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
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
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
