# complex anaysis help

• Jan 25th 2010, 02:33 PM
dori1123
complex anaysis help
Let $w,z \in \mathbb{C}$.
If $|w|<1$ and $|z| \leq 1$, show that $|\frac{z-w}{1-\bar wz}| \leq 1$. When does equality hold?
• Jan 25th 2010, 02:41 PM
Drexel28
Quote:

Originally Posted by dori1123
Let $w,z \in \mathbb{C}$.
If $|w|<1$ and $|z| \leq 1$, show that $|\frac{z-w}{1-\bar wz}| \leq 1$. When does equality hold?

Rudin, huh? What have you tried?
• Jan 25th 2010, 03:25 PM
dori1123
I know $|\frac{z-w}{1- \bar wz}|=\frac{|z-w|}{|1-\bar wz|}$ and $|z-w| \leq |z|+|w| \leq 1+1 =2$, not sure what to do with $|1 - \bar wz|$, I'm trying to work this way, not sure if it works... $|1 - \bar wz|=|-(\bar wz - 1)|=|(- \bar w)(z - \frac{1}{\bar w})|=|(- \bar w)(z - \frac{w}{w \bar w})|=|\bar w||z - \frac{w}{w \bar w}|=|\bar w||z - \frac{w}{|w|^2}|$=...
• Jan 26th 2010, 01:24 AM
Opalg
Quote:

Originally Posted by dori1123
Let $w,z \in \mathbb{C}$.
If $|w|<1$ and $|z| \leq 1$, show that $\Bigl|\frac{z-w}{1-\bar wz}\Bigr| \leq 1$. When does equality hold?

You want to show that $|z-w|\leqslant|1-\bar wz|$. Square both sides and use the fact that $|z|^2 = \bar zz$. Then what you are trying to prove is equivalent to $(z-w)(\bar z -\bar w) \leqslant (1-\bar wz)(1-w\bar z)$. Multiply out both sides and simplify ... .