Mathematics 10 — Exercise 1.2 · Q3 (i)

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.

   $$\frac{1}{5\times i+4}$$

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

   $$4-5\times i$$

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

   $$\left(\frac{4-5\times i}{4-5\times i}\right)\times \left(\frac{1}{5\times i+4}\right)$$

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

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

5. Calculate the numerator first.

6. Multiplication by 1 leaves the conjugate unchanged.

   $$4-5\times i=4-5\times i$$

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.

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

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

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

10. Real × real: multiply the numerical coefficients.

    $$4\times 4=16$$

11. Real × imaginary: multiply the numerical coefficients.

    $$4\times \left(-1\right)\times 5\times i=-20\times i$$

12. Imaginary × real: multiply the numerical coefficients.

    $$5\times 4\times i=20\times i$$

13. Imaginary × imaginary: multiply the numerical coefficients.

    $$5\times \left(-5\right)\times ii=-25\times i^2$$

14. Substitute these four products into the expansion.

    $$16-20\times i+20\times i-25\times i^2$$

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

    $$-25\times i^2=\left(-25\right)\times \left(-1\right)$$

16. Multiply its coefficient by −1.

    $$\left(-25\right)\times \left(-1\right)=25$$

17. Rewrite the expansion after this sign calculation.

    $$16-20\times i+20\times i+25$$

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

    $$16+25+i\times \left(-20+20\right)$$

19. Calculate the sum of the real terms.

    $$16+25=41$$

20. Calculate the sum of the imaginary coefficients.

    $$-20+20=0$$

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

    $$0\times i+41$$

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

    $$41=41$$

23. Put the calculated numerator above the calculated denominator.

    $$\frac{4-5\times i}{41}$$

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

    $$\frac{4}{41}+\frac{-5\times i}{41}$$

25. Move i outside its numerical fraction.

    $$\frac{4}{41}+\left(-\frac{5}{41}\right)\times i$$

26. Simplify the real coefficient fraction.

    $$\frac{4}{41}=\frac{4}{41}$$

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

    $$\frac{\left(-5\right)\times 1}{1\times 41}$$

28. Calculate these two products separately.

    $$-\frac{5}{41}$$

29. Calculate this operation before combining the remaining terms.

    $$\left(-5\right)\times \frac{1}{41}=-\frac{5}{41}$$

30. Simplify the imaginary coefficient fraction.

    $$-\frac{5}{41}=-\frac{5}{41}$$

31. Combine the reduced real and imaginary terms.

    $$\frac{4}{41}-\frac{5\times i}{41}$$

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

    $$1=1$$

Answer

The final answer is shown below.