Question

Let $Z=x+iy$ and $\omega =\frac{1-iZ}{Z-i}$. If $|\omega |=1$, show that $Z$ is purely real.

Answer 1

$|w|=1$

$|\frac{1-iz}{z-i}|=1$

$|1-iz|=|z-i|$

$|1-i(x+iy)|=|x+i(y-1)|$, where $z=x+iy$.

$\sqrt{(1+y{)}^2+(-x{)}^2}=\sqrt{x^2+(y-1{)}^2}$

$(1+y{)}^2+x^2=x^2+(y-1{)}^2$

$y=0$

$z=x+i0=x$, which is purely real.