Proof of Schwarz lemma 5

Question

I dont understand the following proof of Schwarz lemma: Schwarz Lemma In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$". Why does $g(z)$ have a local maximum then?

Answer 1

Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.

It is already proved that $|g(z)|\leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.

Answer 2

On that page, $D$ is defined to be the open complex unit disc: $D=\{z\in {\mathbb C}: |z|<1\}$ and $g$ is defined as $$\left\{ g = \begin{eqnarray} \frac{f(z)}{z} \quad \mbox{ if } f(z) \not= 0 \\ f'(0) \quad \mbox{ otherwise } \end{eqnarray} \right. $$

So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.

Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.