JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Complex Numbers

If $\frac{Z-\alpha }{Z+\alpha }$ is purely imaginary it equals the negative of its own conjugate, so $\frac{Z-\alpha }{Z+\alpha }=-\frac{\bar{Z}-\alpha }{\bar{Z}+\alpha }$, remembering $\alpha$ is real. Cross-multiplying and cancelling, the mixed terms drop out and you are left with $Z\bar{Z}={\alpha }^2$, that is $|Z{|}^2={\alpha }^2$. With $|Z|=2$ this gives ${\alpha }^2=4$, so $\alpha =2$ for the value listed.