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# Complex numbers 1

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Basic concepts of Complex numbers ( part 1 )

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### Complex numbers 1

1. 1. Complex Numbers N. B. Vyas Department of Mathematics,Atmiya Institute of Tech. and Science, Rajkot (Guj.) N. B. Vyas Complex Numbers
2. 2. Definition of Complex NumberComplex NumbersA number of the form √ + iy, where x and y are xreal numbers and i = −1 is called a complexnumber and is denoted by z. It is also denotedby an ordered pair(x, y). N. B. Vyas Complex Numbers
3. 3. Definition of Complex NumberComplex NumbersA number of the form √ + iy, where x and y are xreal numbers and i = −1 is called a complexnumber and is denoted by z. It is also denotedby an ordered pair(x, y).Thus z = x + iy or z = (x, y) N. B. Vyas Complex Numbers
4. 4. Definition of Complex Number The set of complex numbers is denoted by . N. B. Vyas Complex Numbers
5. 5. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then N. B. Vyas Complex Numbers
6. 6. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then x is called the real part of z and denoted by Re(z). N. B. Vyas Complex Numbers
7. 7. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then x is called the real part of z and denoted by Re(z). y is called the imaginary part of z and is denoted by Im(z). N. B. Vyas Complex Numbers
8. 8. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then x is called the real part of z and denoted by Re(z). y is called the imaginary part of z and is denoted by Im(z). If x = 0 and y = 0 then z = 0 + iy = iy is called a purely imaginary number. N. B. Vyas Complex Numbers
9. 9. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then x is called the real part of z and denoted by Re(z). y is called the imaginary part of z and is denoted by Im(z). If x = 0 and y = 0 then z = 0 + iy = iy is called a purely imaginary number. If x = 0 and y = 0 then z = x + i0 = x is called a real number. N. B. Vyas Complex Numbers
10. 10. Definition of Complex Number The set of complex numbers is denoted by . If z = x + iy is a complex number, then x is called the real part of z and denoted by Re(z). y is called the imaginary part of z and is denoted by Im(z). If x = 0 and y = 0 then z = 0 + iy = iy is called a purely imaginary number. If x = 0 and y = 0 then z = x + i0 = x is called a real number. If x = 0 and y = 0 then z = 0 + i.0 = 0 is the zero complex number. N. B. Vyas Complex Numbers
11. 11. Definition of Conjugate Complex NumberConjugate Complex NumbersIf two complex numbers diﬀer only in the sign of theimaginary part then they are called conjugatecomplex numbers. N. B. Vyas Complex Numbers
12. 12. Definition of Conjugate Complex NumberConjugate Complex NumbersIf two complex numbers diﬀer only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x + iy and x − iy are conjugate complexnumbers. N. B. Vyas Complex Numbers
13. 13. Definition of Conjugate Complex NumberConjugate Complex NumbersIf two complex numbers diﬀer only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x + iy and x − iy are conjugate complexnumbers. The conjugate of a complex number z is denoted by z . ¯ N. B. Vyas Complex Numbers
14. 14. Definition of Conjugate Complex NumberConjugate Complex NumbersIf two complex numbers diﬀer only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x + iy and x − iy are conjugate complexnumbers. The conjugate of a complex number z is denoted by z . ¯ The conjugate of real number is the real number itself. N. B. Vyas Complex Numbers
15. 15. Properties of Complex Numbers 1 If x + iy = 0 then x = 0, y = 0 N. B. Vyas Complex Numbers
16. 16. Properties of Complex Numbers 1 If x + iy = 0 then x = 0, y = 0 2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2 N. B. Vyas Complex Numbers
17. 17. Properties of Complex Numbers 1 If x + iy = 0 then x = 0, y = 0 2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2 3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2 N. B. Vyas Complex Numbers
18. 18. Properties of Complex Numbers 1 If x + iy = 0 then x = 0, y = 0 2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2 3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2 4 Sum, diﬀerence and quotient(division) of any two complex numbers is a complex number. N. B. Vyas Complex Numbers
19. 19. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then their N. B. Vyas Complex Numbers
20. 20. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) N. B. Vyas Complex Numbers
21. 21. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 ) N. B. Vyas Complex Numbers
22. 22. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 )Diﬀerence: z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) N. B. Vyas Complex Numbers
23. 23. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 )Diﬀerence: z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) = (x1 − x2 ) + i(y1 − y2 ) N. B. Vyas Complex Numbers
24. 24. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 )Diﬀerence: z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) = (x1 − x2 ) + i(y1 − y2 )Product: z1 .z2 = (x1 + iy1 ).(x2 + iy2 ) N. B. Vyas Complex Numbers
25. 25. Properties of Complex NumbersIf z1 = x1 + iy1 and z2 = x2 + iy2 then theirSum: z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 )Diﬀerence: z1 − z2 = (x1 + iy1 ) − (x2 + iy2 ) = (x1 − x2 ) + i(y1 − y2 )Product: z1 .z2 = (x1 + iy1 ).(x2 + iy2 ) = (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 ) N. B. Vyas Complex Numbers
26. 26. Polar form of a Complex NumberLet P (x, y) be the point which representsz = (x, y) = x + iy N. B. Vyas Complex Numbers
27. 27. Polar form of a Complex NumberLet P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ. N. B. Vyas Complex Numbers
28. 28. Polar form of a Complex NumberLet P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M . N. B. Vyas Complex Numbers
29. 29. Polar form of a Complex NumberLet P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M .x = OM = rcosθ. y = P M = rsinθ. N. B. Vyas Complex Numbers
30. 30. Polar form of a Complex NumberLet P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M .x = OM = rcosθ. y = P M = rsinθ.∴ z = x + iy = r(cosθ + isinθ) N. B. Vyas Complex Numbers
31. 31. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. N. B. Vyas Complex Numbers
32. 32. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z N. B. Vyas Complex Numbers
33. 33. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z Geometrically, |z| is the distance of point z from the origin. θ is called the argument of z or amplitude of z. N. B. Vyas Complex Numbers
34. 34. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z Geometrically, |z| is the distance of point z from the origin. θ is called the argument of z or amplitude of z. It is denoted by argz or Ampz N. B. Vyas Complex Numbers
35. 35. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z Geometrically, |z| is the distance of point z from the origin. θ is called the argument of z or amplitude of z. It is denoted by argz or Ampz y ∴ θ = argz = tan−1 x N. B. Vyas Complex Numbers
36. 36. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z Geometrically, |z| is the distance of point z from the origin. θ is called the argument of z or amplitude of z. It is denoted by argz or Ampz y ∴ θ = argz = tan−1 x θ is the directed angle from the positive X-axis to OP. N. B. Vyas Complex Numbers
37. 37. Polar form of a Complex Number r is called the absolute value or the modulus of z and is denoted by |z|. √ ∴ r = |z| = x2 + y 2 = z.¯ z Geometrically, |z| is the distance of point z from the origin. θ is called the argument of z or amplitude of z. It is denoted by argz or Ampz y ∴ θ = argz = tan−1 x θ is the directed angle from the positive X-axis to OP. The value of θ lies in the interval −π < θ ≤ π is called principal value. N. B. Vyas Complex Numbers
38. 38. Multiplication and Division of ComplexNumbers in Polar FormLet z1 = r1 (cosθ1 + isinθ1 ) and z2 = r2 (cosθ2 + isinθ2 )be two complex numbers in polar form. Then N. B. Vyas Complex Numbers
39. 39. Multiplication and Division of ComplexNumbers in Polar FormLet z1 = r1 (cosθ1 + isinθ1 ) and z2 = r2 (cosθ2 + isinθ2 )be two complex numbers in polar form. Then Product: z1 z2 = r1 (cosθ1 + isinθ1 )r2 (cosθ2 + isinθ2 ) N. B. Vyas Complex Numbers
40. 40. Multiplication and Division of ComplexNumbers in Polar FormLet z1 = r1 (cosθ1 + isinθ1 ) and z2 = r2 (cosθ2 + isinθ2 )be two complex numbers in polar form. Then Product: z1 z2 = r1 (cosθ1 + isinθ1 )r2 (cosθ2 + isinθ2 )= r1 r2 [(cosθ1 cosθ2 − sinθ1 sinθ2 ) + i(cosθ1 sinθ2 +sinθ1 cosθ2 )] N. B. Vyas Complex Numbers
41. 41. Multiplication and Division of ComplexNumbers in Polar FormLet z1 = r1 (cosθ1 + isinθ1 ) and z2 = r2 (cosθ2 + isinθ2 )be two complex numbers in polar form. Then Product: z1 z2 = r1 (cosθ1 + isinθ1 )r2 (cosθ2 + isinθ2 )= r1 r2 [(cosθ1 cosθ2 − sinθ1 sinθ2 ) + i(cosθ1 sinθ2 +sinθ1 cosθ2 )]= r1 r2 [cos(θ1 + θ2 ) + isin(θ1 + θ2 )] N. B. Vyas Complex Numbers
42. 42. Multiplication and Division of ComplexNumbers in Polar FormLet z1 = r1 (cosθ1 + isinθ1 ) and z2 = r2 (cosθ2 + isinθ2 )be two complex numbers in polar form. Then Product: z1 z2 = r1 (cosθ1 + isinθ1 )r2 (cosθ2 + isinθ2 )= r1 r2 [(cosθ1 cosθ2 − sinθ1 sinθ2 ) + i(cosθ1 sinθ2 +sinθ1 cosθ2 )]= r1 r2 [cos(θ1 + θ2 ) + isin(θ1 + θ2 )]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1 r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively. N. B. Vyas Complex Numbers
43. 43. Multiplication and Division of ComplexNumbers in Polar Form z1 r1 (cosθ1 + isinθ1 ) Division: = z2 r2 (cosθ2 + isinθ2 ) N. B. Vyas Complex Numbers
44. 44. Multiplication and Division of ComplexNumbers in Polar Form z1 r1 (cosθ1 + isinθ1 ) Division: = z2 r2 (cosθ2 + isinθ2 ) r1 (cosθ1 + isinθ1 )(cosθ2 − isinθ2 ) = r2 (cosθ2 + isinθ2 )(cosθ2 − isinθ2 ) N. B. Vyas Complex Numbers
45. 45. Multiplication and Division of ComplexNumbers in Polar Form z1 r1 (cosθ1 + isinθ1 ) Division: = z2 r2 (cosθ2 + isinθ2 ) r1 (cosθ1 + isinθ1 )(cosθ2 − isinθ2 ) = r2 (cosθ2 + isinθ2 )(cosθ2 − isinθ2 ) r1= [(cosθ1 cosθ2 + sinθ1 sinθ2 ) + i(sinθ1 cosθ2 − r2cosθ1 sinθ2 )] N. B. Vyas Complex Numbers
46. 46. Multiplication and Division of ComplexNumbers in Polar Form z1 r1 (cosθ1 + isinθ1 ) Division: = z2 r2 (cosθ2 + isinθ2 ) r1 (cosθ1 + isinθ1 )(cosθ2 − isinθ2 ) = r2 (cosθ2 + isinθ2 )(cosθ2 − isinθ2 ) r1= [(cosθ1 cosθ2 + sinθ1 sinθ2 ) + i(sinθ1 cosθ2 − r2cosθ1 sinθ2 )] r1= [cos(θ1 − θ2 ) + isin(θ1 − θ2 )] r2 N. B. Vyas Complex Numbers
47. 47. Multiplication and Division of ComplexNumbers in Polar Form z1 r1 (cosθ1 + isinθ1 ) Division: = z2 r2 (cosθ2 + isinθ2 ) r1 (cosθ1 + isinθ1 )(cosθ2 − isinθ2 ) = r2 (cosθ2 + isinθ2 )(cosθ2 − isinθ2 ) r1= [(cosθ1 cosθ2 + sinθ1 sinθ2 ) + i(sinθ1 cosθ2 − r2cosθ1 sinθ2 )] r1= [cos(θ1 − θ2 ) + isin(θ1 − θ2 )] r2which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the r1quotient of their moduli ( i.e. ) and diﬀerence of r2their arguments( i.e. θ1 − θ2 ) respectively. N. B. Vyas Complex Numbers
48. 48. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! N. B. Vyas Complex Numbers
49. 49. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! N. B. Vyas Complex Numbers
50. 50. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! N. B. Vyas Complex Numbers
51. 51. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! Now θ = iθ N. B. Vyas Complex Numbers
52. 52. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! Now θ = iθ (iθ)2 (iθ)3 (iθ)4 eiθ = 1 + iθ + + + ... 2! 3! 4! N. B. Vyas Complex Numbers
53. 53. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! Now θ = iθ (iθ)2 (iθ)3 (iθ)4 eiθ = 1 + iθ + + + ... 2! 3! 4! θ2 θ3 θ4 eiθ = 1 + iθ − −i + . . . { i2 = −1, i3 = −i, i4 = 1 2! 3! 4! N. B. Vyas Complex Numbers
54. 54. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! Now θ = iθ (iθ)2 (iθ)3 (iθ)4 eiθ = 1 + iθ + + + ... 2! 3! 4! θ2 θ3 θ4 eiθ = 1 + iθ − −i + . . . { i2 = −1, i3 = −i, i4 = 1 2! 3! 4! θ2 θ4 θ3 θ5 eiθ = 1 − + − ... + i θ − + − ... 2! 4! 3! 5! N. B. Vyas Complex Numbers
55. 55. Exponential form of Complex Numbers We know that θ3 θ5 θ7 sinθ = θ − + − ... 3! 5! 7! θ2 θ4 θ6 cosθ = 1 − + − . . . and 2! 4! 6! θ2 θ3 θ4 eθ = 1 + θ + + + ... 2! 3! 4! Now θ = iθ (iθ)2 (iθ)3 (iθ)4 eiθ = 1 + iθ + + + ... 2! 3! 4! θ2 θ3 θ4 eiθ = 1 + iθ − −i + . . . { i2 = −1, i3 = −i, i4 = 1 2! 3! 4! θ2 θ4 θ3 θ5 eiθ = 1 − + − ... + i θ − + − ... 2! 4! 3! 5! eiθ = (cosθ + isinθ) N. B. Vyas Complex Numbers
56. 56. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then N. B. Vyas Complex Numbers
57. 57. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then 1 Triangle Inequality: |z1 + z2 | ≤ |z1 | + |z2 | N. B. Vyas Complex Numbers
58. 58. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then 1 Triangle Inequality: |z1 + z2 | ≤ |z1 | + |z2 | 2 |z1 − z2 | ≥ ||z1 | − |z2 || N. B. Vyas Complex Numbers
59. 59. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then 1 Triangle Inequality: |z1 + z2 | ≤ |z1 | + |z2 | 2 |z1 − z2 | ≥ ||z1 | − |z2 || 3 Parellelogram equality: |z1 + z2 |2 + |z1 − z2 |2 = 2(|z1 |2 + |z2 |2 ) N. B. Vyas Complex Numbers
60. 60. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then 1 Triangle Inequality: |z1 + z2 | ≤ |z1 | + |z2 | 2 |z1 − z2 | ≥ ||z1 | − |z2 || 3 Parellelogram equality: |z1 + z2 |2 + |z1 − z2 |2 = 2(|z1 |2 + |z2 |2 ) 4 |z1 z2 | = |z1 ||z2 | N. B. Vyas Complex Numbers
61. 61. Laws of Complex NumbersIf z1 and z2 are two complex numbers, then 1 Triangle Inequality: |z1 + z2 | ≤ |z1 | + |z2 | 2 |z1 − z2 | ≥ ||z1 | − |z2 || 3 Parellelogram equality: |z1 + z2 |2 + |z1 − z2 |2 = 2(|z1 |2 + |z2 |2 ) 4 |z1 z2 | = |z1 ||z2 | z1 |z1 | 5 = z2 |z2 | N. B. Vyas Complex Numbers
62. 62. Ex 3 + 2i1 Find complex conjugate of 1−i N. B. Vyas Complex Numbers
63. 63. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ N. B. Vyas Complex Numbers
64. 64. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ∴ z = x + iy = r(cosθ + isinθ) N. B. Vyas Complex Numbers
65. 65. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ∴ z = x + iy = r(cosθ + isinθ) zn = rn (cosθ + isinθ)n N. B. Vyas Complex Numbers
66. 66. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ∴ z = x + iy = r(cosθ + isinθ) zn = rn (cosθ + isinθ)n = rn (cosnθ + isinnθ) N. B. Vyas Complex Numbers
67. 67. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ ∴ z = x + iy = r(cosθ + isinθ) zn = rn (cosθ + isinθ)n = rn (cosnθ + isinnθ)E.g.: N. B. Vyas Complex Numbers
68. 68. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ ∴ z = x + iy = r(cosθ + isinθ) zn = rn (cosθ + isinθ)n = rn (cosnθ + isinnθ)E.g.: (cosθ + isinθ)2 = cos2θ + isin2θ N. B. Vyas Complex Numbers
69. 69. TheoremDeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ ∴ z = x + iy = r(cosθ + isinθ) zn = rn (cosθ + isinθ)n = rn (cosnθ + isinnθ)E.g.: (cosθ + isinθ)2 = cos2θ + isin2θ (cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ N. B. Vyas Complex Numbers
70. 70. ExamplesEx. Simplify 2 (cos2θ + isin2θ) 3 (cosθ − isinθ)2 1 (cos3θ − isin3θ)2(cos5θ − isin5θ) 3 N. B. Vyas Complex Numbers
71. 71. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 N. B. Vyas Complex Numbers
72. 72. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 N. B. Vyas Complex Numbers
73. 73. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 N. B. Vyas Complex Numbers
74. 74. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 N. B. Vyas Complex Numbers
75. 75. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ ∴ 1 + i 3 = 2(cos π + isin π ) 3 3 N. B. Vyas Complex Numbers
76. 76. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 3 3 3 N. B. Vyas Complex Numbers
77. 77. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 N. B. Vyas Complex Numbers
78. 78. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 N. B. Vyas Complex Numbers
79. 79. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 = 290 (cos30π + isin30π) + 290 (cos30π − isin30π) N. B. Vyas Complex Numbers
80. 80. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 = 290 (cos30π + isin30π) + 290 (cos30π − isin30π) = 290 (2cos30π) N. B. Vyas Complex Numbers
81. 81. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 = 290 (cos30π + isin30π) + 290 (cos30π − isin30π) = 290 (2cos30π) = 291 cos30π N. B. Vyas Complex Numbers
82. 82. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 = 290 (cos30π + isin30π) + 290 (cos30π − isin30π) = 290 (2cos30π) = 291 cos30π = 291 (1) = 291 N. B. Vyas Complex Numbers
83. 83. Examples √ √Ex. Evaluate (1 + i 3)90 + (1 − i 3)90 √Sol. Let x = 1 and y = 3 √ √ r = x2 + y 2 = 1 + 3 = 4 = 2 √ θ = tan −1 y = tan−1 3 = π x 1 3 √ √ ∴ 1 + i 3 = 2(cos π + isin π ) and 1 − i 3 = 2(cos π − isin π ) 3 √ √ 3 3 3 ∴ (1 + i 3)90 + (1 − i 3)90 90 90 = 2(cos π + isin π ) 3 3 + 2(cos π − isin π ) 3 3 = 290 (cos30π + isin30π) + 290 (cos30π − isin30π) = 290 (2cos30π) = 291 cos30π = 291 (1) = 291 N. B. Vyas Complex Numbers
84. 84. Example n 1 + sinθ + icosθEx. Evaluate 1 + sinθ − icosθ N. B. Vyas Complex Numbers
85. 85. Sol. We have sin2 θ + cos2 θ = 1 N. B. Vyas Complex Numbers
86. 86. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ N. B. Vyas Complex Numbers
87. 87. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) N. B. Vyas Complex Numbers
88. 88. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ N. B. Vyas Complex Numbers
89. 89. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) N. B. Vyas Complex Numbers
90. 90. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) = (sinθ + icosθ)(sinθ − icosθ + 1) N. B. Vyas Complex Numbers
91. 91. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) = (sinθ + icosθ)(sinθ − icosθ + 1) 1 + sinθ + icosθ ∴ = sinθ + icosθ 1 + sinθ − icosθ N. B. Vyas Complex Numbers
92. 92. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) = (sinθ + icosθ)(sinθ − icosθ + 1) 1 + sinθ + icosθ ∴ = sinθ + icosθ 1 + sinθ − icosθ = cos π − θ + isin 2 π 2 −θ N. B. Vyas Complex Numbers
93. 93. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) = (sinθ + icosθ)(sinθ − icosθ + 1) 1 + sinθ + icosθ ∴ = sinθ + icosθ 1 + sinθ − icosθ = cos π − θ + isin 2 π 2 −θ n 1 + sinθ + icosθ π π n ∴ = cos 2 − θ + isin 2 −θ 1 + sinθ − icosθ N. B. Vyas Complex Numbers
94. 94. Sol. We have sin2 θ + cos2 θ = 1 ∴ 1 = sin2 θ + cos2 θ = sin2 θ − i2 cos2 θ = (sinθ − icosθ)(sinθ + icosθ) Now 1 + sinθ + icosθ = (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ) = (sinθ + icosθ)(sinθ − icosθ + 1) 1 + sinθ + icosθ ∴ = sinθ + icosθ 1 + sinθ − icosθ = cos π − θ + isin 2 π 2 −θ n 1 + sinθ + icosθ π π n ∴ = cos 2 − θ + isin 2 −θ 1 + sinθ − icosθ π π = cos n 2 −θ + isin n 2 −θ N. B. Vyas Complex Numbers
95. 95. ExampleEx. If z1 = eiθ1 , z2 = eiθ2 and z1 − z2 = 0 prove that 1 1 − =0 z1 z2 N. B. Vyas Complex Numbers
96. 96. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 N. B. Vyas Complex Numbers
97. 97. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2 N. B. Vyas Complex Numbers
98. 98. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 N. B. Vyas Complex Numbers
99. 99. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0 N. B. Vyas Complex Numbers
100. 100. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0 N. B. Vyas Complex Numbers
101. 101. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0Comparing real and imaginary parts on both the sides, we get N. B. Vyas Complex Numbers
102. 102. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0 N. B. Vyas Complex Numbers
103. 103. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0 1 1 1 1 Now − = − z1 z2 cosθ1 + isinθ1 cosθ2 + isinθ2 N. B. Vyas Complex Numbers
104. 104. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0 1 1 1 1 Now − = − z1 z2 cosθ1 + isinθ1 cosθ2 + isinθ2 cosθ1 − isinθ1 cosθ2 − isinθ2 = − cos2 θ1 + sin2 θ1 cos2 θ2 + sin2 θ2 N. B. Vyas Complex Numbers
105. 105. ExampleSol. Here z1 = eiθ1 = cosθ1 + isinθ1 z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1 ) − (cosθ2 + isinθ2 ) = 0(cosθ1 − cosθ2 ) + i(sinθ1 − sinθ2 ) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0 1 1 1 1 Now − = − z1 z2 cosθ1 + isinθ1 cosθ2 + isinθ2 cosθ1 − isinθ1 cosθ2 − isinθ2 = − cos2 θ1 + sin2 θ1 cos2 θ2 + sin2 θ2 N. B. Vyas Complex Numbers
106. 106. nEx. (1 + i)n + (1 − i)n = 2 2 +1 cos nπ 4 N. B. Vyas Complex Numbers
107. 107. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. N. B. Vyas Complex Numbers
108. 108. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem N. B. Vyas Complex Numbers
109. 109. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n N. B. Vyas Complex Numbers
110. 110. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n N. B. Vyas Complex Numbers
111. 111. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n 1 θ θ (cosθ + isinθ) n = cos + isin n n N. B. Vyas Complex Numbers
112. 112. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n 1 θ θ (cosθ + isinθ) n = cos + isin n n The remaining roots may be obtained by periodic nature of the trigonometric functions, i.e., N. B. Vyas Complex Numbers
113. 113. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n 1 θ θ (cosθ + isinθ) n = cos + isin n n The remaining roots may be obtained by periodic nature of the trigonometric functions, i.e., cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ N. B. Vyas Complex Numbers
114. 114. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n 1 θ θ (cosθ + isinθ) n = cos + isin n n The remaining roots may be obtained by periodic nature of the trigonometric functions, i.e., cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ Hence the general form of complex number cosθ + isinθ form rest of all the roots. N. B. Vyas Complex Numbers
115. 115. Root of a Complex Number De Moivre’s theorem is useful for ﬁnding the roots of a complex number. If n is any positive integer, then by De Moivre’s theorem θ θ n θ θ cos + isin = cos n + isin n = cosθ + isinθ n n n n θ θ Thus cos + isin is one of the nth root of cosθ + isinθ, i.e., n n 1 θ θ (cosθ + isinθ) n = cos + isin n n The remaining roots may be obtained by periodic nature of the trigonometric functions, i.e., cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ Hence the general form of complex number cosθ + isinθ form rest of all the roots. N. B. Vyas Complex Numbers
116. 116. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n N. B. Vyas Complex Numbers
117. 117. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n N. B. Vyas Complex Numbers
118. 118. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). N. B. Vyas Complex Numbers
119. 119. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: N. B. Vyas Complex Numbers
120. 120. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n N. B. Vyas Complex Numbers
121. 121. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n N. B. Vyas Complex Numbers
122. 122. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n 4π + θ 4π + θ k = 2, cos + isin n n N. B. Vyas Complex Numbers
123. 123. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n 4π + θ 4π + θ k = 2, cos + isin n n ...... N. B. Vyas Complex Numbers
124. 124. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n 4π + θ 4π + θ k = 2, cos + isin n n ...... ...... N. B. Vyas Complex Numbers
125. 125. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n 4π + θ 4π + θ k = 2, cos + isin n n ...... ...... 2(n − 1)π + θ 2(n − 1)π + θ k = n − 1, cos + isin n n N. B. Vyas Complex Numbers
126. 126. Root of a Complex Number 1 1 (cosθ + isinθ) n = [cos(2kπ + θ) + isin(2kπ + θ)] n 2kπ + θ 2kπ + θ = cos + isin n n 1 which gives all the roots of (cosθ + isinθ) n for k = 0, 1, 2, . . . (n − 1). This roots are as follows: θ θ For k = 0, cos + isin n n 2π + θ 2π + θ k = 1, cos + isin n n 4π + θ 4π + θ k = 2, cos + isin n n ...... ...... 2(n − 1)π + θ 2(n − 1)π + θ k = n − 1, cos + isin n n The further values of k will give the same roots as above in order. N. B. Vyas Complex Numbers