Question:
If $\frac{\mathrm{z}-\alpha}{\mathrm{z}+\alpha}(\alpha \in \mathrm{R})$ is a purely imaginary number and $|z|=2$, then a value of $\alpha$ is :
Correct Option: , 2
Solution:
$\frac{\mathrm{z}-\alpha}{\mathrm{z}+\alpha}+\frac{\overline{\mathrm{z}}-\alpha}{\overline{\mathrm{z}}+\alpha}=0$
$z \bar{z}+z \alpha-\alpha \bar{z}-\alpha^{2}+z \bar{z}-z \alpha+\bar{z} \alpha-\alpha^{2}=0$
$|z|^{2}=\alpha^{2}, \quad a=\pm 2$
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