Complex numbers 2
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The document describes the expansion of sin nθ and cos nθ in powers of sinθ and cosθ using De Moivre's theorem and the binomial theorem. It shows that cos nθ can be expressed as the sum of terms involving nC0cosnθ, nC2cosn-2θsin2θ, etc. and sin nθ can be expressed as the sum of terms involving nC1cosn-1θsinθ, nC3cosn-3θsin3θ, etc. The expansions are obtained by equating the real and imaginary parts of (cosθ + i sinθ)n.
![Hyperbolic Identities
Identities of Hyperbolic functions can be derived
from the identities of circular functions by
replacing x by ix
Now sin2 x + cos2 x = 1
sin2 (ix) + cos2 (ix) = 1 ⇒
[i sinh(x)]2 + [cosh(x)]2 = 1
N. B. Vyas Complex Numbers - 2](https://image.slidesharecdn.com/complexnumbers2-120305224718-phpapp01/85/Complex-numbers-2-113-320.jpg)
![Hyperbolic Identities
Identities of Hyperbolic functions can be derived
from the identities of circular functions by
replacing x by ix
Now sin2 x + cos2 x = 1
sin2 (ix) + cos2 (ix) = 1 ⇒
[i sinh(x)]2 + [cosh(x)]2 = 1
⇒ cosh2 (x) − sinh2 (x) = 1
N. B. Vyas Complex Numbers - 2](https://image.slidesharecdn.com/complexnumbers2-120305224718-phpapp01/85/Complex-numbers-2-114-320.jpg)
![Hyperbolic Identities
Identities of Hyperbolic functions can be derived
from the identities of circular functions by
replacing x by ix
Now sin2 x + cos2 x = 1
sin2 (ix) + cos2 (ix) = 1 ⇒
[i sinh(x)]2 + [cosh(x)]2 = 1
⇒ cosh2 (x) − sinh2 (x) = 1
Similarly we can obtain
N. B. Vyas Complex Numbers - 2](https://image.slidesharecdn.com/complexnumbers2-120305224718-phpapp01/85/Complex-numbers-2-115-320.jpg)
![Hyperbolic Identities
Identities of Hyperbolic functions can be derived
from the identities of circular functions by
replacing x by ix
Now sin2 x + cos2 x = 1
sin2 (ix) + cos2 (ix) = 1 ⇒
[i sinh(x)]2 + [cosh(x)]2 = 1
⇒ cosh2 (x) − sinh2 (x) = 1
Similarly we can obtain
sech2 (x) = 1 − tanh2 (x)
N. B. Vyas Complex Numbers - 2](https://image.slidesharecdn.com/complexnumbers2-120305224718-phpapp01/85/Complex-numbers-2-116-320.jpg)
![Hyperbolic Identities
Identities of Hyperbolic functions can be derived
from the identities of circular functions by
replacing x by ix
Now sin2 x + cos2 x = 1
sin2 (ix) + cos2 (ix) = 1 ⇒
[i sinh(x)]2 + [cosh(x)]2 = 1
⇒ cosh2 (x) − sinh2 (x) = 1
Similarly we can obtain
sech2 (x) = 1 − tanh2 (x)
cosech2 (x) = coth2 (x) − 1
N. B. Vyas Complex Numbers - 2](https://image.slidesharecdn.com/complexnumbers2-120305224718-phpapp01/85/Complex-numbers-2-117-320.jpg)
Complex numbers 2
- 1. Complex Numbers - 2 N. B. Vyas Department of Mathematics, Atmiya Institute of Tech. and Science, Rajkot (Guj.) N. B. Vyas Complex Numbers - 2
- 2. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ N. B. Vyas Complex Numbers - 2
- 3. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer N. B. Vyas Complex Numbers - 2
- 4. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n N. B. Vyas Complex Numbers - 2
- 5. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ N. B. Vyas Complex Numbers - 2
- 6. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) N. B. Vyas Complex Numbers - 2
- 7. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + N. B. Vyas Complex Numbers - 2
- 8. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) N. B. Vyas Complex Numbers - 2
- 9. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ N. B. Vyas Complex Numbers - 2
- 10. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) N. B. Vyas Complex Numbers - 2
- 11. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) + i ( n C1 cosn−1 θ sinθ N. B. Vyas Complex Numbers - 2
- 12. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) + i ( n C1 cosn−1 θ sinθ − n C3 cosn−3 θ sin3 θ + . . .) N. B. Vyas Complex Numbers - 2
- 13. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) + i ( n C1 cosn−1 θ sinθ − n C3 cosn−3 θ sin3 θ + . . .) By comparing real and imaginary parts on both sides, we get N. B. Vyas Complex Numbers - 2
- 14. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) + i ( n C1 cosn−1 θ sinθ − n C3 cosn−3 θ sin3 θ + . . .) By comparing real and imaginary parts on both sides, we get cos nθ = n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . . N. B. Vyas Complex Numbers - 2
- 15. Expansion Using De Moivre’s Theorem Method of Expansion of Sin nθ and Cos nθ in powers of sinθ cosθ, where n is a positive integer (cosnθ + isinnθ) = (cosθ + isinθ)n = n C0 cosn θ + n C1 cosn−1 θ (isinθ) + n C2 cosn−2 θ (isinθ)2 + . . . + n Cn (isinθ)n (Using Binomial theorem) = ( n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . .) + i ( n C1 cosn−1 θ sinθ − n C3 cosn−3 θ sin3 θ + . . .) By comparing real and imaginary parts on both sides, we get cos nθ = n C0 cosn θ − n C2 cosn−2 θ sin2 θ + . . . sin nθ = n C1 cosn−1 θ sinθ − n C3 cosn−3 θ sin3 θ + . . . N. B. Vyas Complex Numbers - 2
- 16. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ N. B. Vyas Complex Numbers - 2
- 17. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then N. B. Vyas Complex Numbers - 2
- 18. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 Also = cosθ − isinθ z N. B. Vyas Complex Numbers - 2
- 19. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = z z N. B. Vyas Complex Numbers - 2
- 20. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z N. B. Vyas Complex Numbers - 2
- 21. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 ∴z+ = z N. B. Vyas Complex Numbers - 2
- 22. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 ∴ z + = 2cosθ ⇒ cosθ = z N. B. Vyas Complex Numbers - 2
- 23. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z N. B. Vyas Complex Numbers - 2
- 24. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 and z n + n = z N. B. Vyas Complex Numbers - 2
- 25. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 and z n + n = 2cos nθ ⇒ cos nθ = z N. B. Vyas Complex Numbers - 2
- 26. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z N. B. Vyas Complex Numbers - 2
- 27. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z cosn θ = N. B. Vyas Complex Numbers - 2
- 28. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z N. B. Vyas Complex Numbers - 2
- 29. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z similarly sinθ = N. B. Vyas Complex Numbers - 2
- 30. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z 1 1 similarly sinθ = z− and sin nθ = 2i z N. B. Vyas Complex Numbers - 2
- 31. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z 1 1 1 1 similarly sinθ = z− and sin nθ = zn − n 2i z 2i z N. B. Vyas Complex Numbers - 2
- 32. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z 1 1 1 1 similarly sinθ = z− and sin nθ = zn − n 2i z 2i z sinn θ = N. B. Vyas Complex Numbers - 2
- 33. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z 1 1 1 1 similarly sinθ = z− and sin nθ = zn − n 2i z 2i z 1 1 n sinn θ = z− (2i)n z N. B. Vyas Complex Numbers - 2
- 34. Expansion Using De Moivre’s Theorem Method of Expansion of Sinn θ and Cosn θ Let z = cosθ + isinθ then z n = cos nθ + i sin nθ 1 1 Also = cosθ − isinθ ∴ n = cos nθ − i sin nθ z z 1 1 1 ∴ z + = 2cosθ ⇒ cosθ = z+ z 2 z 1 1 1 and z n + n = 2cos nθ ⇒ cos nθ = zn + n z 2 z 1 1 n cosn θ = n z + 2 z 1 1 1 1 similarly sinθ = z− and sin nθ = zn − n 2i z 2i z 1 1 n sinn θ = z− (2i)n z N. B. Vyas Complex Numbers - 2
- 35. Examples Ex. Prove that cos4θ = cos4 θ − 6cos2 θsin2 θ + sin4 θ N. B. Vyas Complex Numbers - 2
- 36. Examples Ex. Using De Moivre’s theorem prove the following: N. B. Vyas Complex Numbers - 2
- 37. Examples Ex. Using De Moivre’s theorem prove the following: (i) cos6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 N. B. Vyas Complex Numbers - 2
- 38. Examples Ex. Using De Moivre’s theorem prove the following: (i) cos6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 (ii) sin6θ = 3 sin2θ − 4 sin3 2θ N. B. Vyas Complex Numbers - 2
- 39. Examples Ex. Using De Moivre’s theorem prove the following: (i) cos6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 (ii) sin6θ = 3 sin2θ − 4 sin3 2θ 6 tanθ − 20 tan3 θ + 6 tan5 θ (iii) tan6θ = 1 − 15 tan2 θ + 15 tan4 θ − tan6 θ N. B. Vyas Complex Numbers - 2
- 40. Examples Ex. Prove that 1 cos8 θ = 7 (cos8θ +8cos6θ +28cos4θ +56cos2θ +35) 2 N. B. Vyas Complex Numbers - 2
- 41. Examples Sol.: We know that, N. B. Vyas Complex Numbers - 2
- 42. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z N. B. Vyas Complex Numbers - 2
- 43. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z = 8 C0 z 8 + N. B. Vyas Complex Numbers - 2
- 44. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 1 = 8 C 0 z 8 + 8 C1 z 7 + z N. B. Vyas Complex Numbers - 2
- 45. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z N. B. Vyas Complex Numbers - 2
- 46. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 8 5 1 C3 z + z N. B. Vyas Complex Numbers - 2
- 47. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 8 5 1 8 4 1 C3 z + C4 z + z z N. B. Vyas Complex Numbers - 2
- 48. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z N. B. Vyas Complex Numbers - 2
- 49. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z N. B. Vyas Complex Numbers - 2
- 50. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z = z8 + N. B. Vyas Complex Numbers - 2
- 51. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z = z 8 + 8z 6 + N. B. Vyas Complex Numbers - 2
- 52. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z = z 8 + 8z 6 + 28z 4 + N. B. Vyas Complex Numbers - 2
- 53. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z = z 8 + 8z 6 + 28z 4 + 56z 2 + N. B. Vyas Complex Numbers - 2
- 54. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z 56 = z 8 + 8z 6 + 28z 4 + 56z 2 + 70 + + z2 N. B. Vyas Complex Numbers - 2
- 55. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z 56 28 = z 8 + 8z 6 + 28z 4 + 56z 2 + 70 + + + z2 z4 N. B. Vyas Complex Numbers - 2
- 56. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z 56 28 8 1 = z 8 + 8z 6 + 28z 4 + 56z 2 + 70 + + 4+ 6+ 8 z2 z z z N. B. Vyas Complex Numbers - 2
- 57. Examples Sol.: We know that, 8 8 1 (2cosθ) = z+ z 2 8 8 8 7 1 8 6 1 = C 0 z + C1 z + C2 z + z z 3 4 5 8 5 1 8 4 1 8 3 1 C3 z + C4 z + C5 z + z z z 6 7 8 8 2 1 8 1 8 1 C6 z + C7 z + C8 z z z 56 28 8 1 = z 8 + 8z 6 + 28z 4 + 56z 2 + 70 + + 4+ 6+ 8 z2 z z z N. B. Vyas Complex Numbers - 2
- 58. Examples 1 = z8 + + z8 N. B. Vyas Complex Numbers - 2
- 59. Examples 1 1 = z8 + + 8 z6 + + z8 z6 N. B. Vyas Complex Numbers - 2
- 60. Examples 1 1 1 = z8 + + 8 z6 + + 28 z 4 + + z8 z6 z4 N. B. Vyas Complex Numbers - 2
- 61. Examples 1 1 1 = z8 + + 8 z6 + 6 + 28 z 4 + + z8 z z4 1 56 z 2 + 2 + 70 z N. B. Vyas Complex Numbers - 2
- 62. Examples 1 1 1 = z8 + + 8 z6 + 6 + 28 z 4 + + z8 z z4 1 56 z 2 + 2 + 70 z = 2cos8θ + N. B. Vyas Complex Numbers - 2
- 63. Examples 1 1 1 = z8 + + 8 z6 + 6 + 28 z 4 + + z8 z z4 1 56 z 2 + 2 + 70 z = 2cos8θ + 16cos6θ + N. B. Vyas Complex Numbers - 2
- 64. Examples 1 1 1 = z8 + + 8 z 6 + 6 + 28 z 4 + 4 + z8 z z 1 56 z 2 + 2 + 70 z = 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + N. B. Vyas Complex Numbers - 2
- 65. Examples 1 1 1 = z8 + + 8 z 6 + 6 + 28 z 4 + 4 + z8 z z 1 56 z 2 + 2 + 70 z = 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70 N. B. Vyas Complex Numbers - 2
- 66. Examples 1 1 1 = z8 + + 8 z 6 + 6 + 28 z 4 + 4 + z8 z z 1 56 z 2 + 2 + 70 z = 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70 1 ∴ cos8 θ = 7 2 N. B. Vyas Complex Numbers - 2
- 67. Examples 1 1 1 = z8 + + 8 z 6 + 6 + 28 z 4 + 4 + z8 z z 1 56 z 2 + 2 + 70 z = 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70 1 ∴ cos8 θ = 7 (cos8θ +8cos6θ +28cos4θ +56cos2θ +35) 2 N. B. Vyas Complex Numbers - 2
- 68. Circular Functions of a Complex Numbers eix = cosx + isinx N. B. Vyas Complex Numbers - 2
- 69. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx N. B. Vyas Complex Numbers - 2
- 70. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. N. B. Vyas Complex Numbers - 2
- 71. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. By adding, we get them, we get N. B. Vyas Complex Numbers - 2
- 72. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. By adding, we get them, we get eix + e−ix cosx = 2 N. B. Vyas Complex Numbers - 2
- 73. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. By adding, we get them, we get eix + e−ix eix − e−ix cosx = and sinx = 2 2i N. B. Vyas Complex Numbers - 2
- 74. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. By adding, we get them, we get eix + e−ix eix − e−ix cosx = and sinx = 2 2i are known as circular functions and are true for all real values of x. N. B. Vyas Complex Numbers - 2
- 75. Circular Functions of a Complex Numbers eix = cosx + isinx and e−ix = cosx − isinx are known as Euler’s formula. By adding, we get them, we get eix + e−ix eix − e−ix cosx = and sinx = 2 2i are known as circular functions and are true for all real values of x. They are also known as Euler’s exponential form of cosines and sines. N. B. Vyas Complex Numbers - 2
- 76. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), N. B. Vyas Complex Numbers - 2
- 77. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) N. B. Vyas Complex Numbers - 2
- 78. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) and hyperbolic tangent of x is defined by tanh(x) N. B. Vyas Complex Numbers - 2
- 79. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) and hyperbolic tangent of x is defined by tanh(x) and are defined respectively as N. B. Vyas Complex Numbers - 2
- 80. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) and hyperbolic tangent of x is defined by tanh(x) and are defined respectively as ex − e−x sinh(x) = 2 N. B. Vyas Complex Numbers - 2
- 81. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) and hyperbolic tangent of x is defined by tanh(x) and are defined respectively as ex − e−x sinh(x) = 2 ex + e−x cosh(x) = 2 N. B. Vyas Complex Numbers - 2
- 82. Hyperbolic functions The hyperbolic sine of x is denoted by sinh(x), the hyperbolic cosine of x is defined by cosh(x) and hyperbolic tangent of x is defined by tanh(x) and are defined respectively as ex − e−x sinh(x) = 2 ex + e−x cosh(x) = 2 ex − e−x and tanh(x) = x ;x R e + e−x N. B. Vyas Complex Numbers - 2
- 83. Hyperbolic functions The reciprocals of these functions are defined as below N. B. Vyas Complex Numbers - 2
- 84. Hyperbolic functions The reciprocals of these functions are defined as below 2 cosech(x) = x e − e−x N. B. Vyas Complex Numbers - 2
- 85. Hyperbolic functions The reciprocals of these functions are defined as below 2 cosech(x) = x e − e−x 2 sech(x) = x e − e−x N. B. Vyas Complex Numbers - 2
- 86. Hyperbolic functions The reciprocals of these functions are defined as below 2 cosech(x) = x e − e−x 2 sech(x) = x e − e−x ex + e−x and coth(x) = x e − e−x N. B. Vyas Complex Numbers - 2
- 87. Relation between Circular & Hyperbolic functions eix − e−ix We know that sinx = 2i N. B. Vyas Complex Numbers - 2
- 88. Relation between Circular & Hyperbolic functions eix − e−ix We know that sinx = 2i replacing x by ix N. B. Vyas Complex Numbers - 2
- 89. Relation between Circular & Hyperbolic functions eix − e−ix We know that sinx = 2i replacing x by ix ei(ix) − e−i(ix) e−x − ex sin(ix) = = 2i 2i N. B. Vyas Complex Numbers - 2
- 90. Relation between Circular & Hyperbolic functions eix − e−ix We know that sinx = 2i replacing x by ix ei(ix) − e−i(ix) e−x − ex sin(ix) = = 2i 2i x −x x −x e −e e −e =− =i 2i 2 N. B. Vyas Complex Numbers - 2
- 91. Relation between Circular & Hyperbolic functions eix − e−ix We know that sinx = 2i replacing x by ix ei(ix) − e−i(ix) e−x − ex sin(ix) = = 2i 2i x −x x −x e −e e −e =− =i 2i 2 ∴ sin(ix) = isinh(x) N. B. Vyas Complex Numbers - 2
- 92. Relation between Circular & Hyperbolic functions Similarly, we can prove N. B. Vyas Complex Numbers - 2
- 93. Relation between Circular & Hyperbolic functions Similarly, we can prove cos(ix) = cosh(x) N. B. Vyas Complex Numbers - 2
- 94. Relation between Circular & Hyperbolic functions Similarly, we can prove cos(ix) = cosh(x) tan(ix) = i tanh(x) N. B. Vyas Complex Numbers - 2
- 95. Relation between Circular & Hyperbolic functions Similarly, we can prove cos(ix) = cosh(x) tan(ix) = i tanh(x) cosec(ix) = −i cosech(x) N. B. Vyas Complex Numbers - 2
- 96. Relation between Circular & Hyperbolic functions Similarly, we can prove cos(ix) = cosh(x) tan(ix) = i tanh(x) cosec(ix) = −i cosech(x) sec(ix) = sech(x) N. B. Vyas Complex Numbers - 2
- 97. Relation between Circular & Hyperbolic functions Similarly, we can prove cos(ix) = cosh(x) tan(ix) = i tanh(x) cosec(ix) = −i cosech(x) sec(ix) = sech(x) cot(ix) = −i coth(x) N. B. Vyas Complex Numbers - 2
- 98. Relation between Circular & Hyperbolic functions Also we can replace x by ix in N. B. Vyas Complex Numbers - 2
- 99. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get N. B. Vyas Complex Numbers - 2
- 100. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get sin i(ix) = isinh(ix) N. B. Vyas Complex Numbers - 2
- 101. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get sin i(ix) = isinh(ix) ⇒ sin(−x) = isinh(ix) N. B. Vyas Complex Numbers - 2
- 102. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get sin i(ix) = isinh(ix) ⇒ sin(−x) = isinh(ix) ⇒ −sinx = isinh(ix) N. B. Vyas Complex Numbers - 2
- 103. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get sin i(ix) = isinh(ix) ⇒ sin(−x) = isinh(ix) ⇒ −sinx = isinh(ix) ⇒ i2 sinx = isinh(ix) N. B. Vyas Complex Numbers - 2
- 104. Relation between Circular & Hyperbolic functions Also we can replace x by ix in sin(ix) = isinh(x), we get sin i(ix) = isinh(ix) ⇒ sin(−x) = isinh(ix) ⇒ −sinx = isinh(ix) ⇒ i2 sinx = isinh(ix) ∴ sinh(ix) = isinx N. B. Vyas Complex Numbers - 2
- 105. Relation between Circular & Hyperbolic functions Similarly, we can prove N. B. Vyas Complex Numbers - 2
- 106. Relation between Circular & Hyperbolic functions Similarly, we can prove cosh(ix) = cos(x) N. B. Vyas Complex Numbers - 2
- 107. Relation between Circular & Hyperbolic functions Similarly, we can prove cosh(ix) = cos(x) tanh(ix) = i tan(x) N. B. Vyas Complex Numbers - 2
- 108. Relation between Circular & Hyperbolic functions Similarly, we can prove cosh(ix) = cos(x) tanh(ix) = i tan(x) cosech(ix) = −i cosec(x) N. B. Vyas Complex Numbers - 2
- 109. Relation between Circular & Hyperbolic functions Similarly, we can prove cosh(ix) = cos(x) tanh(ix) = i tan(x) cosech(ix) = −i cosec(x) sech(ix) = sec(x) N. B. Vyas Complex Numbers - 2
- 110. Relation between Circular & Hyperbolic functions Similarly, we can prove cosh(ix) = cos(x) tanh(ix) = i tan(x) cosech(ix) = −i cosec(x) sech(ix) = sec(x) coth(ix) = −i cot(x) N. B. Vyas Complex Numbers - 2
- 111. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix N. B. Vyas Complex Numbers - 2
- 112. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 N. B. Vyas Complex Numbers - 2
- 113. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 sin2 (ix) + cos2 (ix) = 1 ⇒ [i sinh(x)]2 + [cosh(x)]2 = 1 N. B. Vyas Complex Numbers - 2
- 114. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 sin2 (ix) + cos2 (ix) = 1 ⇒ [i sinh(x)]2 + [cosh(x)]2 = 1 ⇒ cosh2 (x) − sinh2 (x) = 1 N. B. Vyas Complex Numbers - 2
- 115. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 sin2 (ix) + cos2 (ix) = 1 ⇒ [i sinh(x)]2 + [cosh(x)]2 = 1 ⇒ cosh2 (x) − sinh2 (x) = 1 Similarly we can obtain N. B. Vyas Complex Numbers - 2
- 116. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 sin2 (ix) + cos2 (ix) = 1 ⇒ [i sinh(x)]2 + [cosh(x)]2 = 1 ⇒ cosh2 (x) − sinh2 (x) = 1 Similarly we can obtain sech2 (x) = 1 − tanh2 (x) N. B. Vyas Complex Numbers - 2
- 117. Hyperbolic Identities Identities of Hyperbolic functions can be derived from the identities of circular functions by replacing x by ix Now sin2 x + cos2 x = 1 sin2 (ix) + cos2 (ix) = 1 ⇒ [i sinh(x)]2 + [cosh(x)]2 = 1 ⇒ cosh2 (x) − sinh2 (x) = 1 Similarly we can obtain sech2 (x) = 1 − tanh2 (x) cosech2 (x) = coth2 (x) − 1 N. B. Vyas Complex Numbers - 2
- 118. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix N. B. Vyas Complex Numbers - 2
- 119. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = N. B. Vyas Complex Numbers - 2
- 120. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) N. B. Vyas Complex Numbers - 2
- 121. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = N. B. Vyas Complex Numbers - 2
- 122. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = N. B. Vyas Complex Numbers - 2
- 123. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) N. B. Vyas Complex Numbers - 2
- 124. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = N. B. Vyas Complex Numbers - 2
- 125. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) N. B. Vyas Complex Numbers - 2
- 126. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = N. B. Vyas Complex Numbers - 2
- 127. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) N. B. Vyas Complex Numbers - 2
- 128. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) tanh(3x) = N. B. Vyas Complex Numbers - 2
- 129. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) N. B. Vyas Complex Numbers - 2
- 130. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) sinh(x) = N. B. Vyas Complex Numbers - 2
- 131. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) x 2 tanh sinh(x) = 2 2 x 1 − tanh 2 N. B. Vyas Complex Numbers - 2
- 132. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) x 2 tanh sinh(x) = 2 2 x , cosh(x) = 1 − tanh 2 N. B. Vyas Complex Numbers - 2
- 133. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) x x 2 tanh 1 + tanh2 sinh(x) = 2 2 2 x , cosh(x) = 2 x 1 − tanh 1 − tanh 2 2 N. B. Vyas Complex Numbers - 2
- 134. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) x x 2 tanh 1 + tanh2 sinh(x) = 2 2 2 x , cosh(x) = 2 x 1 − tanh 1 − tanh 2 2 tanh(x) = N. B. Vyas Complex Numbers - 2
- 135. Hyperbolic Identities Also sin2x = 2 sinx cosx by replacing x by ix we get sinh(2x) = 2 sinh(x) cosh(x) Similarly cosh(2x) = cosh2 (x) + sinh2 (x) = 2cosh2 (x) − 1 = 1 + 2sinh2 (x) sinh(3x) = 3sinh(x)+4sinh3 (x) , cosh(3x) = 4cosh3 (x)−3cosh(x) 3tanh(x) + tanh3 (x) tanh(3x) = 1 + 3tanh2 (x) x x 2 tanh 1 + tanh2 sinh(x) = 2 2 2 x , cosh(x) = 2 x 1 − tanh 1 − tanh 2 2 x 2tanh tanh(x) = 2 x 1 + tanh2 2 N. B. Vyas Complex Numbers - 2
- 136. Logarithm of a Complex Number If z = ew , then we write w = lnz, called the natural logarithm of z. Thus the natural logarithmic function is the inverse of the exponential function and can be defined by w = lnz = ln(rei(θ+2kπ) ) = lnr + i(θ + 2kπ) ∀z , logz = ln|z| + iarg(z) N. B. Vyas Complex Numbers - 2