Find sup$\{|f′(3)| : f$ maps $Ω$ analytically into the unit disk $\}.$

Question

Let

$Ω=\{z=x+iy∈C : |y|<x\}.$

Find sup$\{|f′(3)| : f$ maps $Ω$ analytically into the unit disk $\}.$

Okay. So I can find a conformal map from $Ω\rightarrow \mathbb{D}$. I used the map $f(z) = \frac{1 - z^2}{1 + z^2}$. But I'm not sure how to proceed (Want to use thwarts but can't.

Thanks

Answer 1

For any simply connected domain $\Omega$ and any point $a\in \Omega$, the supremum $$\sup \{|f'(a)| : |f|\le 1 \text{ in $\Omega$}\} $$ is attained by the conformal map of $\Omega$ onto unit disk such that $f(a)=0$.

Proof: let $g$ be the conformal map described above. Let $f$ be any other map as above. Then $f\circ g^{-1}$ is a map from disk to itself, so by the Schwarz lemma its derivative at $0$ is at most $1$ in absolute value. By the chain rule, this gives $|f'(a)| \le |g'(a)|$ as claimed.

Thus, all you need to do is to find conformal map that sends $3$ to $0$, which your current one doesn't. A standard way is to post-compose with a Möbius transformation of the disk. But in this case it's easier to pre-compose with a transformation of $\Omega$ that sends $3$ to a point that goes to $0$ under the map you already have.