Mathematics 10 — Exercise Worked examples · Example 13

Full question

For the given z₁,z₂, verify the displayed conjugate property.

Step-by-step solution

1. Start with the original quotient. For an inverse, the numerator is 1.

   $$\frac{4\times i+5}{2\times i+3}$$

2. Find the conjugate of the denominator: keep its real part and reverse the imaginary sign.

   $$3-2\times i$$

3. Multiply by the conjugate over itself; this fraction equals 1.

   $$\left(\frac{3-2\times i}{3-2\times i}\right)\times \left(\frac{4\times i+5}{2\times i+3}\right)$$

4. Multiply numerators together and denominators together. Keep their brackets intact.

   $$\frac{\left(3-2\times i\right)\times \left(4\times i+5\right)}{\left(3-2\times i\right)\times \left(2\times i+3\right)}$$

5. Calculate the numerator first.

6. Distribute the first bracket: multiply its real term by the whole second bracket, then its imaginary term by that bracket.

   $$4\times i\times \left(3-2\times i\right)+5\times \left(3-2\times i\right)$$

7. Distribute each of those products. Keep all four multiplications visible.

   $$5\times 3+5\times \left(-1\right)\times 2\times i+4\times 3\times i+4\times \left(-2\right)\times ii$$

8. Real × real: multiply the numerical coefficients.

   $$5\times 3=15$$

9. Real × imaginary: multiply the numerical coefficients.

   $$5\times \left(-1\right)\times 2\times i=-10\times i$$

10. Imaginary × real: multiply the numerical coefficients.

    $$4\times 3\times i=12\times i$$

11. Imaginary × imaginary: multiply the numerical coefficients.

    $$4\times \left(-2\right)\times ii=-8\times i^2$$

12. Substitute these four products into the expansion.

    $$15-10\times i+12\times i-8\times i^2$$

13. In the last term replace i² by −1.

    $$-8\times i^2=\left(-8\right)\times \left(-1\right)$$

14. Multiply its coefficient by −1.

    $$\left(-8\right)\times \left(-1\right)=8$$

15. Rewrite the expansion after this sign calculation.

    $$15-10\times i+12\times i+8$$

16. Group the real terms and factor i out of the imaginary terms.

    $$15+8+i\times \left(-10+12\right)$$

17. Calculate the sum of the real terms.

    $$15+8=23$$

18. Calculate the sum of the imaginary coefficients.

    $$-10+12=2$$

19. Insert both calculated sums in a + bi form.

    $$2\times i+23$$

20. Now calculate the denominator independently.

21. Distribute the first bracket: multiply its real term by the whole second bracket, then its imaginary term by that bracket.

    $$2\times i\times \left(3-2\times i\right)+3\times \left(3-2\times i\right)$$

22. Distribute each of those products. Keep all four multiplications visible.

    $$3\times 3+3\times \left(-1\right)\times 2\times i+2\times 3\times i+2\times \left(-2\right)\times ii$$

23. Real × real: multiply the numerical coefficients.

    $$3\times 3=9$$

24. Real × imaginary: multiply the numerical coefficients.

    $$3\times \left(-1\right)\times 2\times i=-6\times i$$

25. Imaginary × real: multiply the numerical coefficients.

    $$2\times 3\times i=6\times i$$

26. Imaginary × imaginary: multiply the numerical coefficients.

    $$2\times \left(-2\right)\times ii=-4\times i^2$$

27. Substitute these four products into the expansion.

    $$9-6\times i+6\times i-4\times i^2$$

28. In the last term replace i² by −1.

    $$-4\times i^2=\left(-4\right)\times \left(-1\right)$$

29. Multiply its coefficient by −1.

    $$\left(-4\right)\times \left(-1\right)=4$$

30. Rewrite the expansion after this sign calculation.

    $$9-6\times i+6\times i+4$$

31. Group the real terms and factor i out of the imaginary terms.

    $$9+4+i\times \left(-6+6\right)$$

32. Calculate the sum of the real terms.

    $$9+4=13$$

33. Calculate the sum of the imaginary coefficients.

    $$-6+6=0$$

34. Insert both calculated sums in a + bi form.

    $$0\times i+13$$

35. The imaginary coefficient in the denominator is zero, so the denominator is real.

    $$13=13$$

36. Put the calculated numerator above the calculated denominator.

    $$\frac{2\times i+23}{13}$$

37. Separate the real term and imaginary term into two fractions with the same denominator.

    $$\frac{23}{13}+\frac{2\times i}{13}$$

38. Move i outside its numerical fraction.

    $$\frac{23}{13}+\left(\frac{2}{13}\right)\times i$$

39. Simplify the real coefficient fraction.

    $$\frac{23}{13}=\frac{23}{13}$$

40. Simplify the imaginary coefficient fraction.

    $$\frac{2}{13}=\frac{2}{13}$$

41. Combine the reduced real and imaginary terms.

    $$\frac{2\times i}{13}+\frac{23}{13}$$

42. Take the conjugate of that result by changing the sign of its imaginary coefficient.

    $$\frac{23}{13}-\frac{2\times i}{13}$$

43. Now combine the conjugates of the original numbers independently.

44. Start with the original quotient. For an inverse, the numerator is 1.

    $$\frac{5-4\times i}{3-2\times i}$$

45. Find the conjugate of the denominator: keep its real part and reverse the imaginary sign.

    $$2\times i+3$$

46. Multiply by the conjugate over itself; this fraction equals 1.

    $$\left(\frac{5-4\times i}{3-2\times i}\right)\times \left(\frac{2\times i+3}{2\times i+3}\right)$$

47. Multiply numerators together and denominators together. Keep their brackets intact.

    $$\frac{\left(5-4\times i\right)\times \left(2\times i+3\right)}{\left(3-2\times i\right)\times \left(2\times i+3\right)}$$

48. Calculate the numerator first.

49. Distribute the first bracket: multiply its real term by the whole second bracket, then its imaginary term by that bracket.

    $$-4\times i\times \left(2\times i+3\right)+5\times \left(2\times i+3\right)$$

50. Distribute each of those products. Keep all four multiplications visible.

    $$5\times 3+5\times 2\times i-4\times 3\times i-4\times 2\times ii$$

51. Real × real: multiply the numerical coefficients.

    $$5\times 3=15$$

52. Real × imaginary: multiply the numerical coefficients.

    $$5\times 2\times i=10\times i$$

53. Imaginary × real: multiply the numerical coefficients.

    $$-4\times 3\times i=-12\times i$$

54. Imaginary × imaginary: multiply the numerical coefficients.

    $$-4\times 2\times ii=-8\times i^2$$

55. Substitute these four products into the expansion.

    $$15+10\times i-12\times i-8\times i^2$$

56. In the last term replace i² by −1.

    $$-8\times i^2=\left(-8\right)\times \left(-1\right)$$

57. Multiply its coefficient by −1.

    $$\left(-8\right)\times \left(-1\right)=8$$

58. Rewrite the expansion after this sign calculation.

    $$15+10\times i-12\times i+8$$

59. Group the real terms and factor i out of the imaginary terms.

    $$15+8+i\times \left(10-12\right)$$

60. Calculate the sum of the real terms.

    $$15+8=23$$

61. Calculate the sum of the imaginary coefficients.

    $$10-12=-2$$

62. Insert both calculated sums in a + bi form.

    $$23-2\times i$$

63. Now calculate the denominator independently.

64. Distribute the first bracket: multiply its real term by the whole second bracket, then its imaginary term by that bracket.

    $$-2\times i\times \left(2\times i+3\right)+3\times \left(2\times i+3\right)$$

65. Distribute each of those products. Keep all four multiplications visible.

    $$3\times 3+3\times 2\times i-2\times 3\times i-2\times 2\times ii$$

66. Real × real: multiply the numerical coefficients.

    $$3\times 3=9$$

67. Real × imaginary: multiply the numerical coefficients.

    $$3\times 2\times i=6\times i$$

68. Imaginary × real: multiply the numerical coefficients.

    $$-2\times 3\times i=-6\times i$$

69. Imaginary × imaginary: multiply the numerical coefficients.

    $$-2\times 2\times ii=-4\times i^2$$

70. Substitute these four products into the expansion.

    $$9+6\times i-6\times i-4\times i^2$$

71. In the last term replace i² by −1.

    $$-4\times i^2=\left(-4\right)\times \left(-1\right)$$

72. Multiply its coefficient by −1.

    $$\left(-4\right)\times \left(-1\right)=4$$

73. Rewrite the expansion after this sign calculation.

    $$9+6\times i-6\times i+4$$

74. Group the real terms and factor i out of the imaginary terms.

    $$9+4+i\times \left(6-6\right)$$

75. Calculate the sum of the real terms.

    $$9+4=13$$

76. Calculate the sum of the imaginary coefficients.

    $$6-6=0$$

77. Insert both calculated sums in a + bi form.

    $$0\times i+13$$

78. The imaginary coefficient in the denominator is zero, so the denominator is real.

    $$13=13$$

79. Put the calculated numerator above the calculated denominator.

    $$\frac{23-2\times i}{13}$$

80. Separate the real term and imaginary term into two fractions with the same denominator.

    $$\frac{23}{13}+\frac{-2\times i}{13}$$

81. Move i outside its numerical fraction.

    $$\frac{23}{13}+\left(-\frac{2}{13}\right)\times i$$

82. Simplify the real coefficient fraction.

    $$\frac{23}{13}=\frac{23}{13}$$

83. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{\left(-2\right)\times 1}{1\times 13}$$

84. Calculate these two products separately.

    $$-\frac{2}{13}$$

85. Calculate this operation before combining the remaining terms.

    $$\left(-2\right)\times \frac{1}{13}=-\frac{2}{13}$$

86. Simplify the imaginary coefficient fraction.

    $$-\frac{2}{13}=-\frac{2}{13}$$

87. Combine the reduced real and imaginary terms.

    $$\frac{23}{13}-\frac{2\times i}{13}$$

88. Both results are equal.

    $$\frac{23}{13}-\frac{2\times i}{13}=\frac{23}{13}-\frac{2\times i}{13}$$

Answer

The final answer is shown below.