Mathematics 10 — Exercise 1.2 · Q3 (iii)

Full question

Find the multiplicative inverse of the following complex number.

Step-by-step solution

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

   $$\frac{1}{7-3\times i}$$

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

   $$3\times i+7$$

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

   $$\left(\frac{3\times i+7}{3\times i+7}\right)\times \left(\frac{1}{7-3\times i}\right)$$

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

   $$\frac{3\times i+7}{\left(7-3\times i\right)\times \left(3\times i+7\right)}$$

5. Calculate the numerator first.

6. Multiplication by 1 leaves the conjugate unchanged.

   $$3\times i+7=3\times i+7$$

7. Now calculate the denominator independently.

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

   $$-3\times i\times \left(3\times i+7\right)+7\times \left(3\times i+7\right)$$

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

   $$7\times 7+7\times 3\times i-3\times 7\times i-3\times 3\times ii$$

10. Real × real: multiply the numerical coefficients.

    $$7\times 7=49$$

11. Real × imaginary: multiply the numerical coefficients.

    $$7\times 3\times i=21\times i$$

12. Imaginary × real: multiply the numerical coefficients.

    $$-3\times 7\times i=-21\times i$$

13. Imaginary × imaginary: multiply the numerical coefficients.

    $$-3\times 3\times ii=-9\times i^2$$

14. Substitute these four products into the expansion.

    $$49+21\times i-21\times i-9\times i^2$$

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

    $$-9\times i^2=\left(-9\right)\times \left(-1\right)$$

16. Multiply its coefficient by −1.

    $$\left(-9\right)\times \left(-1\right)=9$$

17. Rewrite the expansion after this sign calculation.

    $$49+21\times i-21\times i+9$$

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

    $$49+9+i\times \left(21-21\right)$$

19. Calculate the sum of the real terms.

    $$49+9=58$$

20. Calculate the sum of the imaginary coefficients.

    $$21-21=0$$

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

    $$0\times i+58$$

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

    $$58=58$$

23. Put the calculated numerator above the calculated denominator.

    $$\frac{3\times i+7}{58}$$

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

    $$\frac{7}{58}+\frac{3\times i}{58}$$

25. Move i outside its numerical fraction.

    $$\frac{7}{58}+\left(\frac{3}{58}\right)\times i$$

26. Simplify the real coefficient fraction.

    $$\frac{7}{58}=\frac{7}{58}$$

27. Simplify the imaginary coefficient fraction.

    $$\frac{3}{58}=\frac{3}{58}$$

28. Combine the reduced real and imaginary terms.

    $$\frac{3\times i}{58}+\frac{7}{58}$$

29. A multiplicative inverse must give 1 when multiplied by the original.

    $$1=1$$

Answer

The final answer is shown below.