Mathematics 10 — Exercise Skilled practice · Page 13 (iii)

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.

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

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

   $$i+4$$

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

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

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

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

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.

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

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

   $$2\times 4+2\times 1\times i+3\times 4\times i+3\times 1\times ii$$

8. Real × real: multiply the numerical coefficients.

   $$2\times 4=8$$

9. Real × imaginary: multiply the numerical coefficients.

   $$2\times 1\times i=2\times i$$

10. Imaginary × real: multiply the numerical coefficients.

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

11. Imaginary × imaginary: multiply the numerical coefficients.

    $$3\times 1\times ii=3\times i^2$$

12. Substitute these four products into the expansion.

    $$8+2\times i+12\times i+3\times i^2$$

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

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

14. Multiply its coefficient by −1.

    $$3\times \left(-1\right)=-3$$

15. Rewrite the expansion after this sign calculation.

    $$8+2\times i+12\times i-3$$

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

    $$8-3+i\times \left(2+12\right)$$

17. Calculate the sum of the real terms.

    $$8-3=5$$

18. Calculate the sum of the imaginary coefficients.

    $$2+12=14$$

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

    $$14\times i+5$$

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.

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

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

    $$4\times 4+4\times 1\times i-4\times i-1\times ii$$

23. Real × real: multiply the numerical coefficients.

    $$4\times 4=16$$

24. Real × imaginary: multiply the numerical coefficients.

    $$4\times 1\times i=4\times i$$

25. Imaginary × real: multiply the numerical coefficients.

    $$-4\times i=-4\times i$$

26. Imaginary × imaginary: multiply the numerical coefficients.

    $$-1\times ii=-i^2$$

27. Substitute these four products into the expansion.

    $$16+4\times i-4\times i-i^2$$

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

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

29. Multiply its coefficient by −1.

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

30. Rewrite the expansion after this sign calculation.

    $$16+4\times i-4\times i+1$$

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

    $$16+1+i\times \left(4-4\right)$$

32. Calculate the sum of the real terms.

    $$16+1=17$$

33. Calculate the sum of the imaginary coefficients.

    $$4-4=0$$

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

    $$0\times i+17$$

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

    $$17=17$$

36. Put the calculated numerator above the calculated denominator.

    $$\frac{14\times i+5}{17}$$

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

    $$\frac{5}{17}+\frac{14\times i}{17}$$

38. Move i outside its numerical fraction.

    $$\frac{5}{17}+\left(\frac{14}{17}\right)\times i$$

39. Simplify the real coefficient fraction.

    $$\frac{5}{17}=\frac{5}{17}$$

40. Simplify the imaginary coefficient fraction.

    $$\frac{14}{17}=\frac{14}{17}$$

41. Combine the reduced real and imaginary terms.

    $$\frac{14\times i}{17}+\frac{5}{17}$$

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

    $$\frac{5}{17}-\frac{14\times i}{17}$$

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

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

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

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

    $$4-i$$

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

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

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

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

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.

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

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

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

51. Real × real: multiply the numerical coefficients.

    $$2\times 4=8$$

52. Real × imaginary: multiply the numerical coefficients.

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

53. Imaginary × real: multiply the numerical coefficients.

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

54. Imaginary × imaginary: multiply the numerical coefficients.

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

55. Substitute these four products into the expansion.

    $$8-2\times i-12\times i+3\times i^2$$

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

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

57. Multiply its coefficient by −1.

    $$3\times \left(-1\right)=-3$$

58. Rewrite the expansion after this sign calculation.

    $$8-2\times i-12\times i-3$$

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

    $$8-3+i\times \left(-2-12\right)$$

60. Calculate the sum of the real terms.

    $$8-3=5$$

61. Calculate this operation before combining the remaining terms.

    $$\left(-1\right)\times 12=-12$$

62. Calculate the sum of the imaginary coefficients.

    $$-2-12=-14$$

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

    $$5-14\times i$$

64. Now calculate the denominator independently.

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

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

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

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

67. Real × real: multiply the numerical coefficients.

    $$4\times 4=16$$

68. Real × imaginary: multiply the numerical coefficients.

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

69. Imaginary × real: multiply the numerical coefficients.

    $$4\times i=4\times i$$

70. Imaginary × imaginary: multiply the numerical coefficients.

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

71. Substitute these four products into the expansion.

    $$16-4\times i+4\times i-i^2$$

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

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

73. Multiply its coefficient by −1.

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

74. Rewrite the expansion after this sign calculation.

    $$16-4\times i+4\times i+1$$

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

    $$16+1+i\times \left(-4+4\right)$$

76. Calculate the sum of the real terms.

    $$16+1=17$$

77. Calculate the sum of the imaginary coefficients.

    $$-4+4=0$$

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

    $$0\times i+17$$

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

    $$17=17$$

80. Put the calculated numerator above the calculated denominator.

    $$\frac{5-14\times i}{17}$$

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

    $$\frac{5}{17}+\frac{-14\times i}{17}$$

82. Move i outside its numerical fraction.

    $$\frac{5}{17}+\left(-\frac{14}{17}\right)\times i$$

83. Simplify the real coefficient fraction.

    $$\frac{5}{17}=\frac{5}{17}$$

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

    $$\frac{\left(-14\right)\times 1}{1\times 17}$$

85. Calculate these two products separately.

    $$-\frac{14}{17}$$

86. Calculate this operation before combining the remaining terms.

    $$\left(-14\right)\times \frac{1}{17}=-\frac{14}{17}$$

87. Simplify the imaginary coefficient fraction.

    $$-\frac{14}{17}=-\frac{14}{17}$$

88. Combine the reduced real and imaginary terms.

    $$\frac{5}{17}-\frac{14\times i}{17}$$

89. Both results are equal.

    $$\frac{5}{17}-\frac{14\times i}{17}=\frac{5}{17}-\frac{14\times i}{17}$$

Answer

The final answer is shown below.