Let z and w are two complex number. Let z and w are two complex number. Prove that

$$\left|\frac{z-w}{1-z ̅w}\right| =1 \text{ if } |z|=1 \text{ but } w \neq z$$

 A: If $|z|=1$ then $|z|^2=z\bar z=1\implies |z\bar z|=|z||\bar z|=|\bar z| = 1$
Applying this result to $|z-w|$ gives
$|z-w|=|z-\frac{w\bar z}{\bar z}|=\frac{|z\bar z-w\bar z|}{|\bar z|}=\frac{|1-w\bar z|}{|\bar z|}=|1-w\bar z|$
Since $w\ne z$, $w\bar z\ne z\bar z=1\implies |1-w\bar z|\ne 0$
Therefore, we can divide both sides by $|1-w\bar z|$ to get $\frac{|z-w|}{|1-w\bar z|}=1$
A: 
$$\left|\frac{z-w}{1-\overline{z}w}\right|=1 \text{ if } |z|=1 \text{ but } w\neq z \text{ and } w,z\in\mathbb{C}$$

Notice:


*

*$$\overline{z}=\frac{|z|^2}{z}=\frac{1^2}{z}=\frac{1}{z}$$


$$\left|\frac{z-w}{1-\overline{z}w}\right|=\left|\frac{z-w}{1-\frac{1}{z}\cdot w}\right|=\left|\frac{z-w}{1-\frac{w}{z}}\right|=\left|\frac{z-w}{\frac{z}{z}-\frac{w}{z}}\right|=$$
$$\left|\frac{z-w}{\frac{z-w}{z}}\right|=\frac{\left|z-w\right|}{\left|\frac{z-w}{z}\right|}=\frac{|z-w|}{\frac{|z-w|}{|z|}}=\frac{|z-w|}{1}\cdot\frac{|z|}{|z-w|}=\frac{|z|}{1}=|z|=1$$
A: As Galc127 noted in the comments, it isn't true. For example, take $z=i$, $w=0$. Then your expression returns $\frac 1 {\sqrt{2}}$
A: It is sufficient to show that $|z-w|^{2}=|1-\bar{z}w|^{2}$. Note that
\begin{align*}
|z-w|^{2} &=(z-w)\overline{(z-w)}=(z-w)(\bar{z}-\bar{w})=|z|^{2}-z\bar{w}-w\bar{z}+|w|^{2}.
\end{align*}
And you can show similarly that
\begin{align*}
|1-\bar{z}w|^{2}&=|z|^{2}-z\bar{w}-\bar{z}w+|w|^{2}
\end{align*}
by using $|z|^{2}=1$. It is an easy computation and I'll leave that part for you to fill in.
