If $z$ is a complex number and $|z| = 1$ and $z^2 \neq 1$. Then $\frac{z}{1-z^2}$ lies on :-

Question

If $z$ is a complex number and $|z| = 1$ and $z^2 \neq 1$. Then $\frac{z}{1-z^2}$ lies on :-

$(a)$ A line not through origin.
$(b)$ $|z| = 2$.
$(c)$ $x$-axis.
$(d)$ $y$-axis.

What I Tried:-

From here in the AOPS site, I understood that just bashing it gives us the solution I am looking for, but I want a solution without bashing.

Can someone help me?

Answer 1

\begin{align*} w=\frac{z}{1-z^2}&=\frac{1}{2}\left[\frac{1}{1-z}-\frac{1}{1+z}\right]\\ &=\frac{1}{2}\left[\frac{\bar{z}}{\bar{z}-1}-\frac{\bar{z}}{\bar{z}+1}\right]\\ &=\frac{\bar{z}}{\bar{z}^2-1}\\ &=-\bar{w}\\ w+\bar{w}&=0. \end{align*} Thus $\Re{(w)}=0$. This means $w$ is purely imaginary.

Answer 2

If $z \in \mathbb{C}$ and $|z| = 1$, the $z$ can be represented in a trigonometic form as

$$ z = e^{i\theta} = \cos{(\theta)} + i\sin{\theta}. $$

Given $z^2 \ne 1$, it means that $\theta \ne \pi n, n\in \mathbb{Z}$.

Then,

$$ \begin{aligned} \frac{z}{1-z^2} &= \frac{e^{i\theta}}{1-e^{2i\theta}} = \frac{\cos{(\theta)} + i\sin{\theta}}{(1 - \cos{(2\theta)}) - i\sin{2\theta}} = \\ &= \frac{(\cos{(\theta)} + i\sin{\theta})((1 - \cos{(2\theta)}) + i\sin{2\theta})}{(1 - \cos{(2\theta)})^2 + \sin^2{2\theta}} \end{aligned} $$

Simplify this expression, extract real and imaginary parts, and you will see that only imaginary part is not 0, which means that $\frac{z}{1-z^2}$ lies on $y$-axis.