TITLE: Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$
QUESTION [7 upvotes]: Prove if $|z| < 1$ and  $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1. 
And given the conditions that cannot be true. However I don't know if this part is right.

REPLY [4 votes]: $$
\frac{|z-w|}{|1-\overline{w}z|}<1
$$
iff
$$
|z-w|<|1-\overline{w}z|
$$
iff
$$
|z-w|^2<|1-\overline{w}z|^2.
$$
Therefore, we must check if
$$
(z-w)(\overline{z}-\overline{w})<(1-\overline{w}z)(1-w\overline{z}).
$$
In other words, if
$$
|z|^2-z\overline{w}-\overline{z}w+|w|^2<1-\overline{w}z-\overline{z}w+|z|^2|w|^2.
$$
Therefore, the statement becomes, if $0\leq a,b<1$, then 
$$
a+b<1+ab.
$$
This, however, is obvious since
$$
0<1-a-b+ab=(1-a)(1-b)
$$
is true.  Now, reverse all the steps to get a proof.