Question

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

• A.
• B. k
• C. None of these

Answer 1

$\alpha =\frac{z-bar\omega }{k^2-z\overline{\omega }}\Rightarrow \overline{\alpha }=\frac{\overline{z}-omega}{k^2-barz\omega }$
But $z\overline{z}=\omega \overline{\omega }=k^2$ Hence
$\Rightarrow \overline{\alpha }=\frac{\frac{k^2}{z}-\frac{k^2}{\overline{\omega }}}{k^2+\frac{k^2}{z}-\frac{k^2}{\overline{\omega }}}=\frac{\overline{\omega }-z}{z\overline{\omega }+k^2}=-\alpha$
$\Rightarrow \alpha +\overline{\alpha }=0$
$\Rightarrow Re\left(\alpha \right)$ = 0