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Q. 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 :

JEE MainJEE Main 2019Complex Numbers and Quadratic Equations

Solution:

$\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 $