Mathematics 10 — Exercise Worked examples · Example 12

Full question

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

Step-by-step solution

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

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

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

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

3. Real × real: multiply the numerical coefficients.

   $$3\times 2=6$$

4. Real × imaginary: multiply the numerical coefficients.

   $$3\times 3\times i=9\times i$$

5. Imaginary × real: multiply the numerical coefficients.

   $$4\times 2\times i=8\times i$$

6. Imaginary × imaginary: multiply the numerical coefficients.

   $$4\times 3\times ii=12\times i^2$$

7. Substitute these four products into the expansion.

   $$6+9\times i+8\times i+12\times i^2$$

8. In the last term replace i² by −1.

   $$12\times i^2=12\times \left(-1\right)$$

9. Multiply its coefficient by −1.

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

10. Rewrite the expansion after this sign calculation.

    $$6+9\times i+8\times i-12$$

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

    $$6-12+i\times \left(9+8\right)$$

12. Calculate the sum of the real terms.

    $$6-12=-6$$

13. Calculate the sum of the imaginary coefficients.

    $$9+8=17$$

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

    $$17\times i-6$$

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

    $$-17\times i-6$$

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

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

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

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

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

19. Real × real: multiply the numerical coefficients.

    $$3\times 2=6$$

20. Real × imaginary: multiply the numerical coefficients.

    $$3\times \left(-1\right)\times 3\times i=-9\times i$$

21. Imaginary × real: multiply the numerical coefficients.

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

22. Imaginary × imaginary: multiply the numerical coefficients.

    $$-4\times \left(-3\right)\times ii=12\times i^2$$

23. Substitute these four products into the expansion.

    $$6-9\times i-8\times i+12\times i^2$$

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

    $$12\times i^2=12\times \left(-1\right)$$

25. Multiply its coefficient by −1.

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

26. Rewrite the expansion after this sign calculation.

    $$6-9\times i-8\times i-12$$

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

    $$6-12+i\times \left(-9-8\right)$$

28. Calculate the sum of the real terms.

    $$6-12=-6$$

29. Calculate this operation before combining the remaining terms.

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

30. Calculate the sum of the imaginary coefficients.

    $$-9-8=-17$$

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

    $$-17\times i-6$$

32. Both results are equal.

    $$-17\times i-6=-17\times i-6$$

Answer

The final answer is shown below.