Mathematics 10 — Exercise 1.2 · Q4 (ii)

Full question

For z₁ = 2 + 5i, z₂ = 1 − 3i, z₃ = 2 + i, verify the stated property by computing both sides.

Step-by-step solution

1. Compute the left side first.

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

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

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

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

4. Real × real: multiply the numerical coefficients.

   $$2\times 1=2$$

5. Real × imaginary: multiply the numerical coefficients.

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

6. Imaginary × real: multiply the numerical coefficients.

   $$5\times 1\times i=5\times i$$

7. Imaginary × imaginary: multiply the numerical coefficients.

   $$5\times \left(-3\right)\times ii=-15\times i^2$$

8. Substitute these four products into the expansion.

   $$2-6\times i+5\times i-15\times i^2$$

9. In the last term replace i² by −1.

   $$-15\times i^2=\left(-15\right)\times \left(-1\right)$$

10. Multiply its coefficient by −1.

    $$\left(-15\right)\times \left(-1\right)=15$$

11. Rewrite the expansion after this sign calculation.

    $$2-6\times i+5\times i+15$$

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

    $$2+15+i\times \left(-6+5\right)$$

13. Calculate the sum of the real terms.

    $$2+15=17$$

14. Calculate the sum of the imaginary coefficients.

    $$-6+5=-1$$

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

    $$17-i$$

16. Now independently compute the right side.

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

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

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

    $$1\times 2+5\times i-3\times 2\times i-3\times 5\times ii$$

19. Real × real: multiply the numerical coefficients.

    $$1\times 2=2$$

20. Real × imaginary: multiply the numerical coefficients.

    $$5\times i=5\times i$$

21. Imaginary × real: multiply the numerical coefficients.

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

22. Imaginary × imaginary: multiply the numerical coefficients.

    $$-3\times 5\times ii=-15\times i^2$$

23. Substitute these four products into the expansion.

    $$2+5\times i-6\times i-15\times i^2$$

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

    $$-15\times i^2=\left(-15\right)\times \left(-1\right)$$

25. Multiply its coefficient by −1.

    $$\left(-15\right)\times \left(-1\right)=15$$

26. Rewrite the expansion after this sign calculation.

    $$2+5\times i-6\times i+15$$

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

    $$2+15+i\times \left(5-6\right)$$

28. Calculate the sum of the real terms.

    $$2+15=17$$

29. Calculate the sum of the imaginary coefficients.

    $$5-6=-1$$

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

    $$17-i$$

31. The results are identical, proving the equality.

    $$17-i=17-i$$

Answer

The final answer is shown below.