§7.5 Biholomorphisms between Annuli and the General Riemann Mapping Theorem

7.5 Biholomorphisms between Annuli and the General Riemann Mapping Theorem

The Riemann Mapping states that any two open null-homotopic proper subsets of $ℂ$ are conformally equivalent. The natural extension to this problem is the existence of such a conformal mapping between regions containing “holes.” The existence of such holes is a question of algebraic topology. One answer to such a classification is given by isomorphisms of the regions’ fundamental groups.

As a matter of fact, any region with one “hole” (doubly connected) will be conformally equivalent to an annular region. This problem will be entertained later below. Another interesting question pertaining to this is the extent to which these biholomorphisms are unique. We provide a result that is of similar spirit.

Definition 7.5.1 (Conformal Modulus of an Annulus).

For $0 < 𝑟 < 𝑅$ , let $𝐴 = {𝑧 \in ℂ : 𝑟 < | 𝑧 | < 𝑅}$ . The conformal modulus of this annulus is

mod 𝐴 = \frac{1}{2 π} log \frac{𝑅}{𝑟} .

The following result should shed light upon why such a classification exists:

Theorem 7.5.2.

Suppose $𝑅_{1}, 𝑅_{2} > 1$ and let

𝐴_{1} = {𝑧 \in ℂ : 1 < | 𝑧 | < 𝑅_{1}} and 𝐴_{2} = {𝑧 \in ℂ : 1 < | 𝑧 | < 𝑅_{2}} .

If $𝜙 : 𝐴_{1} \to 𝐴_{2}$ is a biholomorphism, then $𝑅_{1} = 𝑅_{2} = 𝑅$ and, for some $𝜃 \in ℝ$ , exactly one of the following holds:

$𝜙 (𝑧) = e^{i 𝜃} 𝑧$ , in which case $𝜙$ preserves the two boundary components.
$𝜙 (𝑧) = 𝑅 \frac{e^{i 𝜃}}{𝑧}$ , in which case $𝜙$ interchanges them.

Proof.

Assume that there exists a biholomorphism $𝜙 : 𝐴_{1} \to 𝐴_{2}$ .

Claim.

Exactly one of the following two cases is true:

For any sequence ${𝑧_{𝑗}}_{𝑗 \in ℕ} \subset 𝐴_{1}$ whose accumulation points are all in $𝜕 𝔻$ , $𝑤_{𝑗} = 𝜙 (𝑧_{𝑗})$ has accumulation points in $𝜕 𝔻$ ; and for any sequence ${𝑧_{𝑗}}_{𝑗 \in ℕ} \subset 𝐴_{1}$ with accumulation points all in $𝜕 𝐷 (0, 𝑅_{1})$ , $𝑤_{𝑗} = 𝜙 (𝑧_{𝑗})$ has all accumulation points in $𝜕 𝐷 (0, 𝑅_{2})$ .
For any sequence ${𝑧_{𝑗}}_{𝑗 \in ℕ} \subset 𝐴_{1}$ whose accumulation points are all in $𝜕 𝔻$ , $𝑤_{𝑗} = 𝜙 (𝑧_{𝑗})$ has accumulation points in $𝐷 (0, 𝑅_{2})$ ; and for any sequence ${𝑧_{𝑗}}_{𝑗 \in ℕ} \subset 𝐴_{1}$ with accumulation points all in $𝜕 𝐷 (0, 𝑅_{1})$ , $𝑤_{𝑗} = 𝜙 (𝑧_{𝑗})$ has all accumulation points in $𝜕 𝔻$ .

Proof of the claim.

First we must show that for any $𝑧_{𝑗} \to 𝜕 𝐴_{1}$ (not necessarily convergent but rather simply that $| 𝑧_{𝑗} |$ has all accumulation points in ${1, 𝑅_{1}}$ ), $𝑤_{𝑗}$ has all its accumulation points in $𝜕 𝐴_{2}$ .

This general scenario is obtained from a property known as properness (of $𝜙$ ): for any compact $𝐾 ⋐ 𝐴_{2}$ , its preimage ( $𝜙^{- 1} (𝐾)$ is compact). Obviously properness is true as $𝜙^{- 1}$ is continuous (in general, any homeomorphism is proper) and maps a compact set (here $𝐾$ ) to a compact set.

Then for any convergent subsequence ${𝑤_{𝑗_{𝑘}}}_{𝑘}$ converging to $𝑤_{\infty}$ within $\overline{𝐴_{2}}$ (there exists at least one by Bolzano–Weierstrass), if $𝑤_{\infty} \in 𝐴_{2}$ , then by properness $𝜙^{- 1} ({𝑤_{𝑗_{𝑘}}}_{𝑘} \cup {𝑤_{\infty}}) ⋐ 𝐴_{1}$ is compact. Note that this is equivalent to ${𝑧_{𝑗_{𝑘}}}_{𝑘} \cup 𝜙^{- 1} (𝑤_{\infty})$ being compact, meaning that $𝑧_{𝑗}$ has an accumulation point in $𝐴_{1}$ , which is impossible, hence implying all accumulation points of ${𝑤_{𝑗}}$ lie on the boundary $𝜕 𝐴_{2}$ .

We now show that all sequences approaching the same boundary component of $𝐴_{1}$ are mapped toward the same boundary component of $𝐴_{2}$ . Consider first sequences ${𝑧_{𝑗}}_{𝑗}$ for which $| 𝑧_{𝑗} | \to 1$ , and let $𝐶_{2} = {𝑤 \in 𝐴_{2} : | 𝑤 | = \frac{1 + 𝑅_{2}}{2}}$ . By properness, $𝐶_{1} = 𝜙^{- 1} (𝐶_{2}) ⋐ 𝐴_{1}$ is compact. Hence, for $0 < 𝜀 < dist (𝐶_{1}, 𝜕 𝔻)$ , we have

𝐶_{1} \cap {𝑧 \in 𝐴_{1} : 1 < | 𝑧 | < 1 + 𝜀} = ⌀ .

Suppose there were two sequences ${𝑧_{𝑗}}_{𝑗}$ and ${𝑧_{𝑗}^{'}}_{𝑗}$ such that $| 𝑧_{𝑗} | \to 1$ and $| 𝑧_{𝑗}^{'} | \to 1$ , but $| 𝜙 (𝑧_{𝑗}) | \to 1$ and $| 𝜙 (𝑧_{𝑗}^{'}) | \to 𝑅_{2}$ . For all sufficiently large $𝑗$ , the points $𝑧_{𝑗}$ and $𝑧_{𝑗}^{'}$ both lie in the path-connected annulus ${𝑧 \in 𝐴_{1} : 1 < | 𝑧 | < 1 + 𝜀}$ . Thus they can be joined by a curve contained entirely in this annulus. The image of this curve under $𝜙$ joins points lying in different components of $𝐴_{2} ∖ 𝐶_{2}$ . The modulus of this image then traverses from $\begin{matrix} \end{matrix} < \frac{1 + 𝑅_{2}}{2}$ to $\begin{matrix} \end{matrix} > \frac{1 + 𝑅_{2}}{2}$ . By the Intermediate Value Theorem, there exists a point at which it attains $\frac{1 + 𝑅_{2}}{2}$ . Therefore the image curve must intersect $𝐶_{2}$ , contradicting the choice of $𝜀$ .

Together with the preceding boundary-accumulation argument, exactly one of the following is true:

For every sequence ${𝑧_{𝑗}}_{𝑗} \subset 𝐴_{1}$ with $| 𝑧_{𝑗} | \to 1$ , we have $| 𝜙 (𝑧_{𝑗}) | \to 1$ .
For every sequence ${𝑧_{𝑗}}_{𝑗} \subset 𝐴_{1}$ with $| 𝑧_{𝑗} | \to 1$ , we have $| 𝜙 (𝑧_{𝑗}) | \to 𝑅_{2}$ .

Without loss of generality, we assume the former case (for the second case, replacing $𝜙$ with the inversion $𝑧 \to \frac{𝑅_{2}}{𝜙 (𝑧)}$ reduces it to the first case).

Applying the same argument to a sufficiently thin path-connected annulus near $𝜕 𝐷 (0, 𝑅_{1})$ shows that exactly one of the following is true:

For every sequence ${𝑧_{𝑗}}_{𝑗} \subset 𝐴_{1}$ with $| 𝑧_{𝑗} | \to 𝑅_{1}$ , we have $| 𝜙 (𝑧_{𝑗}) | \to 1$ .
For every sequence ${𝑧_{𝑗}}_{𝑗} \subset 𝐴_{1}$ with $| 𝑧_{𝑗} | \to 𝑅_{1}$ , we have $| 𝜙 (𝑧_{𝑗}) | \to 𝑅_{2}$ .

The first possibility cannot occur (under the previous “without loss of generality” assumption). Indeed, suppose that $| 𝜙 (𝑧_{𝑗}) | \to 1$ whenever either $| 𝑧_{𝑗} | \to 1$ or $| 𝑧_{𝑗} | \to 𝑅_{1}$ . Fix a point $𝑧_{0} \in 𝐴_{1}$ , by definition $| 𝜙 (𝑧_{0}) | > 1$ . Then the set

𝑆 = {𝑧 \in 𝐴_{1} : | 𝜙 (𝑧) | \geq | 𝜙 (𝑧_{0}) |}

is compact neighborhood of $𝑧_{0}$ lying in $𝐴_{1}$ : the closedness is direct from continuity of $𝜙$ . Then $| 𝜙 |$ attains its maximum $𝑀 \geq | 𝜙 (𝑧_{0}) | > 1$ in $𝐾$ ; moreover it is the absolute maximum of $| 𝜙 |$ attained in the interior of $𝐴_{1}$ . Then by the Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ), $𝜙$ would then by constant, which is impossible as it must map surjectively to $𝐴_{2}$ .

This concludes the proof (in the second case recall the inversion we have performed to obtain a dichotomous result). ⁠ $□$

Once again, without loss of generality assume $| 𝑧_{𝑗} | \to 1$ implies $| 𝑤_{𝑗} | \to 1$ and $| 𝑧_{𝑗} | \to 𝑅_{1}$ implies $| 𝑤_{𝑗} | \to 𝑅_{2}$ (by composing with an inversion, if necessary).

Then define the function

\begin{aligned} 𝜓 (𝑧) & = log | 𝑧 | log 𝑅_{2} - log | 𝜙 (𝑧) | log 𝑅_{1} \\ = ℜ𝔢 [log 𝑧 log 𝑅_{2} - log 𝜙 (𝑧) log 𝑅_{1}], \end{aligned}

which is harmonic (as the real part of a holomorphic function). This function extends continuously to $𝜕 𝐴_{1}$ and vanishes thereon.

By the Boundary Maximum Principle (Proposition 7.4.1 $Proposition 7.4.1 Proposition 7.4.1$ ), the function vanishes identically to $0$ . Therefore, $\forall 𝑧 \in 𝐴_{1}$ ,

log | 𝑧 | log 𝑅_{2} = log | 𝜙 (𝑧) | log 𝑅_{1} ⟹ | 𝜙 (𝑧) | = {| 𝑧 |}^{\frac{log 𝑅_{2}}{log 𝑅_{1}}} .

Let $𝐴_{1}^{*} = 𝐴_{1} ∖ ℝ_{< 0}$ be a region on which one can always define a holomorphic branch of $𝜓 (𝑧) = 𝑧^{\frac{log 𝑅_{2}}{log 𝑅_{1}}}$ . Then $𝜓$ is nonvanishing and $𝑔 \equiv \frac{𝜙}{𝜓}$ has a constant modulus of $1$ in $𝐴_{1}^{*}$ . Then for $𝑧 \in 𝐴_{1}^{*}$ ,

\begin{aligned} 𝑔^{'} (𝑧) = \frac{𝜕}{𝜕 𝑧} \frac{{| 𝑔 (𝑧) |}^{2}}{\overline{𝑔 (𝑧)}} & = \frac{1}{{\overline{𝑔 (𝑧)}}^{2}} [\frac{𝜕}{𝜕 𝑧} [{| 𝑔 (𝑧) |}^{2}] \overline{𝑔 (𝑧)} - {| 𝑔 (𝑧) |}^{2} \frac{𝜕}{𝜕 𝑧} \overline{𝑔 (𝑧)}] \\ = \frac{1}{{\overline{𝑔 (𝑧)}}^{2}} [\frac{𝜕}{𝜕 𝑧} [{| 𝑔 (𝑧) |}^{2}] \overline{𝑔 (𝑧)} - {| 𝑔 (𝑧) |}^{2} \overline{\frac{𝜕}{𝜕 \overline{𝑧}} 𝑔 (𝑧)}] \\ = \frac{\frac{𝜕 1}{𝜕 𝑧} \overline{𝑔 (𝑧)}}{{\overline{𝑔 (𝑧)}}^{2}} = 0 . \end{aligned}

Then $𝑔$ must be constant, more concretely a fixed unit value. Then on $𝐴_{1}^{*}$ ,

𝑔 \equiv e^{i 𝜃} ⟹ 𝜙 (𝑧) = e^{i 𝜃} 𝑧^{\frac{log 𝑅_{2}}{log 𝑅_{1}}} .

In order for this to extend to $ℝ_{< 0} \cap 𝐴_{1}$ , $\frac{log 𝑅_{2}}{log 𝑅_{1}}$ must be an integral value; moreover for univalence to hold this must be equal to $1$ . Therefore, $𝑅_{1} = 𝑅_{2} = 𝑅$ .

If no inversion was used, then $𝜙 (𝑧) = e^{i 𝜃} 𝑧$ . If an inversion was used, the boundary-preserving map is $𝑧 \mapsto \frac{𝑅}{𝜙 (𝑧)}$ , so

\frac{𝑅}{𝜙 (𝑧)} = e^{i 𝜃} 𝑧 ⟹ 𝜙 (𝑧) = 𝑅 \frac{e^{- i 𝜃}}{𝑧} .

Renaming the angle gives the second form. ⁠ $□$

Therefore, two annuli are conformally equivalent iff they have the same conformal modulus. We now formalize:

Corollary 7.5.2.1 (Holomorphic Automorphism Group of an Annulus).

For

𝑅 > 1

, let

𝐴_{𝑅} = {𝑧 \in ℂ : 1 < | 𝑧 | < 𝑅}

. Then

Aut (𝐴_{𝑅})

is either a rotation, an inversion, or a composition of both.

We now provide a conceptual proof of the Riemann Mapping Theorem based on the Dirichlet problem (which is in fact more historically aligned with the original approach from Riemann). We hope that the proof we provide generalizes easily into more complex domains for readers.

Lemma 7.5.3 (Polygonal Grid Exhaustion).

Let $𝑈 \subseteq ℂ$ be a region. There is a sequence ${𝑈_{𝑛}}_{𝑛 \in ℕ}$ of bounded, connected grid regions, each the interior of a finite union of closed squares from a square grid (and hence polygonal, with finitely many complement components), such that

\overline{𝑈_{𝑛}} ⋐ 𝑈_{𝑛 + 1}, ⋃_{𝑛 \in ℕ} 𝑈_{𝑛} = 𝑈 .

Moreover, every connected component of $𝜕 𝑈_{𝑛}$ is a polygonal Jordan curve, and every point of $𝜕 𝑈_{𝑛}$ satisfies the exterior segment condition in Theorem 7.3.15 $Theorem 7.3.15 Theorem 7.3.15$ .

If $𝑈$ is $𝑞$ -connected, then the domains $𝑈_{𝑛}$ may all be chosen to be $𝑞$ -connected. Equivalently, each $𝑈_{𝑛}$ has precisely $𝑞 - 1$ holes, or $𝑞$ holes in $\hat{ℂ}$ .

Proof.

Choose $𝑧_{0} \in 𝑈$ . If $𝑈 = ℂ$ , the interiors of the closed squares centered at $𝑧_{0}$ with side lengths $2 𝑛$ already give the required exhaustion. Suppose therefore that $𝑈 \neq ℂ$ . For $𝑧 \in ℂ$ , define

𝛿 (𝑧) = inf ({| 𝑧 - 𝑎 | : 𝑎 \in ℂ ∖ 𝑈}) .

Claim.

There is a compact connected exhaustion

{𝐾_{𝑛}}_{𝑛 \in ℕ}

𝑈

such that

𝐾_{𝑛} ⋐ \overset{⚬}{𝐾_{𝑛 + 1}}

Proof of the claim.

The triangle inequality gives $| 𝛿 (𝑧) - 𝛿 (𝜁) | \leq | 𝑧 - 𝜁 |$ , so $𝛿$ is continuous. Choose $𝑛_{0} \in ℕ$ such that $| 𝑧_{0} | < 𝑛_{0}$ and $\frac{1}{𝑛_{0}} < 𝛿 (𝑧_{0})$ , and set

𝐸_{𝑛} = {𝑧 \in ℂ : | 𝑧 | \leq 𝑛_{0} + 𝑛 and 𝛿 (𝑧) \geq \frac{1}{𝑛_{0} + 𝑛}} .

Each $𝐸_{𝑛}$ is compact. Moreover, the two defining inequalities for $𝐸_{𝑛}$ are strict relative to the bounds defining $𝐸_{𝑛 + 1}$ , so $𝐸_{𝑛} \subset \overset{⚬}{𝐸_{𝑛 + 1}}$ .

Let $𝐾_{𝑛}$ be the connected component of $𝐸_{𝑛}$ containing $𝑧_{0}$ . It is compact, and $𝐾_{𝑛} \subset 𝐾_{𝑛 + 1}$ . In fact, if $𝑧 \in 𝐾_{𝑛}$ , then some disk $𝐷 (𝑧, 𝑟)$ lies in $𝐸_{𝑛 + 1}$ . The connected set $𝐾_{𝑛} \cup 𝐷 (𝑧, 𝑟)$ contains $𝑧_{0}$ , so it lies in $𝐾_{𝑛 + 1}$ . Hence a neighborhood of $𝐾_{𝑛}$ lies in $𝐾_{𝑛 + 1}$ and $𝐾_{𝑛} ⋐ \overset{⚬}{𝐾_{𝑛 + 1}}$ .

Finally, fix $𝑧 \in 𝑈$ and choose a path $𝛾 : [0, 1] \to 𝑈$ from $𝑧_{0}$ to $𝑧$ . Its image is compact, so

𝑀 = max_{0 \leq 𝑡 \leq 1} | 𝛾 (𝑡) | < \infty and 𝑚 = min_{0 \leq 𝑡 \leq 1} 𝛿 (𝛾 (𝑡)) > 0 .

For sufficiently large $𝑛$ , $𝑛_{0} + 𝑛 > 𝑀$ and $\frac{1}{𝑛_{0} + 𝑛} < 𝑚$ , whence $𝛾 ([0, 1]) \subset 𝐸_{𝑛}$ . Thus $𝑧 \in 𝐾_{𝑛}$ , proving $⋃_{𝑛} 𝐾_{𝑛} = 𝑈$ . ⁠ $□$

Claim.

The region

𝑈

admits a nested exhaustion

{𝐴_{𝑛}}_{𝑛 \in ℕ}

by bounded connected grid regions with finitely many complementary components.

Proof of the claim.

Set

𝜎_{𝑛} = inf ({| 𝑧_{2} - 𝑧_{1} | : 𝑧_{2} \in 𝐾_{𝑛} and 𝑧_{1} \in ℂ ∖ \overset{⚬}{𝐾_{𝑛 + 1}}}) > 0 .

Consider the grid of compact squares

𝑄_{𝑛}^{(𝑗, 𝑘)} = {𝑥 + i 𝑦 : 𝑗 \frac{𝜎_{𝑛}}{2} \leq 𝑥 \leq (𝑗 + 1) \frac{𝜎_{𝑛}}{2}, 𝑘 \frac{𝜎_{𝑛}}{2} \leq 𝑦 \leq (𝑘 + 1) \frac{𝜎_{𝑛}}{2}}

and let

{𝒢︀}_{𝑛} = {𝑄_{𝑛}^{(𝑗, 𝑘)} : (𝑗, 𝑘) \in ℤ^{2} and 𝑄_{𝑛}^{(𝑗, 𝑘)} \cap 𝐾_{𝑛} \neq ⌀} .

This collection is finite. Each square has diagonal $\frac{𝜎_{𝑛}}{\sqrt{2}} < 𝜎_{𝑛}$ , so

\tilde{𝐾_{𝑛}} = ⋃_{𝑄 \in {𝒢︀}_{𝑛}} 𝑄

is a compact superset of $𝐾_{𝑛}$ contained in $\overset{⚬}{𝐾_{𝑛 + 1}}$ .

Furthermore, $𝐾_{𝑛} ⋐ \overset{⚬}{\tilde{𝐾_{𝑛}}}$ . Indeed, if a point of $𝐾_{𝑛}$ lay on a grid edge or vertex of $𝜕 \tilde{𝐾_{𝑛}}$ , every square adjacent across that edge or at that vertex would also meet $𝐾_{𝑛}$ and hence belong to ${𝒢︀}_{𝑛}$ , a contradiction. Let $𝐴_{𝑛}$ be the connected component of $\overset{⚬}{\tilde{𝐾_{𝑛}}}$ containing $𝐾_{𝑛}$ . Then

\overline{𝐴_{𝑛}} \subseteq \tilde{𝐾_{𝑛}} \subset 𝐾_{𝑛 + 1} ⋐ 𝐴_{𝑛 + 1} .

Thus ${𝐴_{𝑛}}$ is the required grid exhaustion. ⁠ $□$

Claim.

The grid exhaustion may be chosen so that every boundary component is a polygonal Jordan curve and distinct boundary components are disjoint.

Proof of the claim.

We resolve the possible “pinches” (see Figure 7.2 $Figure 7.2 Figure 7.2$ ) in each $\overline{𝐴_{𝑛}}$ . Let

𝜀_{𝑛} = dist (\overline{𝐴_{𝑛}}, ℂ ∖ 𝐴_{𝑛 + 1})

and let $𝛿_{𝑛}$ be the side length of the square used in the grid constructing $\overline{𝐴_{𝑛}}$ . Moreover, let

𝜂_{𝑛} = \frac{𝛿_{𝑛}}{⌊ 4 max {1, \frac{𝛿_{𝑛}}{𝜀_{𝑛}}} ⌋ + 1} .

It follows that the resultant value is a factor of $𝛿_{𝑛}$ , and satisfies $𝜂_{𝑛} < \frac{𝛿_{𝑛}}{4}$ and $𝜂_{𝑛} < \frac{𝜀_{𝑛}}{4}$ .

Figure 7.2: By choosing sufficiently small

𝜂 ∣ 𝛿

, we obtain a grid such that the boundary curves are disjoint.

Note that if a boundary component of $\overline{𝐴_{𝑛}}$ is not a Jordan curve, then in some $2$ -by- $2$ subgrid, exactly two opposite squares will be in $\overline{𝐴_{𝑛}}$ whilst the other two are not (refer to the circled area in Figure 7.2 $Figure 7.2 Figure 7.2$ ). Let $𝑆_{𝑛}$ be the set of grid vertices in $\overline{𝐴_{𝑛}}$ . Then

𝑅_{𝑛} = \overline{𝐴_{𝑛}} \cup ⋃_{𝑧 \in 𝑆_{𝑛}} [square of side length 2 𝜂_{𝑛} centered at 𝑧]

is a connected polygonal compact set (which can be considered to be generated with a grid of side length $𝜂_{𝑛}$ ) whose boundary curves are disjoint. From $𝜂_{𝑛} < \frac{𝜀_{𝑛}}{4}$ , this set lies in $𝐴_{𝑛 + 1}$ , since the maximum distance between the previous grid and this new set is $𝜂_{𝑛} \sqrt{2} < 𝜀_{𝑛}$ . Since $𝜂_{𝑛} < \frac{𝛿_{𝑛}}{4}$ , no such problematic $2$ -by- $2$ subgrid may exist. Thus, setting $𝑉_{𝑛} = \overset{⚬}{𝑅_{𝑛}}$ , we obtain

\overline{𝑉_{𝑛}} = 𝑅_{𝑛} ⋐ 𝑉_{𝑛 + 1} .

The regions ${𝑉_{𝑛}}$ still exhaust $𝑈$ , and every component of $𝜕 𝑉_{𝑛}$ is a polygonal Jordan curve. ⁠ $□$

If $𝑈$ has arbitrarily many holes, set $𝑈_{𝑛} = 𝑉_{𝑛}$ ; this proves the first assertion. It remains to preserve a prescribed finite connectivity.

Claim.

𝑈

𝑞

-connected, the grid regions may all be chosen

𝑞

-connected.

Proof of the claim.

Write $𝐶_{0}, \dots, 𝐶_{𝑞 - 1}$ for the components of $\hat{ℂ} ∖ 𝑈$ , with $\infty \in 𝐶_{0}$ . After omitting finitely many terms, the $𝐶_{𝑗}$ lie in distinct components of $\hat{ℂ} ∖ \overline{𝑉_{𝑛}}$ . Indeed, fix $𝑗 \neq 𝑘$ and suppose they were never separated. If $𝑌_{𝑛}$ denotes the component of $\hat{ℂ} ∖ \overline{𝑉_{𝑛}}$ containing both, then the sets $\overline{𝑌_{𝑛}}$ are decreasing compact connected sets and

𝐶_{𝑗} \cup 𝐶_{𝑘} \subseteq 𝑌 = ⋂_{𝑛} \overline{𝑌_{𝑛}} .

If $𝑧 \in 𝑈$ , then $𝑧 \in 𝑉_{𝑁}$ for some $𝑁$ and hence $𝑧 \notin \overline{𝑌_{𝑛}}$ for $𝑛 > 𝑁$ . Therefore $𝑌 \subseteq \hat{ℂ} ∖ 𝑈$ , contradicting that $𝐶_{𝑗}$ and $𝐶_{𝑘}$ are distinct components. Every pair is thus separated at some finite stage; since there are finitely many pairs, one common tail separates them all.

Let $𝑊_{𝑛, 𝑗}$ be the component of $\hat{ℂ} ∖ \overline{𝑉_{𝑛}}$ containing $𝐶_{𝑗}$ , and fill every other complementary component by setting

𝑃_{𝑛} = \hat{ℂ} ∖ ⋃_{𝑗 = 0}^{𝑞 - 1} 𝑊_{𝑛, 𝑗}, 𝑈_{𝑛} = \overset{⚬}{𝑃_{𝑛}} .

Then $𝑃_{𝑛} ⋐ 𝑈$ : every point outside $𝑈$ belongs to some $𝐶_{𝑗} \subset 𝑊_{𝑛, 𝑗}$ , while $𝑃_{𝑛}$ is compact. The previous claim ensures that the boundary components of $𝑉_{𝑛}$ can be assumed to be pairwise disjoint, and filling components only removes some of those boundary curves. Hence $\hat{ℂ} ∖ 𝑈_{𝑛}$ has exactly the $𝑞$ components corresponding to $𝑊_{𝑛, 0}, \dots, 𝑊_{𝑛, 𝑞 - 1}$ .

Finally, $\overline{𝑊_{𝑛 + 1, 𝑗}} \subseteq 𝑊_{𝑛, 𝑗}$ . Indeed,

\overline{𝑊_{𝑛 + 1, 𝑗}} \subseteq \hat{ℂ} ∖ 𝑉_{𝑛 + 1} \subseteq \hat{ℂ} ∖ \overline{𝑉_{𝑛}},

and the left-hand side is connected and contains $𝐶_{𝑗}$ , so it lies in the component $𝑊_{𝑛, 𝑗}$ . Since there are only finitely many $𝑗$ ,

⋃_{𝑗} \overline{𝑊_{𝑛 + 1, 𝑗}} \subseteq ⋃_{𝑗} 𝑊_{𝑛, 𝑗} .

Taking complements gives

𝑃_{𝑛} \subseteq \hat{ℂ} ∖ ⋃_{𝑗} \overline{𝑊_{𝑛 + 1, 𝑗}} = \overset{⚬}{𝑃_{𝑛 + 1}} = 𝑈_{𝑛 + 1} .

Since $𝑃_{𝑛}$ is compact, $𝑃_{𝑛} ⋐ 𝑈_{𝑛 + 1}$ . Also $𝑉_{𝑛} \subseteq 𝑈_{𝑛}$ , so ${𝑈_{𝑛}}$ exhausts $𝑈$ . ⁠ $□$

Claim.

Every point of

𝜕 𝑈_{𝑛}

satisfies the exterior segment condition.

Proof of the claim.

Notice that each connected boundary component a polygonal Jordan curve. At an interior point of an edge, choose a short segment in the exterior half-plane. At a vertex, of the four generating grid squares around it, at least one of them is not in

𝑈_{𝑛}

; then choose the segment between the vertex and the center of this square. In either case the segment meets

\overline{𝑈_{𝑛}}

only at its initial point. ⁠

□

⁠ $□$

Theorem 5.3.2 (Riemann Mapping Theorem).

Let

𝑈 \subset ℂ

(a proper subset, in other words,

𝑈 \neq ℂ

) be a simply connected (nonempty) open region. Let

𝑧_{0} \in 𝑈

be arbitrary. Then there exists a unique biholomorphism

𝑓 : 𝑈 \to 𝔻

such that

𝑓 (𝑧_{0}) = 0

and

𝑓^{'} (𝑧_{0}) \in ℝ_{> 0}

Proof.

First consider the case where $𝑈$ is a bounded region whose boundary is sufficiently regular (in the sense that each point admits a barrier) and $0 \in 𝑈$ . Then there is a solution $𝑢$ to the Dirichlet Problem (Theorem 7.4.2 $Theorem 7.4.2 Theorem 7.4.2$ ) in $𝑈$ with the continuous boundary function $𝑧 \mapsto - log | 𝑧 |$ . Obviously, the function $𝑢$ cannot be equivalent to $- log | 𝑧 |$ in the interior since the latter would approach $\infty$ at $0$ . Let

Φ (𝑧) = log | 𝑧 | + 𝑢 (𝑧),

then $Φ$ vanishes on $𝜕 𝑈$ and approaches $- \infty$ near $0$ . Then for some $𝜀 > 0$ ,

Φ (𝑧) < 0

on $\overline{𝐷 (0, 𝜀)} ∖ {0}$ . On $𝑈 ∖ \overline{𝐷 (0, 𝜀)}$ , the function $Φ$ is harmonic, is zero on the portion of the boundary contained in $𝜕 𝑈$ , and is negative on $𝜕 𝐷 (0, 𝜀)$ . Thus Proposition 7.4.1 $Proposition 7.4.1 Proposition 7.4.1$ first gives $Φ \leq 0$ there. Combining this with the strict inequality near $0$ gives $Φ \leq 0$ throughout $𝑈 ∖ {0}$ . If $Φ$ vanished at an interior point, the Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ ) applied on the connected region $𝑈 ∖ {0}$ would force $Φ$ to be constant, contradicting $Φ (𝑧) \to - \infty$ as $𝑧 \to 0$ . Therefore, $Φ < 0$ throughout $𝑈 ∖ {0}$ .

Let $𝜓 (𝑧) = 𝑢_{𝑥}^{'} (𝑧) - i 𝑢_{𝑦}^{'} (𝑧)$ , which is complex-harmonic ( $\nabla^{2} 𝜓 \equiv 0$ ). By elliptic regularity, $𝜓$ is $𝐶^{\infty}$ , and moreover,

\frac{𝜕 𝜓}{𝜕 \overline{𝑧}} \equiv \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} + i \frac{𝜕}{𝜕 𝑦}) (𝑢_{𝑥}^{'} - i 𝑢_{𝑦}^{'}) \equiv \frac{1}{2} (𝑢_{𝑥 𝑥}^{″} + 𝑢_{𝑦 𝑦}^{″}) = 0 .

Therefore, $𝜓$ is holomorphic on the simply connected set $𝑈$ , thus the function

𝑓 (𝑧) = \int_{0}^{𝑧} 𝜓 (𝜁) d 𝜁 + 𝑢 (0)

is well-defined and holomorphic. Moreover,

\begin{aligned} ℜ𝔢 𝑓 (𝑧) & = \int_{0}^{𝑧} [𝑢_{𝑥}^{'} (𝜁) d (ℜ𝔢 (𝜁)) + 𝑢_{𝑦}^{'} (𝜁) d (ℑ𝔪 (𝜁))] + 𝑢 (0) \\ = \int_{0}^{𝑧} d (𝑢 (𝜁)) + 𝑢 (0) = 𝑢 (𝑧) . \end{aligned}

Therefore, $𝑓$ is a holomorphic function whose real part is $𝑢$ , unique up to an additive imaginary constant. Define the holomorphic function

𝐹 (𝑧) = 𝑧 exp [𝑓 (𝑧)] .

Observe that $𝐹 (𝑈) \subseteq 𝔻$ since

| 𝐹 (𝑧) | = | 𝑧 | exp (ℜ𝔢 𝑓 (𝑧)) = exp (Φ (𝑧)) < 1

for $𝑧 \neq 0$ , while $𝐹 (0) = 0$ . It has precisely one zero, and since

𝐹^{'} (0) = {exp [𝑓 (𝑧)] (1 + 𝑧 𝑓^{'} (𝑧)) |}_{𝑧 = 0} = exp (𝑢 (0)) > 0,

it is a simple zero. Moreover, $| 𝐹 (𝑧) | = exp (Φ (𝑧)) \to 1$ whenever $𝑧 \to 𝜕 𝑈$ from within $𝑈$ . Consequently, $𝐹 : 𝑈 \to 𝔻$ is proper by the same boundary-compactness argument used in the proof of Theorem 7.5.2 $Theorem 7.5.2 Theorem 7.5.2$ .

We next prove that every $𝑤 \in 𝔻$ is assumed exactly once. We aim to use the Argument Principle, but note that $𝐹$ need not be holomorphic on a neighborhood of $\overline{𝑈}$ , and hence we exhaust $𝑈$ with compact sets. Choose $𝜌$ with $| 𝑤 | < 𝜌 < 1$ and set

𝐾_{𝜌} = {𝑧 \in 𝑈 : | 𝐹 (𝑧) | \leq 𝜌} .

Since $\overline{𝐷 (0, 𝜌)} ⋐ 𝔻$ is compact and $𝐾_{𝜌} = 𝐹^{- 1} (\overline{𝐷 (0, 𝜌)})$ , properness gives that $𝐾_{𝜌} ⋐ 𝑈$ is compact. By Lemma 7.5.3 $Lemma 7.5.3 Lemma 7.5.3$ , choose a polygonal exhaustion ${𝑉_{𝑛}}_{𝑛 \in ℕ}$ of $𝑈$ . For all sufficiently large $𝑛$ , $𝐾_{𝜌} ⋐ 𝑉_{𝑛}$ , and hence $| 𝐹 | > 𝜌 > | 𝑤 |$ on $𝜕 𝑉_{𝑛}$ .

On $𝜕 𝑉_{𝑛}$ we have $| \frac{𝑤}{𝐹} | < 1$ . By compactness, the same inequality holds on some open neighborhood of $𝜕 𝑉_{𝑛}$ . Therefore, $1 - \frac{𝑤}{𝐹}$ attains values in $𝐷 (1, 1) \subset ℂ^{*}$ on this neighborhood and $𝜂 = Log (1 - \frac{𝑤}{𝐹})$ is holomorphic there (using the principal branch). Since

𝐹 (𝑧) - 𝑤 = 𝐹 (𝑧) (1 - \frac{𝑤}{𝐹 (𝑧)}),

logarithmic differentiation gives

\frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤} - \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧)} = 𝜂^{'} (𝑧),

where the right-hand side is the derivative of a holomorphic function on a neighborhood of $𝜕 𝑉_{𝑛}$ . Let $𝛾 : [0, 1] \to 𝜕 𝑉_{𝑛}$ be a closed piecewise $𝐶^{1}$ parametrization. Neither $𝐹$ nor $𝐹 - 𝑤$ vanishes on $𝜕 𝑉_{𝑛}$ , so the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ) yields

\begin{aligned} 𝑛_{𝑉_{𝑛}} (𝐹 - 𝑤) - 𝑛_{𝑉_{𝑛}} (𝐹) & = \frac{1}{2 π i} ∳_{𝜕 𝑉_{𝑛}} [\frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤} - \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧)}] d 𝑧 \\ = \frac{1}{2 π i} ∳_{𝜕 𝑉_{𝑛}} 𝜂^{'} (𝑧) d 𝑧 \\ = \frac{1}{2 π i} \int_{0}^{1} 𝜂^{'} (𝛾 (𝑡)) 𝛾^{'} (𝑡) d 𝑡 \\ = \frac{1}{2 π i} [𝜂 (𝛾 (1)) - 𝜂 (𝛾 (0))] = 0 . \end{aligned}

Thus $𝑛_{𝑉_{𝑛}} (𝐹 - 𝑤) = 𝑛_{𝑉_{𝑛}} (𝐹) = 1$ , where zeros are counted with multiplicity. Every zero of $𝐹 - 𝑤$ in all of $𝑈$ belongs to $𝐾_{𝜌} \subset 𝑉_{𝑛}$ , so this is also the total number of solutions of $𝐹 (𝑧) = 𝑤$ in $𝑈$ . Since $𝑤$ was arbitrary, $𝐹$ is bijective; the multiplicity statement also gives $𝐹^{'} \neq 0$ , so $𝐹 : 𝑈 \to 𝔻$ is biholomorphic.

Now lift the regularity restriction upon $𝜕 𝑈$ . By Lemma 7.5.3 $Lemma 7.5.3 Lemma 7.5.3$ , choose a nested exhaustion ${𝑈_{𝑛}}_{𝑛 \in ℕ}$ of $𝑈$ by simply connected polygonal domains. After discarding finitely many initial terms, assume $0 \in 𝑈_{1}$ . Every point of $𝜕 𝑈_{𝑛}$ satisfies the exterior segment condition, so the regular-boundary case gives a biholomorphism $𝐹_{𝑛} : 𝑈_{𝑛} \to 𝔻$ such that $𝐹_{𝑛} (0) = 0$ and $𝐹_{𝑛}' (0) \in ℝ_{> 0}$ .

For fixed $𝑚$ , the tail ${𝐹_{𝑛}}_{𝑛 \geq 𝑚}$ is defined and bounded by $1$ on $𝑈_{𝑚}$ , so it is normal there by Montel’s Theorem (Theorem 5.2.6 $Theorem 5.2.6 Theorem 5.2.6$ ). We construct nested subsequences inductively.

First choose a subsequence ${𝐹_{𝑛_{1, 𝑘}}}_{𝑘 \in ℕ}$ converging locally uniformly on $𝑈_{1}$ .

If for a given $𝑗$ , ${𝐹_{𝑛_{𝑗, 𝑘}}}_{𝑘 \in ℕ}$ converges locally uniformly on $𝑈_{𝑗}$ , discard finitely many terms so that $𝑛_{𝑗, 𝑘} \geq 𝑗 + 1$ . The remaining functions are defined and bounded by $1$ on $𝑈_{𝑗 + 1}$ ; Montel’s Theorem therefore provides a further subsequence ${𝐹_{𝑛_{𝑗 + 1, 𝑘}}}_{𝑘 \in ℕ}$ converging locally uniformly on $𝑈_{𝑗 + 1}$ .

Let $𝐺_{𝑘} = 𝐹_{𝑛_{𝑘, 𝑘}}$ be the diagonal subsequence. For each fixed $𝑗$ , the tail ${𝐺_{𝑘}}_{𝑘 \geq 𝑗}$ is a subsequence of the sequence chosen at stage $𝑗$ , so it converges locally uniformly on $𝑈_{𝑗}$ to a holomorphic limit $Ψ_{𝑗} : 𝑈_{𝑗} \to \overline{𝔻}$ by Weierstrass (Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ). These limits agree on overlaps: if $𝑖 < 𝑗$ , then both $Ψ_{𝑖}$ and ${Ψ_{𝑗} |}_{𝑈_{𝑖}}$ are limits of the same tail on $𝑈_{𝑖}$ . Consequently, they define a holomorphic function $Ψ : 𝑈 \to \overline{𝔻}$ satisfying ${Ψ |}_{𝑈_{𝑗}} = Ψ_{𝑗}$ , and $𝐺_{𝑘} \to Ψ$ locally uniformly on $𝑈$ .

It remains to show that $Ψ$ is a biholomorphism and not a constant limit. Since $𝑈$ is bounded, choose $0 < 𝑟 < 𝑅$ such that

𝐷 (0, 𝑟) \subseteq 𝑈_{1} \subseteq 𝑈 \subseteq 𝐷 (0, 𝑅) .

Let $𝐻_{𝑘} = 𝐺_{𝑘}^{- 1} : 𝔻 \to 𝑈_{𝑛_{𝑘, 𝑘}}$ . Schwarz’s Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ) applied to $𝑧 \mapsto 𝐺_{𝑘} (𝑟 𝑧)$ and $𝑧 \mapsto \frac{𝐻_{𝑘} (𝑧)}{𝑅}$ gives

| {\frac{d}{d 𝑧} (𝐺_{𝑘} (𝑟 𝑧)) |}_{𝑧 = 0} | = 𝑟 | 𝐺_{𝑘}^{'} (0) | \leq 1

and

| {\frac{d}{d 𝑧} (\frac{𝐻_{𝑘} (𝑧)}{𝑅}) |}_{𝑧 = 0} | = \frac{1}{𝑅 𝐺_{𝑘}^{'} (𝐻_{𝑘} (0))} = \frac{1}{𝑅 𝐺_{𝑘}^{'} (0)} \leq 1;

hence

\frac{1}{𝑅} \leq 𝐺_{𝑘}^{'} (0) \leq \frac{1}{𝑟} and 𝑟 \leq 𝐻_{𝑘}^{'} (0) \leq 𝑅 .

Consequently, by the Weierstrass Convergence Theorem (Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ), $Ψ^{'} (0) = {lim}_{𝑘 \to \infty} 𝐺_{𝑘}^{'} (0) \geq \frac{1}{𝑅} > 0$ , so $Ψ$ is nonconstant. Since $Ψ (𝑈) \subseteq \overline{𝔻}$ , the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ) then implies that $Ψ (𝑈) \subset 𝔻$ .

The family ${𝐻_{𝑘}}$ is uniformly bounded by $𝑅$ on $𝔻$ , so Montel’s Theorem provides indices $𝑘_{1} < 𝑘_{2} < \dots$ for which $𝐻_{𝑘_{𝑗}} \to 𝐻$ locally uniformly on $𝔻$ , where $𝐻 : 𝔻 \to \overline{𝑈}$ is holomorphic. Replace ${𝐺_{𝑘}}$ by the corresponding subsequence ${𝐺_{𝑘_{𝑗}}}$ . Since the original sequence ${𝐺_{𝑘}}$ converges locally uniformly to $Ψ$ , this subsequence still converges to $Ψ$ . Relabeling the paired subsequences as ${𝐺_{𝑘}}$ and ${𝐻_{𝑘}}$ , we henceforth have

𝐺_{𝑘} ⇉ Ψ locally uniformly on 𝑈 and 𝐻_{𝑘} ⇉ 𝐻 locally uniformly on 𝔻,

while $𝐺_{𝑘} (𝐻_{𝑘} (𝑤)) = 𝑤$ and $𝐻_{𝑘} (𝐺_{𝑘} (𝑧)) = 𝑧$ whenever the expressions are defined. By Weierstrass convergence, $𝐻_{𝑘}^{'} (0) \geq 𝑟$ for each $𝑘$ gives $𝐻^{'} (0) \geq 𝑟 > 0$ , so $𝐻$ is nonconstant.

We claim that $𝐻 (𝔻) \subset 𝑈$ rather than merely $𝐻 (𝔻) \subseteq \overline{𝑈}$ . Fix $𝑎 \in ℂ ∖ 𝑈$ . Every function $𝐻_{𝑘} - 𝑎$ is nonvanishing on $𝔻$ ; hence Hurwitz’s Theorem (Theorem 3.3.5 $Theorem 3.3.5 Theorem 3.3.5$ ) implies that $𝐻 - 𝑎$ is either nonvanishing or identically zero. The latter is impossible because $𝐻$ is nonconstant. Since $𝑎$ was arbitrary, $𝐻$ omits $ℂ ∖ 𝑈$ , proving the claim.¹

We may now pass to the limit in the inverse identities. Fix $𝑤 \in 𝔻$ . Since $𝐻 (𝑤) \in 𝑈$ , choose a compact neighborhood $𝐿 ∋ 𝐻 (𝑤)$ (with $𝐿 ⋐ 𝑈$ ). Since $𝐻_{𝑘} (𝑤) \to 𝐻 (𝑤)$ , we have $𝐻_{𝑘} (𝑤) \in 𝐿$ for all sufficiently large $𝑘$ . Moreover, the exhaustion ensures that $𝐿$ lies in the domain of $𝐺_{𝑘}$ for all sufficiently large $𝑘$ . Therefore,

\begin{aligned} | 𝐺_{𝑘} (𝐻_{𝑘} (𝑤)) - Ψ (𝐻 (𝑤)) | & \leq | 𝐺_{𝑘} (𝐻_{𝑘} (𝑤)) - Ψ (𝐻_{𝑘} (𝑤)) | \\ \begin{matrix}  \end{matrix} + | Ψ (𝐻_{𝑘} (𝑤)) - Ψ (𝐻 (𝑤)) | \to 0 . \end{aligned}

The first term tends to $0$ because $𝐺_{𝑘} ⇉ Ψ$ on $𝐿$ , while the second tends to $0$ because $𝐻_{𝑘} (𝑤) \to 𝐻 (𝑤)$ and $Ψ$ is continuous. Since $𝐺_{𝑘} (𝐻_{𝑘} (𝑤)) = 𝑤$ for every $𝑘$ , it follows that

Ψ (𝐻 (𝑤)) = 𝑤 .

As $𝑤 \in 𝔻$ was arbitrary, this equality holds throughout $𝔻$ .

Conversely, fix $𝑧 \in 𝑈$ . The exhaustion places $𝑧$ in the domain of $𝐺_{𝑘}$ for all sufficiently large $𝑘$ . Since $𝐺_{𝑘} (𝑧) \to Ψ (𝑧) \in 𝔻$ , choose a compact neighborhood $𝑀 ⋐ 𝔻$ of $Ψ (𝑧)$ ; then $𝐺_{𝑘} (𝑧) \in 𝑀$ for all sufficiently large $𝑘$ . Hence,

\begin{aligned} | 𝐻_{𝑘} (𝐺_{𝑘} (𝑧)) - 𝐻 (Ψ (𝑧)) | & \leq | 𝐻_{𝑘} (𝐺_{𝑘} (𝑧)) - 𝐻 (𝐺_{𝑘} (𝑧)) | \\ \begin{matrix}  \end{matrix} + | 𝐻 (𝐺_{𝑘} (𝑧)) - 𝐻 (Ψ (𝑧)) | \to 0 . \end{aligned}

Here the first term tends to $0$ because $𝐻_{𝑘} \to 𝐻$ uniformly on $𝑀$ , and the second tends to $0$ by the continuity of $𝐻$ . Since $𝐻_{𝑘} (𝐺_{𝑘} (𝑧)) = 𝑧$ for all sufficiently large $𝑘$ , we conclude that

𝐻 (Ψ (𝑧)) = 𝑧 .

As $𝑧 \in 𝑈$ was arbitrary, this equality holds throughout $𝑈$ .

Therefore, $Ψ : 𝑈 \to 𝔻$ is biholomorphic for every bounded simply connected proper region $𝑈$ .

Finally, an arbitrary distinguished point $𝑧_{0}$ is reduced to $0$ by translation. If $𝑈$ is unbounded, apply the square-root and linear-fractional reduction to a bounded simply connected region established in the preceding proof of Theorem 5.3.2 $Theorem 5.3.2 Theorem 5.3.2$ , apply the bounded case above, and compose the resulting maps. A disk automorphism then imposes $Ψ (𝑧_{0}) = 0$ and $Ψ^{'} (𝑧_{0}) > 0$ . Uniqueness follows from the classification of $Aut (𝔻)$ in Theorem 3.5.3 $Theorem 3.5.3 Theorem 3.5.3$ . ⁠ $□$

We now turn to the first multiply connected case. A region $𝑈 \subset \hat{ℂ}$ is said to be doubly connected when $\hat{ℂ} ∖ 𝑈$ has exactly two connected components. When these components are bounded by disjoint Jordan curves, one may picture $𝑈$ as a region with precisely one hole. The Dirichlet problem provides a natural conformal coordinate measuring the position of a point between the two boundary components.

Theorem 7.5.4.

Let $𝑈 \subset ℂ$ be a bounded doubly connected region whose boundary is the disjoint union

𝜕 𝑈 = Γ_{0} \cup Γ_{1}

of two $𝐶^{1}$ Jordan curves, where $Γ_{0}$ is the outer boundary component. Then there exists a unique number $𝑅 > 1$ and a biholomorphism

𝐹 : 𝑈 \to {𝑤 \in ℂ : 1 < | 𝑤 | < 𝑅}

such that

Any sequence in $𝑈$ with all accumulation points in $Γ_{0}$ map to a sequence with all accumulation points in $𝜕 𝐷 (0, 𝑅)$
Any sequence in $𝑈$ with all accumulation points in $Γ_{1}$ map to a sequence with all accumulation points in $𝜕 𝔻$ .

Proof.

Since every point of the two boundary curves admits a barrier (by Theorem 7.3.16 $Theorem 7.3.16 Theorem 7.3.16$ ), Theorem 7.4.2 $Theorem 7.4.2 Theorem 7.4.2$ gives a unique function $𝑢 \in 𝐶 (\overline{𝑈}, ℝ)$ , harmonic in $𝑈$ , such that

{𝑢 |}_{Γ_{0}} \equiv 0 and {𝑢 |}_{Γ_{1}} \equiv 1 .

The Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ ) gives that $0 < 𝑢 < 1$ throughout $𝑈$ . Consider the differential form

𝜔 = - 𝑢_{𝑦}^{'} d 𝑥 + 𝑢_{𝑥}^{'} d 𝑦 .

Since

d 𝜔 = - 𝑢_{𝑦 𝑦}^{″} d 𝑦 \land d 𝑥 + 𝑢_{𝑥 𝑥}^{″} d 𝑥 \land d 𝑦 = 0,

$𝜔$ is a closed differential form in $𝑈$ . Let $𝛾$ be a positively oriented simple closed curve in $𝑈$ winding once around $Γ_{1}$ , and set

𝑇 = - \oint_{𝛾} 𝜔 . (7.5 . 1)

Define the real function

𝜚 (𝑥) = {\begin{cases} exp (- \frac{1}{𝑥}) & if 𝑥 > 0 \\ 0 & if 𝑥 \leq 0 \end{cases} .

Observe that $𝜚 \in 𝐶^{\infty} (ℝ)$ . Then for $0 < 𝑎 < 𝑏 < 1$ , the real function

𝜌 (𝑥) = \frac{𝜚 (\frac{𝑥 - 𝑎}{𝑏 - 𝑎})}{𝜚 (\frac{𝑥 - 𝑎}{𝑏 - 𝑎}) + 𝜚 (\frac{𝑏 - 𝑥}{𝑏 - 𝑎})} where 𝑥 \in ℝ

is smooth in $𝐶^{\infty}$ , vanishes for $𝑥 \leq 𝑎$ , is identically $1$ for $𝑥 \geq 𝑏$ , and has a positive derivative in $(𝑎, 𝑏)$ . Let $𝜒 \equiv 𝜌 \circ 𝑢$ , then on a relatively open annular neighborhood of $Γ_{0}$ in $\overline{𝑈}$ , $𝑢 < 𝑎$ , so $𝜒 \equiv 0$ thereon, while on a relatively open annular neighborhood of $Γ_{1}$ in $\overline{𝑈}$ , $𝑢 > 𝑏$ , so $𝜒 \equiv 1$ thereon. In each neighborhood, respectively choose Jordan curves $𝛾_{0} \subset 𝑈$ and $𝛾_{1} \subset 𝑈$ , such that $𝛾_{1}$ encloses $Γ_{1}$ and is enclosed by $𝛾$ . Then $𝛾_{0}^{+} \cup 𝛾_{1}^{-}$ encloses a doubly connected region $𝑊$ (as does $𝛾 \cup 𝛾_{1}$ ). Since $𝜒 \equiv 0$ in $𝛾_{0}$ and $𝜒 \equiv 1$ in $𝛾_{1}$ , then $𝜒 𝜔$ is $𝐶^{\infty}$ on a neighborhood of $\overline{𝑊}$ and thus Green’s Theorem (Theorem 3.1.2 $Theorem 3.1.2 Theorem 3.1.2$ ) is applicable:

\iint_{𝑊} d 𝜒 𝜔 = ∳_{𝛾_{0}} 𝜒 𝜔 - ∳_{𝛾_{1}} 𝜒 𝜔 = - ∳_{𝛾_{1}} 𝜔 = 𝑇,

because

∳_{𝛾} 𝜔 - ∳_{𝛾_{1}} 𝜔 = \iint_{int 𝛾 ∖ \overline{int 𝛾_{1}}} d 𝜔 = 0 .

Thus,

\begin{aligned} 𝑇 & = \iint_{𝑊} d 𝜒 𝜔 = \iint_{𝑊} d 𝜒 \land 𝜔 + 𝜒 d 𝜔 \\ = \iint_{𝑊} d 𝜒 \land 𝜔 = \iint_{𝑊} 𝜌^{'} \circ 𝑢 d 𝑢 \land 𝜔 \\ = \iint_{𝑊} 𝜌^{'} \circ 𝑢 (𝑢_{𝑥}^{'} d 𝑥 + 𝑢_{𝑦}^{'} d 𝑦) \land (- 𝑢_{𝑦}^{'} d 𝑥 + 𝑢_{𝑥}^{'} d 𝑦) \\ = \iint_{𝑊} 𝜌^{'} \circ 𝑢 {‖ 𝛁 𝑢 ‖}^{2} d 𝑥 \land d 𝑦 . \end{aligned}

Let

𝑞 (𝑧) = 𝜌^{'} \circ 𝑢 (𝑧) {‖ 𝛁 𝑢 (𝑧) ‖}^{2} .

Suppose for contradiction that $𝑞$ were nowhere positive. Then

{‖ 𝛁 𝜒 ‖}^{2} \equiv {(𝜌^{'} \circ 𝑢)}^{2} {‖ 𝛁 𝑢 ‖}^{2} \equiv 𝑞 𝜌^{'} \circ 𝑢 \equiv 0 .

Then $𝛁 𝜒 \equiv 𝟎$ throughout the connected region $𝑈$ , and is thus constant thereon. But $𝜒 \equiv 0$ on $Γ_{0}$ and $𝜒 \equiv 1$ on $Γ_{1}$ : a contradiction. Therefore at some $𝑧^{'} \in 𝑈$ , $𝑞 (𝑧^{'}) > 0$ (then $𝑎 \leq 𝑢 (𝑧^{'}) \leq 𝑏$ , and by construction $𝑧^{'} \in 𝑊$ ). By continuity, $\exists 𝜀 > 0$ such that $\forall 𝑧 \in 𝐷 (𝑧^{'}, 𝜀) \subseteq 𝑊$ ,

𝜌^{'} \circ 𝑢 (𝑧) {‖ 𝛁 𝑢 (𝑧) ‖}^{2} \geq \frac{𝑞 (𝑧^{'})}{2} > 0,

implying from the nonnegativity of $𝑞$ , that $𝑇 \geq \frac{𝑞 (𝑧^{'})}{2} π 𝜀^{2} > 0$ . Hence $𝑇 > 0$ .

The function $𝑔 (𝑧) = 𝑢_{𝑥}^{'} (𝑧) - i 𝑢_{𝑦}^{'} (𝑧)$ is holomorphic (verifiable by using $\frac{𝜕}{𝜕 \overline{𝑧}}$ ). Choose a simple $𝐶^{1}$ arc $𝐿 \subset \overline{𝑈}$ joining $Γ_{1}$ to $Γ_{0}$ , with its interior in $𝑈$ , which crosses $𝛾$ non-tangentially exactly once and such that $𝑈^{*} = 𝑈 ∖ 𝐿$ is simply connected. Fix $𝑧_{*} \in 𝑈^{*}$ and define

ℎ (𝑧) = 𝑢 (𝑧_{*}) + \int_{𝑧_{*}}^{𝑧} 𝑔 (𝜁) d 𝜁 .

Since $𝑈^{*}$ is simply connected, the integral is path-independent and $ℎ$ is a well-defined holomorphic function on $𝑈^{*}$ . Moreover,

\begin{aligned} \oint_{𝛾} 𝑔 (𝑧) d 𝑧 & = \oint_{𝛾} 𝑢_{𝑥}^{'} d 𝑥 + 𝑢_{𝑦}^{'} d 𝑦 + i 𝑢_{𝑥}^{'} d 𝑦 - i 𝑢_{𝑦}^{'} d 𝑥 \\ = i \oint_{𝛾} 𝜔 + \oint_{𝛾} d 𝑢 = - i 𝑇, \end{aligned}

while, for any path $𝜎$ in $𝑈^{*}$ from $𝑧_{*}$ to $𝑧$ ,

ℜ𝔢 ℎ (𝑧) = 𝑢 (𝑧_{*}) + \int_{𝜎} 𝑢_{𝑥}^{'} d 𝑥 + 𝑢_{𝑦}^{'} d 𝑦 = 𝑢 (𝑧_{*}) + \int_{𝑢 (𝜎)} d 𝑢 = 𝑢 (𝑧) .

Thus $ℜ𝔢 ℎ \equiv 𝑢$ . Let $𝛼 = \frac{2 π}{𝑇}$ , $𝑅 = e^{𝛼}$ , and define on $𝑈^{*}$

𝐹 (𝑧) = exp (𝛼 (1 - ℎ (𝑧))) (7.5 . 2)

We claim that $𝐹$ extends holomorphically across $𝐿$ . Fix $𝑝 \in 𝐿 \cap 𝑈$ and choose a disk $𝐷 (𝑝, 𝜀) ⋐ 𝑈$ . Fix $𝑝_{*} \in 𝐷 (𝑝, 𝜀) ∖ 𝐿$ and define on the entire disk

\tilde{ℎ} (𝑧) = ℎ (𝑝_{*}) + \int_{𝑝_{*}}^{𝑧} 𝑔 (𝜁) d 𝜁 .

The integral is taken along any path in $𝐷 (𝑝, 𝜀)$ from $𝑝_{*}$ to $𝑧$ . By simple connectivity of the disk, it is path-independent, so $\tilde{ℎ}$ is holomorphic there. Moreover,

ℜ𝔢 \tilde{ℎ} (𝑧) = 𝑢 (𝑝_{*}) + \int_{𝑝_{*}}^{𝑧} 𝑢_{𝑥}^{'} d 𝑥 + 𝑢_{𝑦}^{'} d 𝑦 = 𝑢 (𝑝_{*}) + \int_{𝑝^{+}}^{𝑧} d 𝑢 = 𝑢 (𝑧) .

Then for any $𝑧 \in 𝐷 (𝑝, 𝜀)$ ,

\begin{aligned} ℎ (𝑧) - \tilde{ℎ} (𝑧) & = 𝑢 (𝑧_{*}) + \int_{𝑧_{*}}^{𝑧} 𝑔 (𝜁) d 𝜁 - (𝑢 (𝑧_{*}) + \int_{𝑧_{*}}^{𝑝_{*}} 𝑔 (𝜁) d 𝜁 + \int_{𝑧_{*}}^{𝑧} 𝑔 (𝜁) d 𝜁) \\ = (\int_{𝑧_{*}}^{𝑧} + \int_{𝑝_{*}}^{𝑧_{*}} + \int_{𝑧}^{𝑧_{*}}) 𝑔 (𝜁) d 𝜁, \end{aligned}

where in the first integral, a curve is chosen in $𝑈^{*}$ ; in the second integral, a curve is chosen in $𝐷 (𝑝, 𝜀)$ ; in the third integral, a curve is chosen in $𝑈^{*}$ . The union, a closed curve, lies in $𝑈$ . Therefore the integral is equivalent to an integer multiple of $\int_{𝛾} 𝑔 (𝜁) d 𝜁 = \pm i 𝑇$ , where $𝛾$ is a Jordan curve in $𝑈$ . Thus $ℎ - \tilde{ℎ} (𝑧) \in i 𝑇 ℤ$ . Observe that $exp (𝛼 (1 - \tilde{ℎ} (𝑧)))$ is holomorphic on $𝐷 (𝑝, 𝜀)$ , and

\frac{𝐹 (𝑧)}{exp (𝛼 (1 - \tilde{ℎ} (𝑧)))} = exp (𝛼 (1 - ℎ (𝑧) + \tilde{ℎ} - 1)) = exp (\frac{2 π}{𝑇} (i 𝑘 𝑇)) = 1,

so $𝐹$ is identically equivalent to a holomorphic function on a neighborhood of every point of $𝐿$ , and thus we may assume $𝐹$ is holomorphic on $𝑈$ . On $𝑈^{*}$ , $ℜ𝔢 ℎ \equiv 𝑢$ , so $| 𝐹 | \equiv exp [𝛼 (1 - 𝑢)]$ . By continuity, this identity holds throughout $𝑈$ . Therefore $1 < | 𝐹 | < 𝑅$ , and

| 𝐹 (𝑧) | \to 𝑅 as 𝑧 \to Γ_{0}, and | 𝐹 (𝑧) | \to 1 as 𝑧 \to Γ_{1} .

It follows that

𝐹 : 𝑈 \to {𝑤 \in ℂ : 1 < | 𝑤 | < 𝑅} .

For any nonempty compact $𝐾 ⋐ {𝑤 \in ℂ : 1 < | 𝑤 | < 𝑅}$ , the preimage $𝐹^{- 1} (𝐾)$ is closed in $𝑈$ by continuity. Since $𝐾$ is compact, the numbers

𝑎 = min_{𝑤 \in 𝐾} | 𝑤 | and 𝑏 = max_{𝑤 \in 𝐾} | 𝑤 |

satisfy $1 < 𝑎 \leq 𝑏 < 𝑅$ . Since $| 𝐹 (𝑧) | \to 𝑅$ as $𝑧 \to Γ_{0}$ , there is a neighborhood $𝑁_{0}$ of $Γ_{0}$ in $\overline{𝑈}$ such that $| 𝐹 (𝑧) | > 𝑏$ for every $𝑧 \in 𝑁_{0} \cap 𝑈$ . Similarly, since $| 𝐹 (𝑧) | \to 1$ as $𝑧 \to Γ_{1}$ , there is a neighborhood $𝑁_{1}$ of $Γ_{1}$ in $\overline{𝑈}$ such that $| 𝐹 (𝑧) | < 𝑎$ for every $𝑧 \in 𝑁_{1} \cap 𝑈$ . Consequently,

𝐹^{- 1} (𝐾) \subseteq \overline{𝑈} ∖ (𝑁_{0} \cup 𝑁_{1}), \overline{𝐹^{- 1} (𝐾)} ⋐ 𝑈

Because $𝐹^{- 1} (𝐾)$ is relatively closed in $𝑈$ and its closure in $ℂ$ lies therein, it is compact in $𝑈$ by boundedness. Since $𝑈$ is bounded, $\overline{𝑈}$ is compact, so $𝐹^{- 1} (𝐾)$ is compact. Hence $𝐹$ is proper.

Fix $𝑤^{'}$ with $1 < | 𝑤^{'} | < 𝑅$ . Then

{𝑤 \in ℂ : \frac{1 + | 𝑤^{'} |}{2} \leq | 𝑤 | \leq \frac{𝑅 + | 𝑤^{'} |}{2}} ⋐ {𝑤 \in ℂ : 1 < | 𝑤 | < 𝑅},

and thus by properness,

𝐾 = {𝑧 \in 𝑈 : \frac{1 + | 𝑤^{'} |}{2} \leq | 𝐹 (𝑧) | \leq \frac{𝑅 + | 𝑤^{'} |}{2}} ⋐ 𝑈

is compact. Moreover, all solutions to $𝐹 (𝑧) = 𝑤$ lie in $𝐾$ . Because

{𝑧 \in ℂ : 𝐹 (𝑧) = 𝑤} \subseteq 𝐹^{- 1} ({𝑤 \in ℂ : \frac{1 + | 𝑤^{'} |}{2} < | 𝑤 | < \frac{𝑅 + | 𝑤^{'} |}{2}}),

it is also true that all solutions to $𝐹 (𝑧) = 𝑤$ lie in $\overset{⚬}{𝐾}$ .

By Lemma 7.5.3 $Lemma 7.5.3 Lemma 7.5.3$ , choose an exhaustion ${𝐷_{𝑛}}_{𝑛 \in ℕ}$ of $𝑈$ by doubly connected grid regions whose two boundary components are disjoint polygonal Jordan curves. Since $𝐾 ⋐ 𝑈$ is compact, choose $𝑛^{'}$ such that $𝐾 ⋐ 𝐷_{𝑛^{'}}$ , and set $\tilde{𝐾} = \overline{𝐷_{𝑛^{'}}}$ . Write

𝜕 \tilde{𝐾} = Γ_{0}^{'} \cup Γ_{1}^{'},

where $Γ_{0}^{'}$ is the outer boundary curve and $Γ_{1}^{'}$ is the inner boundary curve corresponding respectively to $Γ_{0}$ and $Γ_{1}$ . Both curves are piecewise $𝐶^{\infty}$ .

Since the connected component $𝑈 \cap ext (Γ_{0}^{'})$ must map into exactly one of

{𝑤 : | 𝑤 | \geq \frac{𝑅 + | 𝑤^{'} |}{2}} and {𝑤 : | 𝑤 | \leq \frac{1 + | 𝑤^{'} |}{2}}

by continuity and the Intermediate Value Theorem, then and since $| 𝐹 (𝑧) | \to 𝑅$ as $𝑧 \to Γ_{0}$ , it follows that for $𝑧 \in Γ_{0}^{'}$

| 𝐹 (𝑧) | > | 𝑤^{'} | .

Then on $Γ_{0}^{'}$ , $| \frac{𝑤^{'}}{𝐹 (𝑧)} | < 1$ and $ℜ𝔢 [1 - \frac{𝑤^{'}}{𝐹 (𝑧)}] > 0$ . Moreover, this inequality will continue to hold on a neighborhood $𝑉_{0} ⋑ Γ$ by continuity. Since $Log$ is holomorphic on the right half-plane, we have that

\begin{aligned} 𝐹 (𝑧) - 𝑤^{'} = 𝐹 (𝑧) (1 - \frac{𝑤^{'}}{𝐹 (𝑧)}) \\ \begin{matrix}  \end{matrix} ⟹ \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} - \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧)} = \frac{d}{d 𝑧} Log (1 - \frac{𝑤^{'}}{𝐹 (𝑧)}) = 𝜂_{0}^{'} (𝑧), \end{aligned}

the difference between the two logarithmic derivatives, is the derivative of some holomorphic function $𝜂_{0}$ in $𝑉_{0} ⋑ Γ_{0}^{'}$ .

By similar logic, for $𝑧$ in some open neighborhood $𝑉_{1}$ of $Γ_{1}^{'}$ , we must have

| 𝐹 (𝑧) | < | 𝑤^{'} | .

Therefore, $ℜ𝔢 (1 - \frac{𝐹 (𝑧)}{𝑤^{'}}) > 0$ , therefore implying that $Log (1 - \frac{𝐹 (𝑧)}{𝑤^{'}})$ is holomorphic, and

\frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} = \frac{d}{d 𝑧} Log (1 - \frac{𝐹 (𝑧)}{𝑤^{'}}) = 𝜂_{1}^{'} (𝑧)

is the derivative of some holomorphic function $𝜂_{1}$ in $𝑉_{1} ⋑ Γ_{1}^{'}$ .

Then the region $\overset{⚬}{\tilde{𝐾}}$ may be split into two simply connected polygons with positively oriented Jordan boundaries $𝐿_{1}, 𝐿_{2}$ . Without loss of generality we may assume the solutions to $𝐹 (𝑧) = 𝑤^{'}$ do not lie on their shared boundary. The number of solutions to $𝐹 (𝑧) = 𝑤^{'}$ in $\tilde{𝐾}$ , $𝑛_{\tilde{𝐾}} (𝐹 - 𝑤^{'})$ is the sum of the number of solutions in $int (𝐿_{1})$ and $int (𝐿_{2})$ , and by the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ), this is equal to

\begin{aligned} 𝑛_{\tilde{𝐾}} (𝐹 - 𝑤^{'}) & = \frac{1}{2 π i} \oint_{𝐿_{1}} \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} d 𝑧 + \frac{1}{2 π i} \oint_{𝐿_{2}} \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} d 𝑧 \\ = \frac{1}{2 π i} \oint_{𝜕 \tilde{𝐾}} \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} d 𝑧 \\ = \frac{1}{2 π i} (∳_{Γ_{0}^{'}} + ∲_{Γ_{1}^{'}}) \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} d 𝑧 . \end{aligned}

Since

∳_{Γ_{0}^{'}} [\frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} - \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧)}] d 𝑧 = ∳_{Γ_{0}^{'}} 𝜂_{0}^{'} (𝑧) d 𝑧 = 0

and

∲_{Γ_{1}^{'}} \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧) - 𝑤^{'}} d 𝑧 = ∲_{Γ_{1}^{'}} 𝜂_{1}^{'} (𝑧) d 𝑧 = 0,

it follows that

\begin{aligned} 𝑛_{\tilde{𝐾}} (𝐹 - 𝑤^{'}) & = \frac{1}{2 π i} ∳_{Γ_{0}^{'}} \frac{𝐹^{'} (𝑧)}{𝐹 (𝑧)} d 𝑧 \\ = \frac{1}{2 π i} ∳_{Γ_{0}^{'}} \frac{d}{d 𝑧} [𝛼 (1 - ℎ (𝑧))] d 𝑧 \\ = - \frac{𝛼}{2 π i} ∳_{Γ_{0}^{'}} 𝑔 (𝑧) d 𝑧 = 1 . \end{aligned}

Since all zeros of $𝐹 - 𝑤^{'}$ in $𝑈$ lie in $\tilde{𝐾}$ , $𝐹$ attains every value $𝑤^{'} \in {𝑤 \in ℂ : 1 < | 𝑤 | < 𝑅}$ exactly once, thence implying bijectivity and consequently conformality.

Finally, suppose that $𝑈$ were biholomorphic to both ${1 < | 𝑤 | < 𝑅_{1}}$ and ${1 < | 𝑤 | < 𝑅_{2}}$ . Then Theorem 7.5.2 $Theorem 7.5.2 Theorem 7.5.2$ gives $𝑅_{1} = 𝑅_{2}$ , proving uniqueness. ⁠ $□$

Remark.

Note that the theorem proof relied principally on the smoothness and simplicity of the curves. One can also generalize these results by using exhaustion with locally uniform limits.

Moreover, one can further obtain:

If both components of $\hat{ℂ} ∖ 𝑈$ are nondegenerate, then $𝑈$ is biholomorphic to a unique annulus ${1 < | 𝑧 | < 𝑅}$ with $𝑅 > 1$ . (In this case one would use an exhaustion method as we did in the proof of the simply connected Riemann Mapping Theorem within this section)
If exactly one complementary component is a single point, then $𝑈$ is biholomorphic to the punctured disk $𝔻^{*}$ ; equivalently, after inversion, this essentially gives $𝑅 = \infty$ .
Proof.
Letting the relevant point be $𝑧_{0}$ , we find that $𝑈 \cup {𝑧_{0}}$ is a simply connected nonempty region $𝑉 ⊊ ℂ$ . Then by the Riemann Mapping Theorem (Theorem 5.3.2 $Theorem 5.3.2 Theorem 5.3.2$ ) provides a biholomorphism $𝜑$ between $𝑉$ and $𝔻$ such that $𝜑 (𝑧_{0}) = 0$ , then restricting $𝜑$ to $𝑈$ completes this assertion. ⁠ $□$
If both complementary components are single points (one of which being $\infty$ ), then $𝑈$ is biholomorphic to $ℂ^{*}$ .
Proof.
The region is in the form of $ℂ ∖ {𝑧_{0}}$ . Then simply take $𝜑 : 𝑧 \to 𝑧 - 𝑧_{0}$ to be the biholomorphism. ⁠ $□$

For a region $𝑈$ satisfying the first case, define $mod 𝑈 = mod 𝐴 = \frac{1}{2 π} log 𝑅$ , where $𝐴$ is the unique normalized annular region with inner radius $1$ and outer radius $𝑅$ such that a biholomorphism exists mapping $𝐴$ to $𝑈$ . Thus, this ratio is not only a classification of annuli but also nondegenerate doubly connected regions.