1. ## Complex analysis

Hey guys.

I have this problem, I need to show that it's true and I don't have a clue.
I tried to do like alpha = x+yi but it got me nowhere, any ideas?

Thanks.

2. Originally Posted by asi123
Hey guys.

I have this problem, I need to show that it's true and I don't have a clue.
I tried to do like alpha = x+yi but it got me nowhere, any ideas?

Thanks.
For $|\alpha| < 1, \alpha \in \mathbb{C}$ define $f(z) = \frac{z-\alpha}{1 - \overline{\alpha} z}$.

It is defined for $|z|\leq 1$, this is because $|1 - \overline{\alpha}z| \geq \bigg||1| - |\overline{\alpha}||z|\bigg| \geq 1 - |\alpha| > 0$.

Therefore, on the closed disk $|z|\leq 1$ the function $f$ is a continous function that is analytic on the open disk $|z|<1$. By the maximum modulos theorem theorem the largest value that $|f|$ attains happens on the boundary of the disk i.e. on $|z|=1$.

We know that $|f|^2 = f ~ \overline{f}$, therefore, if $|z|= 1$ we have:

$|f(z)|^2 = \frac{z-\alpha}{1-\overline{\alpha}z} \cdot \overline{ \frac{z-\alpha}{1-\overline{\alpha} z} } = \frac{z-\alpha}{1-\overline{\alpha} z} \cdot \frac{\overline{z}-\overline{\alpha}}{1-\alpha \overline{z}} = 1$.

Therefore, $|f| \leq 1$ for all $|z|\leq 1$ and furthermore $|f| = 1$ if and only if $|z|=1$.