Question

If $\left|z\right|=1$ and let $\omega =\frac{(1-z{)}^2}{1-z^2}$, then prove that the locus of $\omega$ is $|z-2|=|z+2|$.

Answer 1

Given $\omega =\frac{(1-z{)}^2}{1-z^2}=\frac{1-z}{1+z}$
$⟹\omega =\frac{z\overline{z}-z}{z\overline{z}+z}=\frac{\overline{z}-1}{\overline{z}+1}=-\left(\frac{\overline{1-z}}{1+z}\right)=-\overline{\omega }\therefore \omega +\overline{\omega }=0⟹\omega$ is purely imaginary.

Hence, $\omega$ lies on the $y$-axis.
Also $|z-2|=|z+2|⟹z$ lies on perpendicular bisector of $2$ and $-2,$ which is the imaginary axis.
Ans: 1