2 > b2 , if | |a > | |b (b) a 2 < b2 , if | |a < | |b If a < x < b and a, b are positive real numbers, then a 2 < x2 < b2 If a < x < b and a is negative number and b is positive number, then (a) 0 ≤ x2 < b2 , if | |b > | |a (b) 0 ≤ x2 ≤ a 2 , if |a | > | |b
  • 133. 5 X 222 1 1 1 x b a, , if a and b have same sign If x ∈[a b, ] ⇒ 1 x ∈ − ∞ , 1 a 1 b, , if a and b have opposite signs a If > 0, then b (a) a > 0, if b > 0 (b) a < 0, if b < 0 If ai > bi > 0, where i = 1, 2, 3, …, n, then a a a1 2 3an > b b b1 2 3bn If ai > bi , where i = 1, 2, 3, …, n, then a1 + a2 + a3 ++ an > b1 + b2 ++ bn If 0 < a <1and n is a positive rational number, then (a) 0 < a n <1 (b) a −n >1 If a and b are positive real numbers such that a < b and if n is any positive rational number, then (a) a n < bn (b) a −n > b−n (c) a1/n < b1/n If a >1and n is any positive rational number, then (a) a n >1 (b) 0 < a −n <1 If 0 < a <1and m, n are positive rational numbers, then (a) m > n ⇒ a m < a n (b) m < n ⇒ a m > a n
  • 134. 5 223 and a2 < b2 < c2 ⇒ 3a2 < a2 + b2 + c2 < 3c2 …(ii) From Eqs. (i) and (ii), we get 2 2 2 2 + b + c 2 a < < viii. ⇒ c a + b + c a X Example 2. The value of x for which ix.12x − 6 < 0,12 − 3x < 0, is (a) φ (b) R (c) R − {0} (d) None of these x.Sol. (a) 12 x − 6 < 0 ⇒ 12 x < 6 ⇒ x < xi.and 12 − 3x < 0 …(i) ⇒ 12 < 3x ⇒ 4 < x xii.⇒ x > 4 …(ii) no real number x satisfying both the inequalities (i) and (ii). So, there is Hence, the given system of inequalities has no solution. xiii.X Example 3. The value of x for which x − 3 x −1 x − 2 − x < − , 2 − x > 2x − 8 10 (a) 1, 3 (b) 1, 3 xiv.(c) R (d) None of these Sol. (b) We have, x − 3 − x < x − 1 − x − 2 4 2 3 ⇒ 3x − 9 − 12 x < 6x − 6 − 4x + 8 xv.⇒ −11x < 11 ⇒ − x < 1 ⇒ x > − 1 …(i) and 2 − x > 2 x − 8 ⇒ −3x > − 10 3 3c a + b + c 3aa1 < a2 < a3 << an −1 < an If a >1and m n, are positive rational numbers, then (a) m > n ⇒ a m > a n (b) m < n ⇒ a m < a n + + < + + < 4 2 3 10
  • 135. 5 X 224 ⇒ x < …(ii) xvi.From Eqs. (i) and (ii), the solution of the given system of inequalities is given by x ∈ − 1, 10 . 3 Generalised Method of Intervals X Example 1. If a, b and c are positive real for Solving Inequalities by Wavy numbers such that a < b < c, then show that Curve Method a 2 a 2 + b2 + c2 c2 < < . (Line Rule) c a + b + c a Sol. We have, c > b > a > 0 Let F x( ) = (x − a1 )k1 (x − a2 )k2(x − an −1 )kn−1 ⇒ c2 > b2 > a2 > 0 (x − an )kn Now, a < b < c ⇒ 3a < a + b + c < 3c where, k1, k2, …, kn ∈Z and a1, a2, …, an are ⇒ 1 < 1 < 1 …(i) fixed real numbers satisfying the condition For solving F x( ) > 0 or F x( ) < 0, consider the following algorithm: ■ We mark the numbers a1, a2,…, an on the number axis and put the plus sign in the interval on the right of the largestof thesenumbers, i.e. on the right of an. ■ Then, we put the plus sign in the interval on the left of an, if kn is an even number and the minus sign, if kn is an odd number. In the next interval, we put a sign according to the following rule: ■ When passing through the point an − 1 the polynomial F x( ) changes sign, if kn − 1 is an odd number.Then, we consider the next interval and put a sign in it using the same rule. ■ Thus, we consider all the intervals. The solution of the inequality F x( )> 0is the union of all intervals in which we have put the plus sign and the solution of the inequality F x( )< 0is the union of all intervals in which we have put the minus sign. Solution of Rational Algebraic Inequation If P x( ) and Q x( ) are polynomial in x, then the inequation and are known as rational algebraic inequations. To solve these inequations, we use the sign method as explained in the following algorithm : Algorithm Step I Obtain P x( ) and Q x( ). Step II Factorise P x( ) and Q x( ) into linear factors. Step III Make the coefficient of x positive in all factors. Step IV Obtain critical points by equating all factors to zero. Step V Plot the critical points on the number line. If there are n critical points, then they divide the number line into (n +1) regions. Step VI In the right most region the expression P x( ) bears positive sign and in other Q x( ) regions the expression bears positive and negative signs depending on the exponents of the factors. Example 4. Solve (x −1)(x − 2 1)( − 2x) > 0. Sol. We have, (x − 1)(x − 2)(1 − 2 x)> 0 ⇒ −(x − 1)(x − 2)(2 x − 1)> 0 ⇒ (x − 1)(x − 2 2)( x − 1)< 0 On number line mark x = , 1, 2 – + – +
  • 136. 5 225 1/2 1 2 When x > 2, all factors (x − 1), (2 x − 1) and (x − 2) are positive. Hence, (x − 1)(x − 2)(2 x − 1)> 0 for x > 2. Now, put positive and negative sign alternatively as shown in figure. Hence, solution set ofor (x − 1)(x − 2)(2 x − 1)< 0 is , ∪ (1 2, ). 2 X Example 5. Solve . Sol. ⇒ ⇒ ⇒ ⇒ ⇒ (7 x − 12)(3x − 5)≤ 0 Sign scheme of (7 x − 12) (3x − 5) is as follows : + – + ⇒ 3 7 Ø x = is not included in the solutions as at x = denominator becomes zero. X Example 6. The value of x for which (x − 2) (3 x − 3) < 0, is (a) (2, 3) (b) [2, 3) Sol. (a) (x − 2)3 (x − 3)< 0 ⇒ (x − 2)(x − 3)< 0 (c) (0, 3) (d) (2, 3] [as (x − 2)2 is positive for all real values of x ≠ 2] –∞ + – + ∞ 2 3 Here, interval is open as x = 2, 3 do not satisfy inequality. i.e. 2 < x < 3 or x ∈(2, 3) Work Book Exercise 5.1 1 The value of x satisfying the inequalities 1 x + ≥ 2 hold x (0, ∞) R φ [0, ∞) x2 2 ≥ 0 x − 1 (1, ∞) [1, ∞) {0} ∪ (1, ∞) None of these 3 (x − 2) (4 x − 3) (3 x − 4) (2 1 − x) ≤ 0 (1, 3) (− ∞, 1) ∪ (3, ∞) (−∞, 1] ∪ [3, ∞) None of the above 4 If c < d, x2 + (c + d x) + cd < 0 (−d, − c] (−d, − c) R φ 2 1, 2 R φ 1, 2 2 Absolute Value of a Real Number The absolute value of a real number ‘x’ is denoted x, if x ≥ 0 by | |x and defined by | |x x, if x < 0 Ø ● (i) |x| is also defined as (ii) If x is positive, then (iii) If x is negative, then e.g. or x ∈ 5/3 12/7 = =− = − = = =±
  • 137. 5 X 226 ( ) b ∞ − − ∪ ∞ [ ) , , 3 2 3 ● (i) a≤ |a| (ii) |ab| = |a||b| (iii) a = |a| b |b| (iv) |a + b|≤ |a| + |b| [triangle inequality] (v) |a − b|≥ |a| − |b| [triangle inequality] (vi)|a + b| = |a| + |b| , iff ab ≥ 0 (vii) |a + b| = |a| − |b|,iff ab ≤ 0 Inequations Containing Absolute Values By definition, | |x < a ⇒ −a < x < a a( > 0) | |x ≤ a ⇒ −a ≤ x ≤ a | |x > a ⇒ x < − a and x > a | |x ≥ a ⇒ x ≤ − a and x ≥ a a ≤ | |x ≤ b ⇒ x ∈ −[ b, − a] ∪ [a b, ] where, a, b > 0 Forms of the Inequations Containing Absolute Values Form 1 The inequation of the form f (| |x ) < g x( ) is equivalent to the collection of systems f x( ) < g x( ), if x ≥ 0 f (−x) < g x( ), if x < 0 Form 2 The inequation of the form | f x( )| < g x( ), is equivalent to the systems f x( ) < g x( ), if g x( ) > 0 f x( ) < g x( ), if g x( ) > 0 Form 3 The inequation of the form | f x( )| > g x( ), is equivalent to the systems f x( ) > g x( ), if g x( ) < 0 f x( ) > g x( ), if g x( ) < 0 Form 4 The inequation of the form | f (| |)|x≥ g x( ) is equivalent to the collection of systems | f x( )| ≥ g x( ), if x ≥ 0 | f (−x)| ≥ g x( ), if x < 0 Form 5 The inequation of the form | f x( )| ≥ | ( )|g x is equivalent to the collection of system f 2 ( )x ≥ g 2 ( )x Form 6 The inequation of the form h x( , | f x( )|) ≥ g x( ) is equivalent to the collection of systems h x f x{ , ( )} ≥ g x( ), if f x( ) ≥ 0 h x{ , − f x( )} ≥ g x( ), if f x( ) < 0 X Example 7. The solution set of |3 − 4x| ≥ 9 is 3 (a) , 2 ∪ [3, ∞) (c) (−∞, )2 ∪ [2, ∞) (d) None of the above Sol. (b) We have, |3 − 4x|≥ 9 ⇒ 3 − 4x ≤ − 9 or 3 − 4x ≥ 9 [since,| |x ≥ a ⇒ x ≤ − a or x ≥ a] ⇒ −4x ≤ − 12 or −4x ≥ 6 ⇒ x ≥ 3 or x [dividing both sides by −4] ⇒ x ∈ −∞ , − 2 3 ∪ [3, ∞) |x + 3| + x X Example 8. The solution set of >1is x + 2 (a)[−5, − 2] ∪ −[ 1, ∞) (b)[−5, − 2) ∪ −[ 1, ∞) (c) (−5, − 2) ∪ −( 1, ∞) (d) None of these have, |x + 3| + x > 1 Sol. (c) We ⇒ − 1> 0 ⇒> 0 Case I When x + 3 ≥ 0, i.e. x ≥ − 3 0 ∴ |x + 3| − 2 > ⇒> 0 x + 1 ⇒> 0 x + 2 ⇒ {x + 1> 0 and x + 2 > 0} or {x + 1< 0 and x + 2 < 0} ⇒ {x > − 1 and x > − 2} or {x < − 1 and x < − 2} ⇒ x > − 1 or x < − 2 ⇒ x ∈ −( 1, ∞) or x ∈ −∞ −( , 2) ⇒ x ∈ −( 3, − 2) ∪ −( 1, ∞) [x ≥ − 3] …(i) Case II When x + 3 < 0, i.e. x < − 3 x + x x x + + + x x + − + x + x x + − +
  • 138. 5 227 ∴ |x + 3| − 2 > 0 x + 2 ⇒ − x − 3 − 2 > 0 x + 2 ⇒ −(x + 5) > 0 x + 2 ⇒ (x + 5) < 0 x + 2 ⇒ {x + 5 < 0 and x + 2 > 0} or {x + 5 > 0 and x + 2 < 0} ⇒ (x < − 5 and x > − 2), which is not possible or {x > − 5 and x < − 2} ⇒ x ∈ −( 5, − 2) …(ii) From Eqs. (i) and (ii), we get x ∈ −( 5, − 2) ∪ −( 1, ∞) Example 9. Solve the inequation 1 − | |x 1. 1 + | |x 2 Sol. The given inequation is equivalent to the collection of systems. ⇒ 1 − 1 +x x 21, if x ≥ 0 |1 +1 x| ≥ 21, if x ≥ 0 ⇒ 1 + 1 −x x 21, if x < 0 |1 −1 x| ≥ 21, if x < 0 1 ≥ 1 , if x ≥ 0 ⇒ 1+1 x 21 ≥ , if x < 0 1− x 2 1 − x ≥ 0, if x ≥ 0 x − 1 ≤ 0, if x ≥ 0 ⇒ 1 + x ⇒ x + 1 1 + x ≥ 0, if x < 0 x + 1 ≤ 0, if x < 0 1 − x x − 1 From Eqs. (i) and (ii), the solution of the given equation is x ∈ −[ 1, 1]. X Example 10. The value of x, |x + 3| > |2x −1| is −2 2 (a) , 4 (b) , 3 3 (c) (0, 1) (d) None of these Sol. (a) Given,|x + 3|>|2 x − 1| On squaring both sides, we get |x + 3|2 >|2 x − 1|2 ⇒ {(x + 3) − (2 x − 1)} {(x + 3) + (2 x − 1)} > 0 ⇒ {(− x + 4)(3x + 2)} > 0 – + – ⇒ 3 Logarithms Let there be a number a > 0, ≠1. A number p is called the logarithm of a number x to the base a, if a p = x and is written as p = log a x. Obviously x must be positive. Ø ● If x < 0, loga x is imaginary and if x = 0, loga x does not exist . ● loga x exists if and only if x, a> 0 and a ≠ 1 . x x x − + ≤ ≥ ≤ ≤ x x x x + − ≤ < ∴ < ≤ − x –1 + + 1 – –1 + + 1 – x ∈ − –2/3 4
  • 139. 5 X 228 Properties of Logarithms i.x. ii. iii. xi. iv. v. X Example 11. The value of x, log (e x − 3) <1is (a) (0, 3) (b) (0, e) (c) (0, e + 3) (d) (3, 3 + e) vi.Sol. (d) From definition of logarithms x − 3 > 0 ⇒ x > 3 …(i) Also, e > 1 given inequality may written as x − 3 < ( )e 1 ⇒ x < 3 + e ⇒ x ∈(3, 3 + e) vii.X Example 12. The value of x, log1 2/ x ≥ log1 3/ x is (a) (0,1] (b) (0,1) (c)[0,1) (d) None of these Sol. (a) Case I When x ≠ 1and x > 0 log1 2/ x ≥ log1 3/ x viii.⇒ log x 2 ≥ log x 3, when x ≠ 1 which is possible, only if 0 < x < 1. Case II When x = 1 ix.log1 2/ x = log1 3/ x, i.e. equality holds. Combining the above cases, 0 < x ≤ 1 or x ∈(0, 1] Work Book Exercise 5.2 If a >1, then (a) log a x > p ⇒ x > a p (b) 0 < log a x < p ⇒ 0 < x < a p If 0 < a <1, then (a) log a x > p ⇒0 < x < a p (b) 0 < log a x < p ⇒ a p < x <1 a loga x = x; a ≠ 0, ±1, x > 0 a logb x = xlogb a ; a > 0, b > 0, ≠1, x > 0 log a a =1, log a 1 = 0; a > 0, ≠1 1 log a x = ; x, a > 0, ≠1 log x a log x log a x = log b x ⋅ log a b = b ; a, b > 0, ≠1, log b a x > 0 For m, n > 0, a > 0, ≠1 (a) log (a m n⋅ ) = log a m + log a n m (b) log a n = log a m − log a n (c) log (a mx ) = x log a m For x > 0, a > 0, ≠1 1 (a) log n ( )x log a x a n (b) log n (xm ) m log a x a n (a) log a x > 0, iff x >1, a >1or 0 < x <1, 0 < a <1 (b) log a x < 0, iff x >1, 0 < a <1or 0 <x <1, a >1 For x > y > 0, (a) log a x > log a y, if a >1 (b) log a x < log a y, if 0 < a <1
  • 140. 5 229 Solve for x, 1 x + 4 > 5 (−∞, 1) R − −[ 9, 1] 2 | x + 2| < 4 (−6 2,) 3 log2 x > 0 (0, ∞) 4 log4 x > 1 (0, ∞) (−6 0,) (− ∞, 0) (4, ∞) (−∞, 9) None of these (−6 2,] (0, 2) (1, ∞) (4, ∞) (−4, ∞) (−∞, 4) 7 (x − 1)2 > 9 (−∞, − 2) ∪ (4, ∞) (4, ∞) (−∞, 2) (−∞, − 2] ∪ (4, ∞) 8 (x + 1)2 < 25 (−6 4,] (4, 6) 9 log x(x + 7) < 0 (−6 4,) (−4 6,) (−1 0,) (−1 1, ) (0, 1) (−∞, 1) 10 x2 + 6 5 log0.2(x + 5) > 0≥ 1 5x (−5, − 4) (−5 4,) (0, 4) (0, 5) 6 log x 0 5. > 2 (−∞, − 3) (−∞, − 3) ∪ (3, ∞) R , 1 (−∞, 1) (−1 0, ) (−1 1, ) (−∞, − 3] ∪ −[ 2 0,) ∪ (0 2,] ∪ [3, ∞) Arithmetico-Geometric Mean Inequality If a1, a2, …, an are n distinct positive real numbers, then a1 + a2 ++ an 1/n > (a a12an ) n i.e. AM > GM Ø If a1 = a2 == an, then AM = GM. Thus, equality occurs when all quantities are equal. X Example 13. If a, b and c are three distinct positive real numbers, then 1 1 1 (a) (a + b + c) + + 9 a b c 1 1 1 (b) (a + b + c) + + 9 a b c 1 1 1 (c) (a + b + c) + + 9 a b c 1 1 1 (d) (a + b + c) + + 9 a b c Sol. (a) We have, AM > GM 1 1 1 + + 1 3/ a + b + c 1 3/ a b c 1 ∴ > (abc) and 3 3 abc ⇒ a + b + c > 3 (abc)1 3/and a1 + b1 + c1 > (abc3)1 3/ ⇒ (a + b + c) 1 + 1 + 1 > 9 a b c Weighted AM and GM Inequalities If a1, a2,, an are n positive real numbers and m1, m2, …, mn are n positive rational numbers, then m a1 1 + m a2 2 ++ m an n m1 + m2 +... + mn 1 > (a1m1 ⋅ a2m2anmn ) m1 + m2 ++ mn i.e. weighted AM > weighted GM X Example 14. If a, b and c are distinct positive integers, then axb − c + bxc − a + cxa − b is (a) > 2(a + b + c) (b) R (c) > (a + b + c) (d) < (a + b + c)
  • 141. 5 X 230 Sol. (c) We have, a⋅ x b − c + b⋅ x c − a + c ⋅ x a − b > a + b + c {(x b − c ) (a x c − a b) (x a − b ) }c 1/ a + b + c ax b − c + bx c − a + cx a − b ⇒ > a + b + c {xa b( − c ) + b c( − a) + c a( − b)}1/ a + b + c ⇒ ax b − c + bx c − a + cx a − b > (a + b + c) Example 15. If the equation x4 − 4x3 + ax2 + bx +1 = 0 has four positive roots, then a b, are (a) a = 4, b = 6 (b) a = − 4, b = 6 (c) a = 2, b = 3 (d) a = 6, b = − 4 Sol. (d) Let α β γ and δ be four roots of the given equation. Then, α + β + γ + δ = 4 and αβγδ = 1 ⇒ AM of α β γ δ = GM of αβ γ δ ⇒α =β =γ =δ ⇒ α = β = γ = δ = 1 [α + β + γ + δ = 4, also αβ + αγ + αδ + βγ + βδ + γδ = a, αβγ + αβδ + βγδ + αγδ = − b] ∴ x4 − 4x3 + ax2 + bx + 1 = x4 − 4x3 + 6x2 − 4x + 1 ⇒ a = 6 and b = − 4
  • 143. 5 X 232 i. ii. X Example 16. If m >1and n ∈ N, then 1m + 2m + 3m ++ nm n +1 m (a) n 2 m m m m m 1 + 2 + 3 ++ n n +1 (b) n 2 1m + 2m + 3m ++ nm (c)≥1 (d)≤1 n Sol. (a) We know that for m > 1, AM of mth power > mth power of AM m m m m m ∴ 1 + 2 + 3 + + n 1 + 2 + 3 + + n ⇒ n 2 If a1, a2, ..., an are n positive distinct real numbers, then a1m + a2m + anm a1 + a2 ++ an m (a) , n n if m < 0 or m >1 a1m + a2m +...+ anm a1 + a2 ++ an m (b) n n if 0 < m <1. i.e. the arithmetic mean of the mth power of n positive quantities is greater than mth power of their arithmetic mean except when m is a positive proper fraction known as mth power mean. (c) If a1, a2 an and b1, b2 bn are rational numbers and m is a rational number, then b a1 1m + b a2 2m ++ b an nm b1 + b2 ++ bn b a + b a + ... + b a m 1 1 2 2 n n , b1 + b2 ++ bn if m < 0 or m >1and b a1 1m + b a2 2m ++ b an nm b1 + b2 +... + bn m b a + b a + + b a 1 1 2 2  n n , if 0 < m <1 b1 + b2 + + bn If a 1, a 2, a 3, …, a n are distinct positive real numbers and p, q, r are natural numbers, then a 1p + q + r + a 2p + q + r ++ a np + q + r n a1p + a2p ++ anp a1q + a2q ++ anq n n a1r + a2r ++ anr n n 1 2 3 m m m m n + + + +  + + + + > + 
  • 144. 5 233 Some Other Standard Inequalities i. an an ii. ) ) iii. X Example 17. If none of b1, b2, , bn is zero, a1 a2 an 2 then + ++ is b1 b2 bn Weierstrass Inequality (a) If a1, a2,, an are n positive real numbers, then for n ≥ 2 , (1 + a1 )(1 + a2 )(1 + an ) >1 + a1 + a2 + + (b) If a1, a2,, an are positive numbers less than unity, then (1− a1 )(1− a2 )(1− an ) >1 − a1 − a2 − − Cauchy-Schwartz Inequality If a1, a2,, an and b1, b2,, bn are 2n real numbers, then (a b1 1 + a b2 2 +...+ a bn n )2 ≤ (a12 + a22 ++ an2 (b12 + b22 ++ bn2 with the equality holding if and only if a1 a2 an = = = b1 b2 bn Tchebychef’s Inequality If a1, a2,, an and b1, b2,, bn are real numbers such that a1 ≤ a2 ≤ a3 ≤≤ an and b1 ≤ b2 ≤≤ bn , then n a b( 1 1 + a b2 2 ++ a bn n ) a b a b a b or n a1 + a2 ++ an b1 + b2 ++ bn n n
  • 145. 5 X 234 Applications of Inequalities to Find the Greatest and Least Values i. ii. iii. X Example 18. The greatest value of x y2 3 when 3x + 4y = 5, is (a) (b) (c) (d) Sol. (b) Let P = x y2 3. Clearly, P is the product of 5 factors such that two of them are equal to x and the remaining 3 are equal to y. Now, 3x + 4y = 5 ⇒ 2 3x + 3 4y = 5 2 3 ⇒ 5 ⇒ 16 x y23 5 ⇒ x y2 3 ≤ 3 3 5 16 or maximum of x y2 3 = 3 16/ If x1, x2,, xn are n positive variables such that x1 + x2 ++ xn = c(constant), then the product x x1 2 xn is greatest when c x1 = x2 == xn = and the greatest value is n c n . n If x1, x2,, xn are positive variables such that x x1 2 xn = c(constant), then the sum x1 + x2 ++ xn is least when x1 = x2 == xn = c1/n and the least value of the sum is n c( 1/n ). If x 1, x 2,, x n are variables and m 1, m 2,, m n are positive real numbers such that x1 + x2 +… + xn = c(constant), then x1 1 ⋅ x2 m2 … xn mn is greatest, when m x1 x2 xn x1 + x2 +… + xn = =… = = m1 m2 mn m1 + m2 +… + mn
  • 146. 5 235 + < λ λ Work Book Exercise 5.3 1 If a, b and c are all positive real numbers, which one of the following is true? (b + c)(c + a)(a + b) ≥ 8abc (b + c)(c + a)(a + b) < 8abc ea + b + + f + gc < 9 (a + b) 1 a b 2 If a 1, a 2, , a n and b 1, b 2, , b n are any two sets of positive real numbers, then a12 + a22 +... + an2 2 b1 b2 bn ≤ (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) ≥ (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) < (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) > (a14 + a24 + + an4 )(b1−2 + b2−2 + + bn−2) 3 If a and b are two positive quantities whose sum is 1 1 λ, then the minimum value of 1 + 1 + is a b λ − λ − 1 + 1 + 4 If a x2 4 + b y2 4 = c 6, then the greatest value of xy is c3 c3 c3 c3 2ab 4ab 5 Which of the following is true, where a, b, c > 0? 2(a3 + b3 + c3) ≥ bc b( + c) + ca c( + a) + ab a( + b) a3 + b3 + c3 (a + b + c) (⋅ a2 + b2 + c2 ) 11 Quadratic Equation with Real Coefficients An equation of the form ax2 + bx + c = 0 …(i) Ø where, a ≠ 0, a, b, c∈R is called a quadratic equation with real coefficients. The quantity D = b2 − 4acis known as the discriminant of the quadratic equation in (i) whose roots are given by −b + b2 − 4ac −b − b2 − 4ac α =and β = 2a 2a The nature of the roots is as given below: (i) If a quadratic equation in x has more than two roots, then it is an identity in x that is a = b = c = 0. (ii) If a =1and b, c∈I and the roots are rational numbers, then these roots must be integers. X (iii) The roots are of the form p ± q p q( , ∈Q), iff a, b and care rational and D is not a perfect square. (iv) The roots are real and distinct, iff D > 0. (v) The roots are real and equal, iff D = 0. (vi) The roots are complex with non-zero imaginarypart, iff D < 0. (vii) The roots are rational, iff a, b and care rational and D is a perfect square. ● If α is a root of the equation f( )x = 0, then the polynomial f( )xis exactly divisible by (x −α) or (x −α) is a factor of f( )x and conversely. ● Every equation of nth degree (n≥ 1) has exactly n roots and if the equation has more than n roots, it is an identity. ● If the coefficients of the equation f( )x = 0 are all real and α + βi is its root, then α − βi is also a root, i.e. imaginary roots occur in conjugate pairs. ● If the coefficients in the equation are all rational and α + β is one of its roots, then α − β is also a root, where α, β ∈Q and β is not a perfect square. > + + + + + + < + + + + + + < + +
  • 147. 5 236 ● If there is any two real numbers ‘a’ and ‘b’ such that f( )a and f b( ) are of opposite signs, then f( )x = 0 must have atleast one real root between‘a’ and ‘b’. ● Every equation f( )x = 0 of odd degree has atleast one real root of a sign opposite to that of its last term. Example 19. The number of values of a for which (a 2 − 3a + 2)x2 + (a 2 − 5a + 6)x + a 2 − 4 = 0 is an identity in x, is (a)0 (b)2 (c)1 (d)3 Sol. (b) It is an identity in x, if a2 − 3 a + 2 = 0, a2 − 5a + 6 = 0, a2 − 4 = 0 ⇒ a = 1, 2 and a = 2, 3 and a = 2, −2 ∴ Equation is identity, if a = 2