Showing an inequality in the complex plane

Question

Suppose $z,w$ lie in the unit disk. want to show

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

try:

Can I assume that $1 - \overline{w}z \neq 0 $? IF so, then I expand the inequality as follows:

$$ |w - z | < |1 - \overline{w}{z}| \iff (w-z)(\overline{w}-\overline{z})< (1 - \overline{w}z)(1- w \overline{z}) \iff$$

$$ \iff |w| - z \overline{w} - \overline{z} w + |z| < 1 - w \overline{z} - \overline{w} z + |w||z| \iff |w| + |z| < 1 + |zw|$$

Hence, the problem reduces to show that $|w| + |z| < 1 + |zw| $. I am stuck here. Perhaps this is not a right approach?

Answer 1

Finally note that $$|w| + |z| < 1 + |wz| \Leftrightarrow (1- |z|)(1 - |w|) >0$$

which is true since $|z|,|w| < 1$

Answer 2

$\bar{w}z\ne 1$ since, if you pass to modulus, $|\bar{w}z|< 1$ ($z$ and $w$ are in the open unit disk, so $|z|<1$ and $|\bar{w}|=|w|<1$).

As regarding your last statement you can observe that:

$$ |w|+|z|-1-|zw|=|w|+|z|-1-|z||w|=(1-|z|)(1-|w|)>0, $$ as $z$ and $w$ are both in the open unit disk ($|z|<1,|w|<1$)

Answer 3

$$\Big|\dfrac{z-a}{1-\overline{a}z}\Big|^2=\Big(\dfrac{z-a}{1-\overline{a}z}\Big)\overline{\Big(\dfrac{z-a}{1-\overline{a}z}\Big)}=\dfrac{|z|^2-\overline{a}z-a\overline{z}+|a|^2}{1-\overline{a}z-a\overline{z}+|a|^2|z|^2}$$ Hence $$\Big|\dfrac{z-a}{1-\overline{a}z}\Big|^2=1-\dfrac{(1-|z|^2)(1-|a|^2)}{|1-\overline{a}z|^2}.$$