§5.4 The Schwarz–Christoffel Transformation

5.4 The Schwarz–Christoffel Transformation

The Riemann Mapping Theorem is elegant in its own simplicity and definitions. However, it is only a theorem that guarantees existence of biholomorphisms. No information whatsoever can be straightforwardly extracted regarding the explicit construction of such biholomorphisms. However, in the explicit case that $𝑈$ is the open interior of a polygon, the result is provided by the Schwarz–Christoffel Transformation.

Let $𝑎_{1} < 𝑎_{2} < \dots < 𝑎_{𝑛}$ be $𝑛 \in ℕ$ distinct real numbers and let $𝑎_{0} = - \infty$ and $𝑎_{𝑛 + 1} = \infty$ . Suppose $𝛼_{1}, 𝛼_{2}, \dots, 𝛼_{𝑛}$ are $𝑛$ positive real numbers satisfying $\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} < 𝑛 - 1$ . Let

𝛽 (𝜁) = {(𝜁 - 𝑎_{1})}^{𝛼_{1} - 1} \dots {(𝜁 - 𝑎_{𝑛})}^{𝛼_{𝑛} - 1} = \prod_{𝑘 = 1}^{𝑛} {(𝜁 - 𝑎_{𝑘})}^{𝛼_{𝑘} - 1},

where the branch of each factor is selected to be

{(𝜁 - 𝑎_{𝑘})}^{𝛼_{𝑘} - 1} = e^{(𝛼_{𝑘} - 1) (log (𝜁 - 𝑎_{𝑘}))},

where the branch of $log$ is selected such that $- \frac{π}{2} < ℑ𝔪 (log) \leq \frac{3 π}{2}$ , holomorphic on $ℂ ∖ i ℝ_{\leq 0}$ (the lower imaginary axis is known as a branch cut). For $𝜁 < 𝑎_{𝑘}$ , the argument of this factor is $π (𝛼_{𝑘} - 1)$ . For $𝜁 < 𝑎_{1}$ ,

arg (𝛽 (𝜁)) = π (- 𝑛 + \sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘}),

achieved by selecting branches of each factor by the method described earlier.

Let $𝑘$ be fixed. If $𝜁 \in (𝑎_{𝑘 - 1}, 𝑎_{𝑘})$ , the branches of all ${(𝜁 - 𝑎_{𝑗})}^{𝛼_{𝑗} - 1}$ where $1 \leq 𝑗 \leq 𝑘 - 1$ have vanishing arguments; hence,

arg (𝛽 (𝜁)) = π (- 𝑛 + 𝑘 - 1 + \sum_{𝑗 = 𝑘}^{𝑛} 𝛼_{𝑗}) .

If $𝜁 > 𝑎_{𝑛}$ , we have

arg (𝛽 (𝜁)) = 0 .

Then, we define $𝑛 + 2$ complex numbers via

𝑤_{𝑘} = \int_{0}^{𝑎_{𝑘}} 𝛽 (𝜁) d 𝜁, (5.4 . 1)

where we integrate over the real line as an improper Riemann integral.

The absolute integrability of $𝛽 (𝜁)$ along the real axis concerns only the convergence at each singularity $𝜁 = 𝑎_{𝑘}$ and the behavior as $𝜁 \to \pm \infty$ . For each fixed $𝑘$ ,

𝛽 (𝜁) = ℎ_{𝑘} (𝜁) {(𝜁 - 𝑎_{𝑘})}^{𝛼_{𝑘} - 1} for ℎ_{𝑘} = \prod_{\begin{matrix} 𝑗 = 1 \\ 𝑗 \neq 𝑘 \end{matrix}}^{𝑛} {(𝜁 - 𝑎_{𝑗})}^{𝛼_{𝑗} - 1} = 𝒪︀ (1)

as $𝜁 \to 𝑎_{𝑘}$ (where $ℎ_{𝑘}$ is holomorphic and nonzero in a compact neighborhood of $𝑎_{𝑘}$ ). Since $𝛼_{𝑘} - 1 > - 1$ ,

| 𝛽 (𝜁) | = 𝒪︀ ({| 𝜁 - 𝑎_{𝑘} |}^{𝛼_{𝑘} - 1}), (5.4 . 2)

and therefore $𝜁 = 𝑎_{𝑘}$ is an integrable singularity. Since $𝛽 (𝜁) \sim 𝜁^{\sum 𝛼_{𝑘} - 𝑛}$ as $𝜁 \to \pm \infty$ and $\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} - 𝑛 < - 1$ , $𝛽$ is integrable on $ℝ$ .

Let

𝑓 (𝑧) = \int_{0}^{𝑧} 𝛽 (𝜁) d 𝜁 . (5.4 . 3)

We first aim to show that this integral is path-independent for $𝑧 \in \overline{ℍ^{+}}$ . This would be a direct consequence of Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), if it weren’t for the singularities along the real line. We invoke a method used in Section 4.5 $Section 4.5 Section 4.5$ , namely the indentation of contours; Let $Γ \subset \overline{ℍ^{+}}$ be an arbitrary piecewise $𝐶^{1}$ Jordan curve. There exists a $𝜀^{'} > 0$ such that for all $0 < 𝜀 < 𝜀^{'}$ , each $\overline{𝐷 (𝑎_{𝑘}, 𝜀)}$ is disjoint. Then the indented contour

\tilde{Γ} = 𝜕 (int Γ ∖ ⋃_{𝑘 = 1}^{𝑛} \overline{𝐷 (𝑎_{𝑘}, 𝜀)})

is a Jordan curve in $\overline{ℍ^{+}}$ , and can be decomposed as $\tilde{Γ} = 𝛾 \cup 𝐶_{𝜀}$ where $𝛾 \subseteq Γ$ and $𝐶_{𝜀} \subseteq ⋃_{𝑘 = 1}^{𝑛} 𝜕 𝐷 (𝑎_{𝑘}, 𝜀)$ . Then by Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ )

\oint_{\tilde{Γ}} 𝛽 (𝜁) d 𝜁 = \int_{𝛾} 𝛽 (𝜁) d 𝜁 + \int_{𝐶_{𝜀}} 𝛽 (𝜁) d 𝜁 = 0 .

As $𝜀 \to 0^{+}$ , we have that

\oint_{Γ} 𝛽 (𝜁) d 𝜁 = lim_{𝜀 \to 0^{+}} \int_{𝛾} 𝛽 (𝜁) d 𝜁 = - lim_{𝜀 \to 0^{+}} \int_{𝐶_{𝜀}} 𝛽 (𝜁) d 𝜁,

where

\begin{aligned} | \int_{𝐶_{𝜀}} 𝛽 (𝜁) d 𝜁 | & \leq \sum_{𝑘 = 1}^{𝑛} \oint_{𝜕 𝐷 (𝑎_{𝑘}, 𝜀)} | 𝛽 (𝜁) | | d 𝜁 | \leq \sum_{𝑘 = 1}^{𝑛} \int_{0}^{2 π} | 𝛽 (𝑎_{𝑘} + 𝜀 e^{i 𝜃}) | 𝜀 d 𝜃 \\ = 𝑛 \int_{0}^{2 π} 𝒪︀ (𝜀^{𝛼_{𝑘}}) d 𝜃 = 𝒪︀ (𝜀^{𝛼_{𝑘}}) \to 0 . (by (5.4.2) (5.4.2) (5.4.2)) \end{aligned}

Hence, $\oint_{Γ} 𝛽 (𝜁) d 𝜁 = 0$ for any piecewise $𝐶^{1}$ Jordan curve $Γ \subset \overline{ℍ^{+}}$ , and the integral in (5.4.3) $(5.4.3) (5.4.3)$ is path-independent for $𝑧 \in \overline{ℍ^{+}}$ .

The holomorphy of $𝑓$ in $ℍ^{+}$ is equally evident: choosing the polygonal chain $0 \to i \to 𝑧$ and applying $\frac{𝜕}{𝜕 \overline{𝑧}}$ gives

\begin{aligned} \frac{𝜕 𝑓}{𝜕 \overline{𝑧}} & = \frac{𝜕}{𝜕 \overline{𝑧}} \int_{i}^{𝑧} 𝛽 (𝜁) d 𝜁 = \frac{𝜕}{𝜕 \overline{𝑧}} (𝑧 - i) \int_{0}^{1} 𝛽 (i + (𝑧 - i) 𝑡) d 𝑡 \\ = (𝑧 - i) \int_{0}^{1} \frac{𝜕}{𝜕 \overline{𝑧}} 𝛽 (i + (𝑧 - i) 𝑡) d 𝑡 = 0, \end{aligned}

since $𝛽 (𝜁)$ is bounded on compact subsets of $ℍ^{+}$ (this is the reason for which we choose $0 \to i \to 𝑧$ , as $𝛽$ may be unbounded near $0$ ). Thus $𝑓$ is holomorphic on $ℍ^{+}$ ; it is also trivial that $𝑓$ is continuous on $\overline{ℍ^{+}}$ .

For $𝑥 \in ℝ$ where $𝑥 \in (𝑎_{𝑘 - 1}, 𝑎_{𝑘})$ (where $1 \leq 𝑘 \leq 𝑛 + 1$ ) and $𝑎_{0} = - \infty$ and $𝑎_{𝑛 + 1} = \infty$ ,

\begin{aligned} 𝑓 (𝑥) & = \int_{0}^{𝑥} 𝛽 (𝑡) d 𝑡 = 𝑤_{𝑘 - 1} + \int_{𝑎_{𝑘 - 1}}^{𝑥} 𝛽 (𝑡) d 𝑡 \\ = 𝑤_{𝑘 - 1} + \int_{𝑎_{𝑘 - 1}}^{𝑥} | 𝛽 (𝑡) | 𝑒^{i arg (𝛽 (𝑡))} d 𝑡 \\ = 𝑤_{𝑘 - 1} + exp [i π (- 𝑛 + 𝑘 - 1 + \sum_{𝑗 = 𝑘}^{𝑛} 𝛼_{𝑗})] \int_{𝑎_{𝑘 - 1}}^{𝑥} | 𝛽 (𝑡) | d 𝑡, \end{aligned}

where each $𝑤_{𝑘 - 1}$ is defined as in (5.4.1) $(5.4.1) (5.4.1)$ (and observe that $𝑤_{𝑘 - 1} = 𝑓 (𝑎_{𝑘 - 1})$ ). Therefore,

arg [𝑓 (𝑥) - 𝑤_{𝑘 - 1}] = π (- 𝑛 + 𝑘 - 1 + \sum_{𝑗 = 𝑘}^{𝑛} 𝛼_{𝑗}) (5.4 . 4)

and

| 𝑓 (𝑥) - 𝑤_{𝑘 - 1} | = \int_{𝑎_{𝑘 - 1}}^{𝑥} | 𝛽 (𝑡) | d 𝑡 (5.4 . 5)

for $𝑥 \in (𝑎_{𝑘 - 1}, 𝑎_{𝑘})$ . Define

𝑙_{𝑘} = \int_{𝑎_{𝑘 - 1}}^{𝑎_{𝑘}} | 𝛽 (𝑡) | d 𝑡 and Δ_{𝑘} = 𝑓 ([𝑎_{𝑘 - 1}, 𝑎_{𝑘}]) (5.4 . 6)

for $1 \leq 𝑘 \leq 𝑛 + 1$ . Note that $Δ_{𝑘}$ is a curve from $𝑤_{𝑘 - 1}$ to $𝑤_{𝑘}$ . By (5.4.4) $(5.4.4) (5.4.4)$ , since $arg [𝑓 (𝑥) - 𝑤_{𝑘 - 1}]$ is constant for $𝑎_{𝑘 - 1} < 𝑥 < 𝑎_{𝑘}$ , $Δ_{𝑘}$ is a straight line segment. Furthermore, by (5.4.5) $(5.4.5) (5.4.5)$ with $𝑥 = 𝑎_{𝑘}$ , $𝑙_{𝑘}$ measures the length of said segment and its slope angle is given by

𝜃_{𝑘} = arg [𝑤_{𝑘} - 𝑤_{𝑘 - 1}] = π (- 𝑛 + 𝑘 - 1 + \sum_{𝑗 = 𝑘}^{𝑛} 𝛼_{𝑗}) .

Therefore, $𝑓$ maps the real axis $ℝ$ to a polygonal chain with vertices at $𝑤_{0}, 𝑤_{1}, \dots, 𝑤_{𝑛}, 𝑤_{𝑛 + 1}$ . We now aim to prove that $𝑤_{0} = 𝑤_{𝑛 + 1}$ , which is equivalent to showing

𝑤_{𝑛 + 1} - 𝑤_{0} = \int_{- \infty}^{\infty} 𝛽 (𝑡) d 𝑡 = 0 .

Let $𝑅^{'} > {max}_{𝑘 \in ℕ_{\leq 𝑛}} | 𝑎_{𝑘} |$ be arbitrary; then $\exists 𝜀^{'} > 0$ such that

⋃_{𝑘 \in ℕ_{\leq 𝑛}} \overline{𝐷 (𝑎_{𝑘}, 𝜀^{'})} \subseteq 𝐷 (0, 𝑅^{'}) \cap ℍ^{+}

and each $\overline{𝐷 (𝑎_{𝑘}, 𝜀^{'})}$ are disjoint. Moreover, $\forall 𝑅 > 𝑅^{'}$ , $\forall 𝜀 \in (0, 𝜀^{'})$ ,

⋃_{𝑘 \in ℕ_{\leq 𝑛}} \overline{𝐷 (𝑎_{𝑘}, 𝜀)} \subseteq 𝐷 (0, 𝑅) \cap ℍ^{+} .

Therefore, we may define a closed (indented semicircle) contour (oriented counterclockwise)

\begin{aligned} Γ & = {𝑅 e^{i 𝜃} : 0 \leq 𝜃 \leq π} \\ \cup ([- 𝑅, 𝑅] ∖ ⋃_{𝑘 \in ℕ_{\leq 𝑛}} (𝑎_{𝑘} - 𝜀, 𝑎_{𝑘} + 𝜀)) \\ \cup ⋃_{𝑘 \in ℕ_{\leq 𝑛}} (𝜕 𝐷 (𝑎_{𝑘}, 𝜀) \cap \overline{ℍ^{+}}) \\ = 𝐶_{𝑅} \cup 𝐼_{𝑅} \cup 𝐶_{𝜀}, \end{aligned}

such that $𝛽$ is continuous on $\overline{int Γ}$ and holomorphic on $int Γ$ . By Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ),

∳_{Γ} 𝛽 (𝜁) d 𝜁 = 0,

and moreover, taking $𝑅 \to \infty$ and $𝜀 \to 0^{+}$ gives

\int_{- \infty}^{\infty} 𝛽 (𝑡) d 𝑡 = - lim_{𝑅 \to \infty} \int_{𝐶_{𝑅}} 𝛽 (𝜁) d 𝜁 - lim_{𝜀 \to 0^{+}} \int_{𝐶_{𝜀}} 𝛽 (𝜁) d 𝜁 .

By (5.4.2) $(5.4.2) (5.4.2)$ , we know that integrals over small indentations vanish; therefore

\begin{aligned} | 𝑤_{𝑛 + 1} - 𝑤_{0} | & = lim_{𝑅 \to \infty} | \int_{𝐶_{𝑅}} 𝛽 (𝜁) d 𝜁 | \leq lim_{𝑅 \to \infty} 𝑅 \int_{0}^{π} | \prod_{𝑘 = 1}^{𝑛} {(𝑅 e^{i 𝜃} - 𝑎_{𝑘})}^{𝛼_{𝑘} - 1} | d 𝜃 \\ = lim_{𝑅 \to \infty} 𝑅 \int_{0}^{π} \prod_{𝑘 = 1}^{𝑛} {| 𝑅 e^{i 𝜃} - 𝑎_{𝑘} |}^{𝛼_{𝑘} - 1} d 𝜃 \\ \leq lim_{𝑅 \to \infty} 𝑅 \int_{0}^{π} \prod_{𝑘 = 1}^{𝑛} max {{(𝑅 + | 𝑎_{𝑘} |)}^{𝛼_{𝑘} - 1}, {| 𝑅 - | 𝑎_{𝑘} | |}^{𝛼_{𝑘} - 1}} d 𝜃 \\ \leq lim_{𝑅 \to \infty} 𝑅 π {𝒪︀ (𝑅)}^{- 𝑛 + \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑗}} = lim_{𝑅 \to \infty} 𝒪︀ (𝑅^{1 - 𝑛 + \sum_{𝑗 = 1}^{𝑛} 𝛼_{𝑗}}) = 0, \end{aligned}

since $\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} < 𝑛 - 1$ (by assumption). Here, we developed the upper bound $max {{(𝑅 + | 𝑎_{𝑘} |)}^{𝛼_{𝑘} - 1}, {| 𝑅 - | 𝑎_{𝑘} | |}^{𝛼_{𝑘} - 1}} d 𝜃$ , as the first part accounts solely for $𝛼_{𝑘} - 1 \geq 0$ and the second for $𝛼_{𝑘} - 1 \leq 0$ (although here it made little difference).

Hence, $𝑤_{0} = 𝑤_{𝑛 + 1}$ and $𝑓$ maps $\hat{ℝ} = ℝ \cup {\infty}$ to a closed polygonal chain with $𝑛 + 1$ sides ( $Δ_{𝑘}$ ). For $𝑘 \in ℕ_{\leq 𝑛}$ , at the vertex $𝑤_{𝑘}$ connecting $Δ_{𝑘}$ and $Δ_{𝑘 + 1}$ , the tangent angle changes from $𝜃_{𝑘}$ to $𝜃_{𝑘 + 1}$ , with an exterior angle of $𝜃_{𝑘 + 1} - 𝜃_{𝑘} = π - π 𝛼_{𝑘}$ . Then at this vertex, the polygon has an interior angle of $π 𝛼_{𝑘}$ .

At the vertex $𝑤_{0} = 𝑤_{𝑛 + 1}$ , the polygon has an interior angle of $π (𝑛 - 1) - \sum_{𝑘 = 1}^{𝑛} π 𝛼_{𝑘} = π (𝑛 - 1 - \sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘}) > 0$ , since the interior angles of the polygon add to $π (𝑛 - 1)$ .

If the values ${𝛼_{𝑘}}$ are chosen such that $\sum_{𝑘 = 1}^{𝑛} π 𝛼_{𝑘} = 𝑛 - 2$ , then the interior angle at vertex $𝑤_{0} = 𝑤_{𝑛 + 1}$ can be considered to be $π$ ; the vertex then is simply a straight line and we have a $𝑛$ -gon instead.

Now assume that the polygonal chain is simple (therefore, we concern ourselves only when there is a polygonal interior).

It remains to show that $𝑓$ maps $ℍ^{+}$ biholomorphically onto the polygon interior. First, the continuity of $𝑓$ at infinity must be established (we have established that the limits exist from only the real line, which is insufficient to prove that a $𝛿$ -infimum exists in every direction). As $| 𝑧 | \to \infty$ (uniformly in all directions),

𝛽 (𝑧) = 𝒪︀ ({| 𝑧 |}^{\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} - 𝑛}) . (5.4 . 7)

Let $𝑧 = 𝑟 e^{i 𝜃} \in \overline{ℍ^{+}}$ , where $0 \leq 𝜃 \leq π$ . Join $𝑧 \in \overline{ℍ^{+}}$ to $𝑟 \in ℝ$ along the circular arc of radius $𝑟$ , and then join $𝑟$ to infinity along the positive real axis. By (5.4.7) $(5.4.7) (5.4.7)$ ,

\begin{aligned} | 𝑤_{𝑛 + 1} - 𝑓 (𝑧) | & \leq | 𝑓 (𝑟) - 𝑓 (𝑧) | + | 𝑤_{𝑛 + 1} - 𝑓 (𝑟) | \\ \leq \int_{0}^{𝜃} | 𝛽 (𝑟 e^{i 𝑡}) | 𝑟 d 𝑡 + \int_{𝑟}^{\infty} | 𝛽 (𝑡) | d 𝑡 \\ \leq 𝒪︀ (𝑟^{\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} - 𝑛 + 1}) + \int_{𝑟}^{\infty} 𝒪︀ (𝑡^{\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} - 𝑛}) d 𝑡 \\ = 𝒪︀ (𝑟^{\sum_{𝑘 = 1}^{𝑛} 𝛼_{𝑘} - 𝑛 + 1}) \to 0 as 𝑟 \to \infty . \end{aligned}

Thus, defining $𝑓 (\infty) = 𝑤_{𝑛 + 1} = 𝑤_{0}$ gives a continuous extension of $𝑓$ to $\overline{ℍ^{+}} \cup {\infty}$ .

Let $𝜓 (𝑧) = \frac{𝑧 - i}{𝑧 + i}$ be the Cayley transform from $ℍ^{+}$ onto $𝔻$ , extended by $𝜓 (\infty) = 1$ , and define

𝑔 = 𝑓 \circ 𝜓^{- 1} .

The preceding continuity at infinity shows that $𝑔 \in 𝐶 (\overline{𝔻})$ , while $𝑔$ is holomorphic on $𝔻$ . Furthermore, $𝑔 |_{𝜕 𝔻}$ maps $𝜕 𝔻$ injectively onto the polygonal boundary $⋃_{𝑘} Δ_{𝑘}$ . Therefore, by Theorem 5.1.5 $Theorem 5.1.5 Theorem 5.1.5$ , $𝑔$ maps $𝔻$ univalently and surjectively onto $int ⋃_{𝑘} Δ_{𝑘}$ . Since $𝜓$ is a biholomorphism, it follows that $𝑓$ maps $ℍ^{+}$ biholomorphically onto the polygon.

The definition in (5.4.3) $(5.4.3) (5.4.3)$ is known as the Schwarz–Christoffel Transformation, which maps the upper-half plane to a polygon with specified angles. The side lengths can be controlled by numerically computing suitable values for ${𝑎_{𝑘}}$ . Note that for any $𝑐 \in ℂ$ , $𝑐^{'} \in ℂ^{*}$ , $𝑧 \mapsto 𝑐 + 𝑐^{'} 𝑓 (𝑧)$ also suffices as such a transformation.

Example 5.4.1.

Let

𝑄 = {𝑧 : max {| ℜ𝔢 𝑧 |, | ℑ𝔪 𝑧 |} < 1}

be a square centered at the origin with side length

2

. Find a biholomorphism between

𝔻

and

𝑄

Figure 5.1: The biholomorphism between

𝔻

and

𝑄

via

𝐹

Solution to Example 5.4.1 $Example 5.4.1 Example 5.4.1$ .

By the Cayley transform, it suffices to construct a biholomorphism from $ℍ^{+}$ to $𝑄$ .

Since the interior angles of a square are $\frac{π}{2}$ and (5.4.3) $(5.4.3) (5.4.3)$ maps to an $𝑛 + 1$ -gon, we define the three exponents

𝛼_{𝑘} = \frac{1}{2} for 𝑘 = 1, 2, 3 .

Then a biholomorphism is given by

𝑓_{𝑎_{1}, 𝑎_{2}, 𝑎_{3}} (𝑧) = \int_{0}^{𝑧} 𝛽 (𝜁) d 𝜁 = \int_{0}^{𝑧} \frac{d 𝜁}{\sqrt{𝜁 - 𝑎_{1}} \sqrt{𝜁 - 𝑎_{2}} \sqrt{𝜁 - 𝑎_{3}}},

where $𝑎_{1} < 𝑎_{2} < 𝑎_{3}, (\pm \infty)$ are finite prevertices on the real line. Observe that by the restrictions set upon $𝛼_{𝑘}$ , any such configuration of $𝑎_{1}, 𝑎_{2}, 𝑎_{3}$ will give a rectangle. Therefore, we want to find a configuration of prevertices that ensures equal side lengths. From the rectangularity of the boundary, it is already known that

𝑙_{1} = 𝑙_{3} and 𝑙_{2} = 𝑙_{4}, (5.4 . 8)

where

\begin{matrix} 𝑙_{1} = \int_{- \infty}^{𝑎_{1}} | 𝛽 (𝑡) | d 𝑡, 𝑙_{2} = \int_{𝑎_{1}}^{𝑎_{2}} | 𝛽 (𝑡) | d 𝑡, \\ 𝑙_{3} = \int_{𝑎_{2}}^{𝑎_{3}} | 𝛽 (𝑡) | d 𝑡, 𝑙_{4} = \int_{𝑎_{3}}^{𝑎_{4}} | 𝛽 (𝑡) | d 𝑡 \end{matrix}

correspond to the side lengths of the rectangle.

Fix $𝑎_{1}, 𝑎_{3}$ . We now aim to find $𝑎_{2}$ in terms of $𝑎_{1}$ and $𝑎_{3}$ such that $𝑙_{1} = 𝑙_{2}$ (which would then imply $𝑙_{1} = 𝑙_{2} = 𝑙_{3} = 𝑙_{4}$ from (5.4.8) $(5.4.8) (5.4.8)$ ). Define a linear affine map that sends $- 1 \to 𝑎_{1}$ , $1 \to 𝑎_{3}$ , namely $𝜑 (𝑧) = \frac{𝑧 (𝑎_{3} - 𝑎_{1})}{2} + \frac{𝑎_{3} + 𝑎_{1}}{2}$ . By a substitution,

\begin{aligned} 𝑙_{1} = \int_{- \infty}^{𝑎_{1}} | 𝛽 (𝑡) | d 𝑡 & = \frac{𝑎_{3} - 𝑎_{1}}{2} \int_{- \infty}^{- 1} \frac{d 𝑡}{\sqrt{| 𝜑 (𝑡) - 𝑎_{1} | | 𝜑 (𝑡) - 𝑎_{2} | | 𝜑 (𝑡) - 𝑎_{3} |}} \\ = \int_{- \infty}^{- 1} \frac{d 𝑡}{\sqrt{(𝑡^{2} - 1) | \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} |}}, \\ = \int_{1}^{\infty} \frac{d 𝑡}{\sqrt{(𝑡^{2} - 1) | \frac{𝑡 (𝑎_{1} - 𝑎_{3})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} |}}, (1) \\ 𝑙_{4} = \int_{𝑎_{3}}^{\infty} | 𝛽 (𝑡) | d 𝑡 & = \frac{𝑎_{3} - 𝑎_{1}}{2} \int_{1}^{\infty} \frac{d 𝑡}{\sqrt{| 𝜑 (𝑡) - 𝑎_{1} | | 𝜑 (𝑡) - 𝑎_{2} | | 𝜑 (𝑡) - 𝑎_{3} |}} \\ = \int_{1}^{\infty} \frac{d 𝑡}{\sqrt{(𝑡^{2} - 1) | \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} |}} . (2) \end{aligned}

Now force $𝑙_{1} = 𝑙_{4}$ .

Our aim is to show that $𝑎_{2}$ lies at the midpoint between $𝑎_{1}$ and $𝑎_{3}$ . For the sake of contradiction, assume $𝑎_{3} + 𝑎_{1} - 2 𝑎_{2} > 0$ . Then for all $𝑡 > 1$ ,

\frac{𝑡 (𝑎_{1} - 𝑎_{3})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} < 𝑎_{1} - 𝑎_{2} < 0,

implying that

| \frac{𝑡 (𝑎_{1} - 𝑎_{3})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} | = \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} - \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} < \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2},

while

| \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} | = \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} + \frac{𝑎_{3} + 𝑎_{1} - 2 𝑎_{2}}{2} > \frac{𝑡 (𝑎_{3} - 𝑎_{1})}{2} .

Then we obtain that (1) is strictly greater than (2). Similarly, $𝑎_{3} + 𝑎_{1} - 2 𝑎_{2} < 0$ forces $𝑙_{1} < 𝑙_{4}$ . Therefore, for this equality to hold, we must have $𝑎_{2} = \frac{𝑎_{1} + 𝑎_{3}}{2}$ .

Then the general biholomorphism between $ℍ^{+}$ and a square is in the form of

𝑐^{'} 𝑓_{𝑎_{1}, \frac{𝑎_{1} + 𝑎_{3}}{2}, 𝑎_{3}} (𝑧) + 𝑐 . (5.4 . 9)

Without loss of generality, assume $𝑎_{1} = - 1, 𝑎_{3} = 1, 𝑎_{2} = 0$ and write $𝑓 (𝑧) = 𝑓_{- 1, 0, 1} (𝑧)$ . Let

\begin{aligned} 𝑙 & = 𝑙_{1} = \dots = 𝑙_{4} = \int_{- 1}^{0} | 𝛽 (𝑡) | d 𝑡 = \int_{0}^{1} \frac{d 𝑡}{\sqrt{𝑡 (1 - 𝑡^{2})}} \\ = \int_{0}^{1} \frac{d 𝑡}{2 \sqrt{𝑡} \sqrt{\sqrt{𝑡} (1 - 𝑡)}} = \frac{1}{2} \int_{0}^{1} 𝑡^{- \frac{3}{4}} {(1 - 𝑡)}^{- \frac{1}{2}} d 𝑡 \\ = \frac{1}{2} Β (\frac{1}{4}, \frac{1}{2}) = \frac{Γ (\frac{1}{4}) Γ (\frac{1}{2})}{2 Γ (\frac{3}{4})} = \frac{{Γ (\frac{1}{4})}^{2}}{2 \sqrt{2 π}} \end{aligned}

by Definition 12.1.9 $Definition 12.1.9 Definition 12.1.9$ , Theorem 12.1.5 $Theorem 12.1.5 Theorem 12.1.5$ (the Gamma and Beta function identities). Traversing $ℝ$ from $- \infty$ to $+ \infty$ , by using (5.4.4) $(5.4.4) (5.4.4)$ ,

\int_{- \infty}^{- 1} 𝛽 = i 𝑙, \int_{- 1}^{0} 𝛽 = - 𝑙, \int_{0}^{1} 𝛽 = - i 𝑙, \int_{1}^{\infty} 𝛽 = 𝑙 .

Hence,

𝑓 (- 1) = 𝑙, 𝑤 (0) = 0, 𝑤 (1) = - i 𝑙, 𝑓 (\infty) = 𝑙 - i 𝑙 = 𝑓 (- \infty) .

The four distinct vertices $0, 𝑙, 𝑙 - i 𝑙, - i 𝑙$ bound

𝑆 = {𝑥 + i 𝑦 : 0 < 𝑥 < 𝑙, - 𝑙 < 𝑦 < 0},

a square of side $𝑙$ . Thus $𝑓 (ℍ^{+}) = 𝑆$ . The center of $𝑆$ is $\frac{𝑙}{2} - i \frac{𝑙}{2}$ . Thus translate and dilate to $𝑄 = (- 1, 1) \times (- 1, 1)$ as in (5.4.9) $(5.4.9) (5.4.9)$ , to obtain

Φ (𝑧) = \frac{2}{𝑙} 𝑓 (𝑧) - 1 + i = \frac{2}{𝑙} \int_{0}^{𝑧} \frac{d 𝜁}{\sqrt{𝜁 + 1} \sqrt{𝜁} \sqrt{𝜁 - 1}} - 1 + i .

Then $Φ : ℍ^{+} \to 𝑄$ is biholomorphic with an extension to $Φ (0) = - 1 + i$ , $Φ (- 1) = 1 + i$ , $Φ (\pm \infty) = 1 - i$ , $Φ (1) = - 1 - i$ .

Finally $𝜓 (𝑧) = \frac{𝑧 - i}{𝑧 + i}$ maps $ℍ^{+}$ onto $𝔻$ with inverse $𝜓^{- 1} (𝑤) = \frac{i (1 + 𝑤)}{1 - 𝑤}$ . Hence

𝐹 (𝑤) = Φ \circ 𝜓^{- 1} (𝑤) = \frac{4 \sqrt{2 𝜋}}{{Γ (\frac{1}{4})}^{2}} \int_{0}^{\frac{i (1 + 𝑤)}{1 - 𝑤}} \frac{d 𝜁}{\sqrt{𝜁 + 1} \sqrt{𝜁} \sqrt{𝜁 - 1}} - 1 + i (5.4 . 10)

gives the required biholomorphism, where the square root branches are selected such that their corresponding logarithms have imaginary parts in $(- \frac{π}{2}, \frac{3 π}{2}]$ . A visualization of this conformal map has been provided in Figure 5.1 $Figure 5.1 Figure 5.1$ . ⁠ $□$