Question

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

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

Answer 1

Answer: (B)

$2$

Answer 2

$\frac{z-\alpha}{z+\alpha} + \frac{\bar{z} -\alpha}{\bar{z} +\alpha} = 0$
$z\bar{z}+z\alpha -\alpha\bar{z} -\alpha^{2} +z\bar{ z } - z\alpha + \bar{z} \alpha-\alpha^{2} = 0$
$\left|z\right|^{2} = \alpha^{2} , a = \pm2$