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Let $D$ be a doubly connected domain in $\mathbb{C}$. It is well known that $D$ is conformal to an annulus $A(r,1)$ for some $0 \leq r < 1$ if $\mathbb{C} \setminus D$ contains at least two points.

What if $\mathbb{C} \setminus D$ contains only one point? Let's say $\mathbb{C} \setminus D =\lbrace a \rbrace$ for some $a \in \mathbb{C}$.

I guess $D$ would then not be conformal to any annulus $ A(r,1)$ with $0 \leq r < 1$.

I know that $D' = D \cup \lbrace a \rbrace$ is also not conformal to the open unit disk. But what would be a valid argumentation/proof for showing that $D$ is not conformal to an annulus? I guess $D$ not being conformal to an annulus implies that $D'$ is not conformal to the open unit disk but this is not the statement I want to show.

Levi
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Arji
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2 Answers2

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Let $D = \mathbb C \setminus \{a\}$. We want to show that $D$ is not conformally equivalent to any annulus. We will achieve this by showing that if a holomorphic map $f: D \rightarrow \mathbb C$ is injective, then its image is unbounded.

Suppose $f: D \rightarrow \mathbb C$ is holomorphic and injective. By the Caserati-Weierstrass Theorem, $a$ is not an essential singularity (because otherwise $f$ would not be injective). Alternatively, you can use Picard's Great Theorem to rule this out.

Case 1. $a$ is a removable singularity. In this case $f$ extends to a holomorphic map $F: \mathbb C \rightarrow \mathbb C$ and has unbounded image by Liouville's Theorem, so also $f$ has unbounded image. So $f$ is not a conformal equivalence between $D$ and an annulus.

Case 2. $a$ is a pole. But then $f$ is unbounded near $a$, so its image is not an annulus.

Levi
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One simple argument is to note that any bounded analytic function on $\mathbb C \setminus \{a\}$ is a constant but that is not true for an annulus: consider $\frac 1 {z-c}$ for suitable $c$.

Kavi Rama Murthy
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