# Using the riemann mapping theorem for the inverse domain

Let the riemann mapping theorem be

Let $$D\subset \mathbb{C}$$ be a simply-connected domain and $$z_0 \in D$$. Then there exists a unique conformal map $$f$$ from $$D$$ onto the open unitdisk $$\mathbb{D} = \lbrace z \in \mathbb{C} \ | \ |z| < 1 \rbrace$$ with $$f(z_0) = 0$$.

A text Im reading now uses this for mapping $$\hat{\mathbb{C}} \setminus \overline{B}$$ conformal onto $$\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$$ with $$f(\infty) = \infty$$ (where $$\hat{\mathbb{C}} = \mathbb{C} \cup \lbrace \infty \rbrace$$ and $$B$$ is a simply-connected domain that contains the point $$0$$). How can one show that this is always allowed by the riemann mapping theorem?

• What is $B$???? – David C. Ullrich Nov 8 at 16:50
• Sorry. I edited the question. – Arjihad Nov 8 at 17:34
• Ok. Not sure what aspect of this you're troubled about. First, $\hat\Bbb C\setmminus \overline B$ is simply connected, just because $\overline B$ is connected... – David C. Ullrich Nov 8 at 20:45
• $\hat{\Bbb C}\setminus\overline B$. – David C. Ullrich Nov 9 at 2:40
• Let $g(z) = 1/z$. Show that $g(\hat {\Bbb C} \setminus \overline B)$ is simply connected. Apply the theorem to it with $z_0 = 0$, then consider $g\circ f\circ g$ – Paul Sinclair Nov 9 at 3:31