Question

If $k>0,|z|=k$ and $w=\frac{z-k}{z+k}$, then $Re\left(w\right)$ equals

• A. 0
• B. k
• C. $\frac{1}{k}$
• D. $-\frac{1}{k}$

Answer 1

The correct option is A

0

$|z|=k\Rightarrow |z{|}^2=k^2\text{or}z\overline{z}=k^2$
Next
$2Re\left(w\right)=w+\overline{w}=\frac{z-k}{z+k}+\frac{\overline{z}-k}{\overline{z}+k}$
Using $\overline{z}=\frac{k^2}{z}$,
we get $2Re\left(w\right)=0$
$\Rightarrow Re\left(w\right)=0$