If z-z+ ( R) is a purely imaginary number and |z|=2, then a value of is:

Question

If $\dfrac{z-\alpha}{z+\alpha}$ ($\alpha\in\mathbb R$) is a purely imaginary number and $|z|=2$, then a value of $\alpha$ is:

Accepted Answer

$${{z - \alpha } \over {z + \alpha }} + {{\overline z  - \alpha } \over {\overline z  + \alpha }} = 0$$

$$z\overline z  + z\alpha  - \alpha \overline z  - {\alpha ^2} + z\overline z  - z\alpha  + \overline z \alpha  - {\alpha ^2} = 0$$

$${\left| z \right|^2} = {\alpha ^2},$$  $$a =  \pm 2$$

Solutions · 1