Existence of conformal mapping between two regions

Question

Let $\Omega$ be a simply connected nonempty open subset of $\mathbb{C}$ such that $0\notin \Omega$. Show that there is a holomorphic bijection $\phi:\Omega \to \Omega'$ where $\Omega' \subset \mathbb{D}$ is simply connected nonempty open set.

Notice: Do NOT use Riemann mapping theorem.

Of course this is trivial by Riemann mapping theorem, but how can I show the existence without using that theorem? I guess this has something to do with complex logarithm since $\Omega$ is simply connected and $0 \notin \Omega$. Does anyone have ideas?

Any hints or advices will help a lot!

Answer 1

You can proceed as follows:

• There exists a holomorphic logarithm on $\Omega$, i.e. a holomorphic function $L: \Omega \to \Bbb C$ such that $e^{L(z)} = z$ for all $z \in \Omega$.
• $h(z) = e^{\frac 12 L(z)}$ is a holomorphic square root on $\Omega$, i.e. $h(z)^2 = z$ for all $z \in \Omega$.
• Let $U = h(\Omega)$. Then $w \in U \Longrightarrow -w \not\in U$.
• Choose any disk $\overline{B_r(w_0)}\subset U$. Then $\overline{B_r(-w_0)} \cap U = \emptyset$, so that $\vert h(z) + w_0 \vert > r$ for all $z \in \Omega$.
• $\phi(z) = \dfrac{r}{h(z)+w_0}$ is injective and $\phi(\Omega) \subset \Bbb D$.

(This is essentially the first part of the proof of the Riemann mapping theorem in “Ahlfors, Lars V. Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable”.)