Question

If $|z|=1,z\ne \pm 1$, then $\frac{z}{1-z^2}$ lies on

• A. a line not passing through the origin.
• B. a circle.
• C. on the real axis
• D. on the imaginary axis

Answer 1

The correct option is D on the imaginary axis

Given $|z|=1,z\ne \pm 1$
Let $\omega =\frac{z}{1-z^2}$
$\Rightarrow \omega =\frac{z}{z\overline{z}-z^2}(\because |z|=1\Rightarrow z\overline{z}=1)\Rightarrow \omega =\frac{1}{\overline{z}-z}$
As $\overline{z}-z$ is purely imaginary, so
$\omega$ lies on the imaginary axis.