Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$ [duplicate]

Solution 1:

$$ \frac{|z-w|}{|1-\overline{w}z|}<1 $$ 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.

Solution 2:

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 $|w|^2<1$.

Solution 3:

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 $|w|<1$