1. **Problem (iii): Simplify** $\left( \frac{1}{1 + i} \right)^2$. 2. Start by simplifying $\frac{1}{1 + i}$. Multiply numerator and denominator by the conjugate of denominator, $1 - i$: $$\frac{1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 - i}{(1 + i)(1 - i)}$$ 3. Compute the denominator: $$(1 + i)(1 - i) = 1 - i + i - i^2 = 1 - (-1) = 2$$ 4. So, $$\frac{1}{1 + i} = \frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2}$$ 5. Now square the result: $$\left( \frac{1}{2} - \frac{i}{2} \right)^2 = \left( \frac{1}{2} \right)^2 - 2 \times \frac{1}{2} \times \frac{i}{2} + \left( \frac{i}{2} \right)^2 = \frac{1}{4} - \frac{i}{2} + \frac{i^2}{4}$$ 6. Since $i^2 = -1$, this becomes: $$\frac{1}{4} - \frac{i}{2} - \frac{1}{4} = - \frac{i}{2}$$ 7. **Final answer:** $$\left( \frac{1}{1 + i} \right)^2 = - \frac{i}{2}$$ --- 8. **Problem (ii) Part 1: Check if** $z_1 - z_2 = \overline{z_1} - \overline{z_2}$ 9. Recall that for complex numbers, conjugation distributes over subtraction: $$\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$$ 10. But $z_1 - z_2$ usually is not equal to $\overline{z_1} - \overline{z_2}$ unless both $z_1, z_2$ are real numbers. 11. **Conclusion:** $$z_1 - z_2 \neq \overline{z_1} - \overline{z_2} \text{ in general}$$ --- 12. **Problem (ii) Part 2: Verify whether** $\frac{z_1}{z_2} = \frac{\overline{z_1}}{z_2}$ 13. Generally, this equality does **not** hold since conjugation applies to numerator alone in RHS but LHS is regular division. 14. For equality, $z_1$ must be equal to $\overline{z_1}$ (i.e., $z_1$ is real). 15. **Conclusion:** $$\frac{z_1}{z_2} \neq \frac{\overline{z_1}}{z_2} \text{ generally}$$