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Q.
If $|z|=1$ and $z \neq \pm 1$, then all the values of $\frac{z}{1-z^2}$ lie on:
Complex Numbers and Quadratic Equations
Solution:
Solution: $\operatorname{As}|z|=1$, we get $z \bar{z}=1$.
Let $ w=\frac{z}{1-z^2}$, then
$w+\bar{w} =\frac{z}{1-z^2}+\frac{\bar{z}}{1-\bar{z}^2}=\frac{z}{1-z^2}+\frac{1 / z}{1-(1 / z)^2} $
$ =\frac{z}{1-z^2}+\frac{z}{z^2-1}=0 $
$\Rightarrow 2 \operatorname{Re}(w) =0 \Rightarrow$ $\operatorname{Re}(w)=0$
Thus, $w$ lies on the $y$-axis.
Alternate Solution
$w=\frac{z}{1-z^2}= \frac{z}{z \bar{z}-z^2}=\frac{1}{\bar{z}-z} $
$ =\frac{1}{-2 i \operatorname{Im}(z)}$
$\Rightarrow w$ is purely imaginary, that is, $w$ lies on the $y$-axis.