Question

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

A)1 

B)2

C)$\sqrt{2}$ 

D)$\frac{1}{2}$

Answer 1

The correct answer is B) 2 

Explanation ::

${\displaystyle \frac{z-\alpha }{z+\alpha }}+{\displaystyle \frac{\bar{z}-\alpha }{\bar{z}+\alpha }}=0$

$z\bar{z}+z\alpha -\alpha \bar{z}-{\alpha }^2+z\bar{z}-z\alpha +\bar{z}\alpha -{\alpha }^2=0$

$|z{|}^2={\alpha }^2,\alpha =\pm 2$