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

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