Question

If $\frac{z-\alpha }{z+\alpha },z$ is a complex number and $\alpha ϵR$ is purely imaginary number and |z| = 2 then $\alpha =$

• A. 1
• B. 4
• C. 8
• D. 2

Answer 1

The correct option is D 2

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