Complex Numbers 2 Question 4
4. If $\frac{z-\alpha}{z+\alpha}(\alpha \in R)$ is a purely imaginary number and $|z|=2$, then a value of $\alpha$ is
(2019 Main, 12 Jan I)
(a) $\sqrt{2}$
(b) $\frac{1}{2}$
(c) 1
(d) 2
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Answer:
Correct Answer: 4. (d)
Solution:
- Since, the complex number $\frac{z-\alpha}{z+\alpha}(\alpha \in R)$ is purely imaginary number, therefore
$$ \begin{array}{rrr} & \frac{z-\alpha}{z+\alpha}+\frac{\bar{z}-\alpha}{\bar{z}+\alpha}=0 & {[\because \alpha \in R]} \\ \Rightarrow & z \bar{z}-\alpha \bar{z}+\alpha z-\alpha^{2}+z \bar{z}-\alpha z+\alpha \bar{z}-\alpha^{2}=0 \\ \Rightarrow & 2|z|^{2}-2 \alpha^{2}=0 & {\left[\because z \bar{z}=|z|^{2}\right]} \\ \Rightarrow & \alpha^{2}=|z|^{2}=4 & {[|z|=2 \text { given }]} \\ \Rightarrow & \alpha= \pm 2 & \end{array} $$
