Question

State true or false:

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

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