Question

If $k>0,\mid z\mid =\mid w\mid =k$ and $\alpha =\frac{z-\overline{w}}{k^2+z\overline{w}}$, then $Re\left(\alpha \right)$ equals

• A. $0$
• B. $\frac{k}{2}$
• C. $k$
• D. None of these

Answer 1

The correct option is A $0$

$\alpha =\frac{z-\overline{w}}{k^2+z\overline{w}}\Rightarrow \overline{\alpha }=\frac{\overline{z}-w}{k^2+\overline{z}w}$
But $z\overline{z}=w\overline{w}=k^2$. Hence,
$\Rightarrow \overline{\alpha =\frac{k^2}{z}-\frac{k^2}{\overline{w}}}{k}^2+\frac{k^2}{z}\frac{k^2}{\overline{w}}=\frac{\overline{w}-z}{z\overline{+}{k}^2}=-\alpha$
$\Rightarrow \alpha +\overline{\alpha }=0$
$\Rightarrow Re\left(\alpha \right)=0$