3 complex numbers part 3 of 3

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MATHS H.O.D.: SUHAG R. KARIYA , BHOPAL, COMPLEX NUMBERS PART 3 OF 3
30 of 38
CCOOMMPPLLEEXX NNUUMMBBEERRSS
Some questions (Assertion–Reason type) are given below. Each question contains Statement – 1 (Assertion) and
Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select
the correct choice :
Choices are :
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.
(B)Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False.
(D) Statement – 1 is False, Statement – 2 is True.
344. Let z = eiθ
= cosθ + isinθ
Statement 1: Value of eiA
.eiB
. eiC
= –1 if A + B + C = π. Statement 2: arg(z) = θ and |z| = 1.
345 Let a1, a2, .... , an ∈R+
Statement–1 : Minimum value of 1 2 n 1 n
2 3 n 1
a a a a
....
a a a a
−
+ + + +
Statement–2 : For positive real numbers, A.M ≥ G.M.
346. Let log
5c 3b a
, log and log
a 5c 3b
     
     
     
then A.P., where a, b, c are in G.P. If a, b, c represents the sides of a
triangle. Then : Statement–1 : Triangle represented by the sides a, b, c will be an isosceles triangle
Statement–2 : b + c < a
347. Let Z1, Z2 be two complex numbers represented by points on the curves |z| = 2 and |z – 3 – 3i| = 2 2 . Then
Statement–1 : min |z1–z2| = 0 and max |z1 – z2| = 6 2
Statement–2 : Two curves |z| = 2 and |z – 3 –3i| = 2 2 touch each other externally
348. Statement–1 : If |z – i| ≤ 2 and z0 = 5 + 3i, then the maximum value of |iz + z0| is 7
Statement–2 : For the complex numbers z1 and z2 |z1 + z2| ≤ |z1| + |z2|
349. Let z1 and z2 be complex number such that 1 2 1 2z z | z | | z |+ = +
Statement–1 : arg 1
2
z
0
z
 
= 
 
Statement–2 : z1, z2 and origin are collinear and z1, z2 are on the same side of origin.
350. Let fourth roots of unity be z1, z2, z3 and z4 respectively
Statement–1 : 2 2 2 2
1 2 3 4z z z z 0+ + + = Statement–2 : z1 + z2 + z3 + z4 = 0.
351. Let z1, z2, . . . , zn be the roots of zn
= 1, n ∈ N.
Statement–1 : z1. z2 . . . zn = (– 1)n
Statement–2 : Product of the roots of the equation anxn
+ an – 1xn – 1
+ an – 2 xn – 2
+ . . . + a1x + a0 = 0, an ≠ 0, is (– 1)n
. 0
n
a
a
.
352. Let z1, z2, z3 and z4 be the complex numbers satisfying z1 – z2 = z4 – z3.
Statement–1 : z1, z2, z3, z4 are the vertices of a parallelogram
Statement–2 : 1 3 2 4
2 2
z z z z+ +
= .
353. Statement–1 : The minimum value of | z | | z i |+ − | is 0.
Statement–2 : For any two complex number z1 and z2, 1 2 1 2z z z z+ ≤ + .
354. Statement–1 : Let z1 and z2 are two complex numbers such that 1 2 1 2| z z | | z z |− = + then the orthocenter
of ∆AOB is 1 2z z
2
+
. (where O is the origin)
Statement–2 : In case of right angled triangle, orthocenter is that point at which triangle is right angled.
355. Statement–1 : If ω is complex cube root of unity then (x – y) (xω – y) (xω2
– y) is equal to x3
+ y2
Statement–2 : If ω is complex cube root of unity then 1 + ω + ω2
= 0 and ω3
= 1
356. Statement-1 : If |z| ≤ 4, then greatest value of |z + 3 – 4i| is 9.
Statement-2 : ∀Z1, Z2 ∈C, |Z1 + Z2| ≤ |Z1| + |Z2|

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357. Statement-1: The slope of line (2 – 3i) z + (2 + 3i) z − 1 = 0 is
2
3
Statement-2:: The slope of line az az b 0+ + = b∈R & a be any non-zero complex. Constant is
Re(a)
Im(a)
−
358. Statement-1: The value of
6
k 1
2 k 2 k
sin icos
7 7=
π π 
− 
 
∑ is i
Statement-2: The roots of the equation zn
= 1 are called the nth roots of unity where
z =
cos2 k 2 k
isin
n n
π π   
+   
   
where k = 0, 1, 2, ... (n − 1)
359. Statement-1: |z1 – a| < a, |z2 – b| < b |z3 – c| < c, where a, b, c are +ve real nos, then |z1 + z2 + z3| is greater than 2|a
+ b + c| Statement-2: |z1 ± z2| ≤ |z1| + |z2|
360. Statement-1: (cos2 + isin2)π
= 1
Statement-2: (cosθ +isinθ)n
= cosnθ + isin nθ it is not true when n is irrational number.
361. Statement-1 : If α1, α2, α3 …. α 8 be the 8th
root of unity, then α1
16
+ α2
16
+ α3
16
+ … + α8
16
= 8
Statement-2 : In case of sum of pth power of nth roots of unity sum = 0 if p ≠ kn where p, k, n are integers sum =
n if p = kn.
362. Statement-1: Locus of z, satisfying the equation |z – 1| + |z – 8| = 16 is an ellipse of eccentricity 7/16
Statement-2:: Sum of focal distances of any point is constant for an ellipse
363. Statement-1: arg 2
1
z
z
 
 
 
= arg z2 – arg z1 & arg zn
= n(argz) Statement-2: If |z| = 1, then arg (z2
+ z ) =
1
2
arg z.
364. Statement-1: If |z − z + i| ≤ 2 then 5 2 | z | 5 2− ≤ ≤ +
Statement-2: If |z − 2 + i| ≤ 2 the z lies inside or on the circle having centre (2, −1) & radius 2.
365. Statement-1: The area of the triangle on argand plane formed by the complex numbers z, iz and z + iz is
1
2
|z|2
Statement-2: The angle between the two complex numbers z and iz is
2
π
.
366. Statement-1: If 1 2
1 2
zz z
k,
zz z
−
=
+
(z1, z2 ≠ 0), then locus of z is circle.
Statement-2 : As, 1
2
z z
z z
−
= λ
−
represents a circle if, λ∉{0, 1}
367. Statement-1: If z1 and z2 are two complex numbers such that |z1| = |z2| + |z1 – z2|, then Im 1
2
z
0
z
 
= 
 
.
Statement-2: arg (z) = 0 ⇒ z is purely real.
368. Statement-1: If α = cos
2 2
isin
7 7
π π   
+   
   
, p = α + α2
+ α4
, q = α3
+ α5
+ α6
, then the equation whose roots
are p and q is x2
+ x + 2 = 0
Statement-2: If α is a root of z7
= 1, then 1 + α + α2
+ …. + α6
= 0.
369. Statement-1: If |z| < 2 1− then |z2
+ 2z cosα| is less than one.
Statement-2: |z1 + z2| < |z1| + |z2| . Also |cosα| ≤ 1.
370. Statement-1: The number of complex number satisfying the equation |z|2
+ P|z| + q = 0 (p, q, ∈ R) is atmost 2.
Statement-2 : Aquadratic equationinwhich all the co-efficients arenon-zero real can have exactlytworoots.
371. Statement-1: If
1
1( 0)β + = β ≠
β
is a complex number, then the maximum value of |β| is
5 1
2
+
.
Statement-2 :: On the locus
1
1β+ =
β
the farthest distance from origin is
5 1
2
+
.

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372. Statement-1: The locus of z moving in the Argand plane such that arg
z 2
z 2 2
− π 
= 
+ 
is a circle.
Statement-2: This is represent a circle, whose centre is origin and radius is 2.
ANSWER
344. B 345. A 346. D 347. A 348. A 349. A 350. B
351. D 352. A 353. D 354. D 355. D 356. A 357.. A
358. A 359. D 360. D 361. A 362. A 363. B 364. A
365. A 366. D 367. A 368. A 369. A 370. D 371. A
372. A
SOLUTION
345. Using AM ≥ GM 1 2 n 1 n
2 3 n 1
a a a a
...
a a a a
−
+ + + + ≥
1/n
1 2 n
2 3 1
a a a
n . ....
a a a
 
 
 
⇒ 1 2 n
2 3 1
a a a
...
a a a
+ + + ≥ n
Hence (A) is correct option.
346. 2log
3b 5c a
log log
5c a 3b
= + ⇒
2
3b 5c a
.
5c a 3b
 
= 
 
⇒ 3b = 5c
Also, b2
= ac ⇒ 9ac = 25c2
or 9a = 25c ∴
9a
5c 3b
5
= = ⇒
a b c
5 3 9/ 5
= = ⇒ b + c < a
∴ (D) is the correct answer
347. From the diagram it is clear that both circles touch each other
externally
∴ min |z1 – z2| = 0
max |z1 – z2| = 36 36 6 2+ = ]
Hence (A) is correct option.
(1, 1)
•(3, 3)
•
|z| = 2 2
|z| = 2
348. |iz + z0| = |i(z – i) – 1 + 5 + 3i| = |i (z–i) + 4 + 3i|
≤ |i| |z – i| + |4 + 3i| ≤ 7 Hence (A) is the correct option.
349. (A) arg (z1) = arg (z2)
∴ arg ( ) ( )1
1 2
2
z
arg z arg z 0
z
 
= − = 
 
.
350. (B) Fourth roots of unity are – 1, 1, – i and i
∴
2 2 2 2
1 2 3 4z z z z 0+ + + = and 1 2 3 4z z z z 0+ + + = .
351. Statement – II is true (a known fact).
Hence if z1, z2, . . . , zn are roots of zn
– 1 = 0, then z1. z2 . . . zn = (– 1)n
.
( )
( ) 11
1
1
n+−
= − ,
which is never equal to (– 1)n
Hence (d) is the correct answer.
352. Both statements – I and II are true and statement – II is the correct reasoning of statement – I, because
1 3 2 4
2 2
z z z z+ +
= ⇒ mid point of join of z1, z3 and z2, z4 are same, which is the necessary and sufficient
condition for a quadrilateral ABCD, when A ≡ A(z1), B ≡ B(z2),
C ≡ C(z3), D ≡ D(z4) to be a parallelogram Hence (A) is the correct answer.
353. | z i z | | z | | i z |+ − ≤ + − ⇒ | z | | z i | | i | 1+ − ≥ =
∴ Hence (d) is the correct answer.
354.
2 2
1 2 1 2| z z | | z z |− = +
⇒ 1 2 1 2z z z z 0+ = ⇒
2 2 2
1 2 1 2z z z z− = +
⇒ ∆AOB is right angled at O.
orthocenter is the origin. ∴ Hence (d) is the correct answer.
355. (D) (x – y) (xω – y) (xω2
– y)
= x3
ω2
– x2
yω – x2
yω2
+ xy2
– x2
yω + xy2
ω + xy2
ω2
– y3
= x3
– y3

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356. Option (A) is correct
Since
|z + 3 – 4i| ≤ |z| + |3-4i| = 9 (∴ |z| ≤ 4).
357. Option (A) is correct.
358.
6
k 1
2 k 2 k
( i) cos isin
7 7=
π π 
− − 
 
∑
= ( )
76
k
k 1
z z
i) z ( i)
1 z<
 −
− = −  
− 
∑ [ z7
= 1]
= (-i) (-1) = i Ans. (A)
359. |z1 + z2 + z3| = |z1 – a + z2 – b + z3 – c + (a + b + c)
≤ |z1 – a| + |z2 – b| + |z3 – c| + |a + b + c| ≤ 2|a + b + c| Ans. (D)
360. (cos2 + i sin2)π
can not be evaluated because demoviers theorem does not hold for irrational index.
‘d’ is correct.
361. 1, α, α2
, … α7
are 8, 8th
root of unity then after raising 16th
power, we get 1, α16
, α32
, α48
… α112
1 + α16
+ α32
+ α48
+ … + α112
Now α8
= 1
So α16
= 1
1 + 1 + 1 + … + 1 = 8
‘A’ is correct.
365. (A)
1
| z | | iz |
2
2
| z |
2
=
z
2
iz
z + iz
366. (D)
1 2
1 2
zz z
k
z z z
−
=
+
⇒
2
1
2
1
z
z
z
k
z
z
z
−
=
+
Clearly, if k ≠ 0, 1; then z would lie on a circle. If k = 1, z would lie on the perpendicular bisector of line segment
joining 2
1
z
z
and 2
1
z
z
−
and represents a point, if k = 0.
367. We have, arg (z) = 0 ⇒ z is purely real. R is true
Also, |z1| = |z2| + |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 = 0
⇒ arg 1
2
z
0
z
 
= 
 
⇒ 1
2
z
z
is purely real.
Im
1
2
z
0
z
 
= 
 
(A)
368. (A)
α is seventh root of unity ⇒ 1 + α + α2
+ … + α6
= 0
⇒ p + q = –1.
pq = α4
+ α6
+ α7
+ α5
+ α7
+ α8
+ α7
+ α9
+ α10
= 3 – 1 = 2.

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∴ x2
+ x + 2 = 0 is the req. equation.
Both A and R are true and R is correct explanation of A.
369. (A)
|z2
+ 2z cosα| < |z2
| + |2z cosα| < |z2
| + 2|z| |cosα|
< (
2
2 1) 2( 2 1) 1− + − < .
( |cosα| ≤ 1).
372.
i /2z 2 z 2 z 2z 2
e i
z 2 z 2
π− − − +
= =
+ +
... (i)
therefore
z 2 z 2 z 2
( 1) i
z 2 z 2 z 2
− − −
= − =−
+ + +
... (ii)
Then adding (i) & (ii)
z 2 z 2
z 2 z 2
− −
+
+ +
= 0
i.e., (z – 2) z 2)+ + (z + 2) ( z - 2) = 0, 2z z - 8 = 0
|z|2
= 4 ∴ x2
+ y2
= 4.
Ans. (a)
Imp. Que. From Competitive Exams
1. The number of real values of a satisfying the equation 01sin22
=+− xaa is
(a) Zero (b) One
(c) Two (d) Infinite
2. For positive integers 21 ,nn the value of the expression 2211
)1()1()1()1( 753 nnnn
iiii +++++++ where 1−=i is a
real number if and only if [IIT 1996]
(a) 121 += nn (b) 121 −= nn
(c) 21 nn = (d) 0,0 21 >> nn
3. Given that the equation ,0)(2
=++++ sirziqpz where srqp ,,, are real and non-zero has a real root, then
(a) sprpqr 22
+= (b) prqprs 22
+=
(c) qspqrs 22
+= (d) rqspqs 22
+=
4. If ,425 −+−=x then the value of the expression 4359 234
+−++ xxxx is [IIT 1972]
(a) 160 (b) 160−
(c) 60 (d) 60−
5. If ))((3 idcibai ++=+ , then +




−
a
b1
tan 




−
c
d1
tan has the value
(a) Inn ∈+ ,2
3
π
π
(b) Inn ∈+ ,
6
π
π
(c) Inn ∈− ,
3
π
π (d) Inn ∈− ,
3
2
π
π
6. If ,sincos,sincos ββαα ibia +=+=
,1andsincos =+++=
b
a
a
c
c
b
ic γγ then )cos()cos()cos( βααγγβ −+−+− is equal to [RPET 2001]
(a) 3/2 (b) – 3/2
(c) 0 (d) 1
7. If ibaniiii +=++++ )1).....(31)(21)(1( , then 2.5.10.... )1( 2
n+ is equal to
[Karnataka CET 2002; Kerala (Engg.) 2002]

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(a) 22
ba − (b) 22
ba +
(c) 22
ba + (d) 22
ba −
8. If z is a complex number, then the minimum value of |1||| −+ zz is [Roorkee 1992]
(a) 1 (b) 0
(c) 1/2 (d) None of these
9. For any two complex numbers 1z and 2z and any real numbers a and b; =++− 2
21
2
21 |)(||)(| azbzbzaz
[IIT 1988]
(a) |)||)(|( 21
22
zzba ++ (b) )|||)(|( 2
2
2
1
22
zzba ++
(c) )|||)(|( 2
2
2
1
22
zzba −+ (d) None of these
10. The locus of z satisfying the inequality >+ |1|log 3/1 z |1|log 3/1 −z is
(a) 0)( <zR (b) 0)( >zR
(c) 0)( <zI (d) None of these
11. If ibaz +=1 and idcz +=2 are complex numbers such that 1|||| 21 == zz and ,0)( 21 =zzR then the pair of
complex numbers icaw +=1 and idbw +=2 satisfies
[IIT 1985]
(a) 1|| 1 =w (b) 1|| 2 =w
(c) ,0)( 21 =wwR (d) All the above
12. Let z and w be two complex numbers such that ,1|| ≤z 1|| ≤w and 2|||| =−=+ wiziwz . Then z is equal to
[IIT 1995]
(a) 1 or i (b) i or i−
(c) 1 or – 1 (d) i or –1
13. The maximum distance from the origin of coordinates to the point z satisfying the equation a
z
z =+
1
is
(a) )1(
2
1 2
aa ++ (b) )2(
2
1 2
aa ++
(c) )4(
2
1 2
aa ++ (d) None of these
14. Find the complex number z satisfying the equations 1
8
4
,
3
5
8
12
=
−
−
=
−
−
z
z
iz
z
[Roorkee
1993]
(a) 6 (b) i86 ±
(c) ii 176,86 ++ (d) None of these
15. If 321 ,, zzz are complex numbers such that == |||| 21 zz =|| 3z ,1
111
321
=++
zzz
then || 321 zzz ++ is
[MP PET 2004; IIT Screening 2000]
(a) Equal to 1 (b) Less than 1
(c) Greater than 3 (d) Equal to 3
16. If iziz 64,610 21 +=+= and z is a complex number such that ,
42
1 π
=







−
−
zz
zz
amp then the value of |97| iz −−
is equal to [IIT 1990]
(a) 2 (b) 22
(c) 23 (d) 32
17. If 321 ,, zzz be three non-zero complex number, such that |||,|, 2112 zbzazz ==≠ and || 3zc = suppose that
0=
bac
acb
cba
, then 







2
3
z
z
arg is equal to

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(a)
2
13
12








−
−
zz
zz
arg (b) 







−
−
13
12
zz
zz
arg
(c)
2
12
13








−
−
zz
zz
arg (d) 







−
−
12
13
zz
zz
arg
18. Let z and w be the two non-zero complex numbers such that |||| wz = and π=+ wargzarg . Then z is
equal to
[IIT 1995; AIEEE 2002]
(a) w (b) w−
(c) w (d) w−
19. If 15|25| ≤− iz , then =− |)(.min)(.max| zampzamp
(a) 




−
5
3
cos 1
(b) 





− −
5
3
cos2 1
π
(c) 





+ −
5
3
cos
2
1π
(d) 





−




 −−
5
3
cos
5
3
sin 11
20. If 21 , zz and 43 , zz are two pairs of conjugate complex numbers, then 







+







3
2
4
1
z
z
arg
z
z
arg equals
(a) 0 (b)
2
π
(c)
2
3π
(d) π
21. Let wz, be complex numbers such that 0=+ wiz and π=zwarg . Then arg z equals [AIEEE
2004]
(a) 4/5π (b) 2/π
(c) 4/3π (d) 4/π
22. If ,.....)1( 2
210
n
n
n
xCxCxCCx ++++=+ then the value of .....6420 +−+− CCCC is
(a) n
2 (b)
2
cos2
πnn
(c)
2
sin2
πnn
(d)
4
cos2 2/ πnn
23. If θθ sincos ix += and φφ sincos iy += , then nmnm
yxyx −−
+ is equal to
(a) )cos( φθ nm +
(b) )cos( φθ nm −
(c) )cos(2 φθ nm +
(d) )cos(2 φθ nm −
24. The value of ∑=






+
8
1
9
2
cos
9
2
sin
r
r
i
r ππ
is
(a) 1− (b) 1
(c) i (d) i−
25. If cba ,, and wvu ,, are complex numbers representing the vertices of two triangles such that rbarc +−= )1(
and rvurw +−= )1( , where r is a complex number, then the two triangles
(a) Have the same area (b) Are similar
(c) Are congruent (d) None of these
26. Suppose 321 ,, zzz are the vertices of an equilateral triangle inscribed in the circle 2|| =z . If ,311 iz +=
then values of 3z and 2z are respectively [IIT 1994]
(a) 31,2 i−− (b) 31,2 i+
(c) 2,31 −+ i (d) None of these

37 of 38
27. If the complex number 21, zz the origin form an equilateral triangle then =+ 2
2
2
1 zz [IIT
1983]
(a) 21 zz (b) 21 zz
(c) 12 zz (d) 2
2
2
1 |||| zz =
28. If at least one value of the complex number iyxz += satisfy the condition 23|2| 2
+−=+ aaz and the
inequality 2
|2| aiz <+ , then
(a) 2>a (b) 2=a
(c) 2<a (d) None of these
29. If z, iz and izz + are the vertices of a triangle whose area is 2 units, then the value of || z is
[RPET 2000]
(a) – 2 (b) 2
(c) 4 (d) 8
30. If 0|||| 22
=++ zzzz , then the locus of z is
(a) A circle (b) A straight line
(c) A pair of straight lines(d) None of these
31. If 0sinsinsincoscoscos =++=++ γβαγβα then γβα 3cos3cos3cos ++ equals to [Karnataka CET
2000]
(a) 0 (b) )cos( γβα ++
(c) )cos(3 γβα ++ (d) )sin(3 γβα ++
32. If ,sincos 22
n
r
i
n
r
zr
αα
+= where r = 1, 2, 3,….,n, then n
n
zzzz ...lim 321
∞→
is equal to [UPSEAT
2001]
(a) αα sincos i+ (b) )2/sin()2/cos( αα i−
(c) 2/αi
e (d) 3 αi
e
33. If the cube roots of unity be ,,,1 2
ωω then the roots of the equation 08)1( 3
=+−x are
[IIT 1979; MNR 1986; DCE 2000; AIEEE 2005]
(a) 2
21,21,1 ωω ++−
(b) 2
21,21,1 ωω −−−
(c) 1,1,1 −−−
(d) None of these
34. If 132
.......,,,,1 −n
ωωωω are the th
nn, roots of unity, then )1).....(1)(1( 12 −
−−− n
ωωω equals
[MNR 1992; IIT 1984; DCE 2001; MP PET 2004]
(a) 0 (b) 1
(c) n (d) 2
n
35. The value of the expression .......)3)(3.(2)2)(2.(1 22
+−−+−− ωωωω
),)().(1(.... 2
ωω −−−+ nnn
where ω is an imaginary cube root of unity, is[IIT 1996]
(a) )43()1(
2
1 2
++− nnnn
(b) )43()1(
4
1 2
++− nnnn
(c) )43()1(
2
1 2
+++ nnnn
(d) )43()1(
4
1 2
+++ nnnn

38 of 38
36. If ,1−=i then
334
2
3
2
1
54








+−+
i
365
2
3
2
1
3








+−+
i
is equal to [IIT 1999]
(a) 31 i− (b) 31 i+−
(c) 3i (d) 3i−
37. If ),7/2sin()7/2cos( ππ ia += then the quadratic equation whose roots are 42
aaa ++=α and 653
aaa ++=β
is
[RPET 2000]
(a) 022
=+− xx (b) 022
=−+ xx
(c) 022
=−− xx (d) 022
=++ xx
38. Let 1z and 2z be nth roots of unity which are ends of a line segment that subtend a right angle at the origin.
Then n must be of the form [IIT Screening 2001; Karnataka 2002]
(a) 4k + 1 (b) 4k + 2
(c) 4k + 3 (d) 4k
39. Let ω is an imaginary cube roots of unity then the value of
.....)12)(12(3)1)(1(2 22
++++++ ωωωω )1)(1)(1( 2
++++ ωω nnn is [Orissa JEE 2002]
(a) n
nn
+




 +
2
2
)1(
(b)
2
2
)1(





 +nn
(c) n
nn
−




 +
2
2
)1(
(d) None of these
40. ω is an imaginary cube root of unity. If =+ m
)1( 2
ω ,)1( 4 m
ω+ then least positive integral value of m is
[IIT Screening 2004]
(a) 6 (b) 5
(c) 4 (d) 3
ANSWER
1 c 2 d 3 d 4 b 5 b
6 d 7 b 8 a 9 b 10 a
11 d 12 c 13 c 14 c 15 a
16 c 17 c 18 d 19 b 20 a
21 c 22 d 23 c 24 d 25 b
26 a 27 a 28 a 29 b 30 c
31 c 32 c 33 b 34 c 35 b
36 c 37 D 38 d 39 a 40 d

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