The Riemann Mapping Theorem states that every simply connected domain which is not equal to the whole plane is conformal to the open unit disk.
My teacher told me casually that doubly connected domains are conformal to an annulus. I would like to know a more precise statement. Is every doubly connected domain $D\subset \mathbb{C}$ conformal to an annulus?
I consider an annulus to be the set $A(r,R) = \lbrace z \in \mathbb{C} \ | \ r < |z| < R \rbrace$. For $0<r<R<\infty$. When I think of a doubly connected domain $D$ then there are two closed sets $A,B$ in $\mathbb{C} \cup \lbrace \infty \rbrace$ such that $A\cup D\cup B$ and without loss of generality $\infty \in B$. I asked myself what if $A$ contains only one point. I found out that $A\cup D$ is simply connected and then there is a conformal map $f$ from $A\cup D$ onto the open unit disk such that $f(A) = \lbrace 0 \rbrace$. So this means $D$ is also conformal to a degenerated kind of annulus; the open unit disk without $0$. The case that $B$ only contains one point implies that $B = \lbrace \infty \rbrace$. In this case we apply $\varphi (z) = \dfrac{1}{z-a}$ for some $a \in A$ and get the case we had before right?
Are my considerations right? Do we have to take other limitations? Is there a doubly connected domain in $\mathbb{C}$ which is not conformal to an annulus like the whole plane is not conformal to the open unit disk?
Best Answer
With the requirement that $r>0$, then for example $\mathbb D^*$ is indeed not conformal to an annulus. Note however than $\mathbb D^*=A(0,1)$. On the other hand, $\mathbb C^*$ is also a doubly connected region, that is not conformal to $\mathbb D^*$ (if it were, it would extend to a conformal map between $\mathbb D$ and $\mathbb C$).
The complete and correct statement is that if $U$ is a doubly connected open subset of $\mathbb C$ such that $\mathbb C \backslash U$ contains at least two points, then there is a unique $r \in [0,1)$ such that $U$ is conformal to $A(r,1)$.
If $r>0$, then the number $\frac{1}{2\pi} \ln r^{-1}$ is called the modulus of $U$; otherwise $U$ is often called an annulus of infinite modulus. In that case, $U$ is biholomorphic to a simply connected domain with one point removed.