Question

If $z=x+iy$ and $w=\frac{(1-iz)}{(z-i)}$ and $\left|w\right|=1$, then prove that $z$ is purely real.

Answer 1

We have,
$\left|w\right|=1⟹\left|\frac{1-iz}{z-i}\right|=1$
or $\frac{|1-iz|}{|z-i|}=1$
or $|1-iz|=|z-i|$ (1)
or $|1-i(x+iy\left)\right|=|x+iy-i|$, where $z=x+iy$
or $|1+y-ix|=|x+i(y-1\left)\right|$
or $\sqrt{(1+y{)}^2+(-x{)}^2}=\sqrt{x^2+(y-1{)}^1}$
or $(1+y{)}^2+x^2=x^2+(y-1{)}^2$
or $y=0$
$⟹z=x+i0=x,$ which is purely real.
Ans: 1