Question

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

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

Answer 1

Answer: (A)

0

$\left|z\right|=\left|w\right|=k$ where $k>0$
$z=k(\cos A+i\sin A)\&w=k(\cos B+i\sin B)$
$\alpha =\frac{z-\overline{w}}{k^2+z\overline{w}}$
$Re\left(\alpha \right)=\frac{\alpha +\overline{\alpha }}{2}=\frac{1}{2}(\frac{z-\overline{w}}{k^2+z\overline{w}}+\frac{\overline{z}-w}{k^2+w\overline{z}})$
$\Rightarrow Re\left(\alpha \right)=\frac{1}{2}\left(\frac{z{k}^2+w{\left|z\right|}^2-\overline{w}{k}^2-\overline{z}{\left|w\right|}^2+\overline{z}{k}^2+\overline{w}{\left|z\right|}^2-w{k}^2-z{\left|w\right|}^2}{(k^2+z\overline{w})(k^2+w\overline{z})}\right)=\frac{1}{2}\left(\frac{z{k}^2+w{k}^2-\overline{w}{k}^2-\overline{z}{k}^2+\overline{z}{k}^2+\overline{w}{k}^2-w{k}^2-z{k}^2}{(k^2+z\overline{w})(k^2+w\overline{z})}\right)$
$\therefore Re\left(\alpha \right)=0$
Ans: A