Question

If $\frac{z-\alpha }{z+\alpha }(\alpha \in R)$ is a purely imaginary number and $|z|=2$, then a value of $\alpha$ is:

• A. $\frac{1}{2}$
• B. $2$
• C. $\sqrt{2}$
  
• D. $1$

Answer 1

The correct option is B $2$

Let $u$ $=\frac{z-\alpha }{z+\alpha }$
As, $u$ is purely imaginary
So, $u+\overline{u}=0$
$\Rightarrow \frac{z-\alpha }{z+\alpha }+\frac{\overline{z}-\alpha }{\overline{z}+\alpha }=0$
$\Rightarrow z\overline{z}+z\alpha -\overline{z}\alpha -{\alpha }^2+z\overline{z}-z\alpha +\overline{z}\alpha -{\alpha }^2=0$
$\Rightarrow 2z\overline{z}-2{\alpha }^2=0$
$\Rightarrow |z{|}^2={\alpha }^2$
$\Rightarrow \alpha =\pm |z|$
$\Rightarrow \alpha =\pm 2$