\begin{array}{rl}a_1 & z\bar{z}=|z|^2 \\ & \text{^^20also^^20},\operatorname{Im}(z)=-\operatorname{Jm}(z)\end{array} \Rightarrow \operatorname{Im}(z)-\frac{\operatorname{Im}(z)}{|z|^{2}}=0 \Rightarrow \quad \operatorname{Im}(z) \cdot\left[1-\frac{1}{|z|^{2}}\right]=0 \Rightarrow \text { either } \operatorname{Im}(z)=0 \text { or } 1-\frac{1}{|z|^{2}}=0 either z is real or IzI^2 =1 i.e., IzI=1 3) c) as we know that - (a \cdot b)^{m}=a^{m} \cdot b^{m} \quad \text { iff atleat one } a>0 \text { or } b>0 but here (-1)^{1 / 2}(-1)^{1 / 2}=(-1 \times-1)^{1 / 2} This step is false hence given statement is wrong. \Rightarrow \quad \operatorname{Im}\left(z+\frac{1}{z}\right)=0 3) b) z+1/z is real.
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