Let $w=\frac{1-iz}{z-i}$, where $z=x+iy$. Given $|w|=1$, prove $z\in\mathbb{R}$.

Question

Let $w=\frac{1-iz}{z-i}$, where $z=x+iy$. Given $|w|=1$ we have to prove $z$ is a purely real number.

What i did was:

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

$w=\frac{-i(z-i)}{(z-i)}$ took $-i$ common

$w=-i$

I am not sure whether it correct or not but how am I supposed to continue forward?

Answer 1

By hypothesis $|1-iz|=|z-i|$. Squaring both sides we get $1+|z|^{2}-2\Re {iz}=|z^{2}+1+2 \Re {iz}$ so $\Re iz =0$. This means $iz =ic$ for some real number $c$ which gives $z=c \in \mathbb R$.

Answer 2

$$w=\frac{1-iz}{z-i} \implies w \bar w=\frac{1-iz}{z-i}\frac{1+i\bar z}{\bar z+i}= \frac{1-iz+i\bar z+z\bar z}{1+z\bar z+iz-i\bar z}=1\implies z=\bar z.$$

Answer 3

WLOG

$w=\cos2t+i\sin2t$

$$\cos2t+i\sin2t=\dfrac{1-iz}{z-i}$$

$$\implies z=\dfrac{1+i(\cos2t+i\sin2t)}{\cos2t+i\sin2t+i}$$

$$=\dfrac{(\cos t-\sin t)^2+i(\cos t-\sin t)(\cos t+\sin t)}{(\cos t-\sin t)(\cos t+\sin t)+i(\cos t+\sin t)^2}$$

$$=\dfrac{\cos t-\sin t}{\cos t+\sin t}$$