If $\frac{z-\alpha}{z+\alpha}(\alpha \in \mathrm{R})$ is a purely imaginary number and $|\mathrm{z}|=2$, then a value of $\alpha$ is :
Correct Option: 1
Let $t=\frac{z-\alpha}{z+\alpha}$
$\because t$ is purely imaginary number.
$\therefore \quad t+\bar{t}=0$
$\Rightarrow \quad \frac{z-\alpha}{z+\alpha}+\frac{\bar{z}-\alpha}{\bar{z}+\alpha}=0$
$\Rightarrow \quad(z-\alpha)(\bar{z}+\alpha)+(\bar{z}-\alpha)(z+\alpha)=0$
$\Rightarrow \quad z \bar{z}-\alpha^{2}+z \bar{z}-\alpha^{2}=0$
$\Rightarrow \quad z \bar{z}-\alpha^{2}=0$
$\Rightarrow \quad|z|^{2}-\alpha^{2}=0$
$\Rightarrow \quad \alpha^{2}=4$
$\Rightarrow \quad \alpha=\pm 2$
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