Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$ 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.
 A: $$
\frac{|z-w|}{|1-\overline{w}z|}<1
$$
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.
A: You're right! After all, since $|w^*|=|w|<1$ and $|z|<1$, then $|zw^*|=|z||w^*|<1=|1|,$ so that's enough.
For the second, you must equivalently show that $|z-w|<|1-zw^*|.$ It suffices to show that $$|z-w|^2<|1-zw^*|^2,$$ or equivalently that $$|1-zw^*|^2-|z-w|^2>0.$$ Now use the given equation, together with the fact that $|z|^2<1$ and $|w|^2<1$.
A: Actually what you are trying to prove that the image of the unit disk $|z|<1$ is the unit disk $|w|<1$
Start by 
$$w=\frac{a-z}{1-\bar{a}z}$$Then 
$$z=\frac{a-w}{1-\bar{a}w}$$
$$\bar{z}=\frac{\bar{a}-\bar{w}}{1-a\bar{w}}$$
$$|z|^2=\frac{|a|^2-a\bar{w}-\bar{a}w+|w|^2}{1-a\bar{w}-\bar{a}w+|a|^2|w|^2}$$
since by assumption $|z|<1$
$$|a|^2-a\bar{w}-\bar{a}w+|w|^2 <1-a\bar{w}-\bar{a}w+|a|^2|w|^2$$
$$|a|^2+|w|^2 <1+|a|^2|w|^2$$
$$(1-|w|^2)(1-|a|^2)>0$$
since $|a|<1$ we must have $|w|<1$
