Let $\displaystyle w,z \in \mathbb{C}$.
If $\displaystyle |w|<1$ and $\displaystyle |z| \leq 1$, show that $\displaystyle |\frac{z-w}{1-\bar wz}| \leq 1$. When does equality hold?
I know $\displaystyle |\frac{z-w}{1- \bar wz}|=\frac{|z-w|}{|1-\bar wz|}$ and $\displaystyle |z-w| \leq |z|+|w| \leq 1+1 =2$, not sure what to do with $\displaystyle |1 - \bar wz| $, I'm trying to work this way, not sure if it works... $\displaystyle |1 - \bar wz|=|-(\bar wz - 1)|=|(- \bar w)(z - \frac{1}{\bar w})|=|(- \bar w)(z - \frac{w}{w \bar w})|=|\bar w||z - \frac{w}{w \bar w}|=|\bar w||z - \frac{w}{|w|^2}|$=...
You want to show that $\displaystyle |z-w|\leqslant|1-\bar wz|$. Square both sides and use the fact that $\displaystyle |z|^2 = \bar zz$. Then what you are trying to prove is equivalent to $\displaystyle (z-w)(\bar z -\bar w) \leqslant (1-\bar wz)(1-w\bar z)$. Multiply out both sides and simplify ... .