Question

If $z=x+iy$ and $\omega =\frac{1-iz}{z-i}$ , then $\left|\omega \right|=1$ implies that, in the complex plane $z$ lies on the real axis.

Answer 1

$\left|\omega \right|=1\Rightarrow \left|\frac{1-iz}{z-i}\right|=1$
$\Rightarrow$ ${|1-zi|}^2={|z-i|}^2$
$\Rightarrow$ $(1-iz)(1+i\overline{z})=(z-i)(\overline{z}+i)$
$\because \overline{iz}=\overline{i}\overline{z}=-i\overline{z}$
$\Rightarrow$ $1-iz+i\overline{z}+z\overline{z}=z\overline{z}+iz-i\overline{z}+1$
$\Rightarrow$ $2i(z-\overline{z})=0\Rightarrow 2i\left(2iy\right)=0\Rightarrow y=0$,
which is the equation of real axis.