JEE PYQ: Complex Numbers Question 43
Question 43 - 2019 (12 Jan Shift 2)
If $\frac{z-\alpha}{z+\alpha}$ $(\alpha \in \mathbb{R})$ is a purely imaginary number and $|z| = 2$, then a value of $\alpha$ is:
(1) $2$
(2) $1$
(3) $\frac{1}{2}$
(4) $\sqrt{2}$
Show Answer
Answer: (1)
Solution
$\frac{z-\alpha}{z+\alpha}$ purely imaginary $\Rightarrow \frac{z-\alpha}{z+\alpha} + \overline{\left(\frac{z-\alpha}{z+\alpha}\right)} = 0$. This gives $|z|^2 = \alpha^2$. So $\alpha^2 = 4 \Rightarrow \alpha = 2$.
BETA
