Prove that this number is less than $1$, a) Prove that, if $z$ and $w$ are complex numbers and $|w| = 1$, then
$$\frac{|z-w|}{|1- \bar z w|} = 1$$
b) Prove that, if $|z|<1$, $|w|<1$, then
$$\frac{|z-w|}{|1- \bar z w|} < 1$$
I have proved part(a), but am stuck on part(b).  I tried using triangle inequalities but am getting not-good-enough upper bounds, e.g., a bound of $2$ instead of $1$.
Any hints or solutions for part(b) are welcome and appreciated.
Thanks,
 A: For part (b), we prove $|1-\overline{z}w|^2 > |z-w|^2$
Left-hand side is:
$\begin{align}
|1-\overline{z}w|^2 = (1-\overline{z}w)(1-\overline{\overline{z}w}) = (1-\overline{z}w)(1-z\overline{w}) &= -\overline{z}w-z\overline{w}+1+z \overline{z}w \overline{w} \\
&= -(\overline{z}w+z\overline{w}) + (1+|z|^2|w|^2)
\end{align}$
Right-hand side is:
$\begin{align}
|z-w|^2 = (z-w)(\overline{z}-\overline{w}) &= -\overline{z}w-z\overline{w}+z\overline{z}+w\overline{w} \\
&= -(\overline{z}w+z\overline{w}) + (|z|^2+|w|^2)
\end{align}$
Now we need only show that $1+|z|^2|w|^2 > |z|^2+|w|^2$
We are given $|z|,|w|<1$ so $|z|^2,|w|^2<1$.
Hence
$\begin{align}
(1-|z|^2)(1-|w|^2) > 0 &\implies \\
1+|z|^2|w|^2-|z|^2-|w|^2 > 0 &\implies \\
1+|z|^2|w|^2 > |z|^2+|w|^2  
\end{align}$
as required.

Notes:


*

*Since $|v|\ge0$ for any complex (or real) v, we can prove desired results for $|*|^2$ instead of $|*|$

*$|v|^2 = v\overline{v}$ is guaranteed to be $\ge 0$

*there was no need to do anything further with the $\overline{z}w+z\overline{w}$ term 

