I'm looking at the following problem from some old qualifying exam I found:
Let $\Omega$ be a simply connected proper open subset of $\Bbb C$, let $a\in\Omega$ and suppose we are given an analytic bijection $f:\Omega\to\mathbb D:=\{z:|z|<1\}$ satisfying $f(a)=0$ and $f'(a)>0$. Prove $$\inf_{z\in\partial\Omega}|z-a|\le\frac1{f'(a)}\le\sup_{z\in\partial\Omega}|z-a|.$$
To be honest, I have no idea where to start with this; my first thought was noticing that $1/f'(a)=(f^{-1})'(0)$, and was hoping to use this to try to apply the maximum modulus principle to some function involving $f'$ or $(f^{-1})'$ and $z-a$, but the first issue I come to before even trying to write a function down was realizing that $f'$ and $(f^{-1})'$ might not even be defined on the respective boundaries, so it doesn't seem like I could use maximum modulus to determine anything about the points on the boundary with respect to those functions.
I also assumed when I saw this that this was just something proved and used in the proof of the Riemann mapping theorem, but it doesn't actually seem to show up there.
Let $m=\inf_{z\in\partial\Omega} |z-a|$ and $M=\sup_{z\in\partial\Omega} |z-a|$.
Let $h_1(z)=f(mz+a)$ for $|z|<1$. It's easy to see that $h_1$ is well-defined($mz+a\in \Omega$) and $|h_1(z)|<1$ for all $|z|<1$ and $h_1(0)=f(a)=0$. Schwarz’s lemma implies that $|h_1’(0)|\leq 1$, which is exactly the left part of the desired inequality.
For the right part, consider $$h_2(z)=\frac{g(z)-a}{M},\ \ |z|<1,$$ where $g=f^{-1}$.