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Conformal Mappings

A bijective holomorphic function \(f: U\rightarrow V\) is called a conformal map or biholomorphism. Given such a mapping \(f\), we call \(U\) and \(V\) are conformally equivalent, or simply biholomorphic.

If A holomorphic function is injective, whose inverse is automatically holomorphic.

Holomorphism of the inverse of a conformal mapping

Assume a holomorphic function \(f:U\rightarrow V\) is injective, then \(f'(z)\neq 0\) for all \(z\in U\). Moreover, the inverse of \(f\) defined on its range is holomorphic.

Using power series expansion and contradiction of injection.

  • Show the first proposition. Show by contrsdiction.

If not, there exists \(z_0\in U\) such that \(f'(z_0)=0\) and assume \(f'(z)\neq 0\) for all \(z\in U\) except \(z_0\). Then expand \(f\) using power series

\[ f(z)-f(z_0)=a(z-z_0)^k +G(z) \]

where \(a\neq 0, k\geq 2\) and \(G(z)\) vanishes to order \(k+1\) at \(z_0\). For sufficiently small \(w\) (significant, for two distinct zeros, not multiple roots), write

\[ f(z)-f(z_0) - w :=F(z)+G(z), \quad F(z):=a(z-z_0)^k-w \]

Choose a small enough circle centered at \(z_0\) such that \(|F(z)|>|G(z)|\). Since \(F(z)\) has at least \(2\) zeros inside the circle, by Rouché Theorem \(f(z)-f(z_0)-w\) has at least \(2\) zeros inside the circle. We claim the zeros of \(f(z)-f(z_0)-w\) is distinct, hence \(f\) is not injective. Indeed, since \(f'(z)\neq z\) for all \(z\neq z_0\), \(z_0\) is not a zero of \(f(z)-f(z_0)-w\), so sufficiently close to \(z_0\), it follows \(f(z)-f(z_0)-w\) has two distinct roots, hence \(f\) is not injective.

  • Prove inverse of \(f\) defined on its range is holomorphic.

Let \(g=f^{-1}\) be the inverse of \(f\) defined on \(f(U)\), which we assume is \(V\). Then for \(w_0\in V\), \(w\) is close to \(w_0\). Write \(w=f(z)\) and \(w_0=f(z_0)\). So

\[ \frac{g(w)-g(w_0)}{w-w_0}=\frac{1}{\frac{f(z)-f(z_0)}{z-z_0}} \]

let \(z\rightarrow z_0\), we have \(w\rightarrow w_0\), so \(g'(w_0)=1/f'(g(w_0))\).

\(\square\)

From the above theorem, we could show that

Corollary: method of proving conformal equivalence of two open sets

Two open sets \(U\) and \(V\) are conformally equivalent if and only if there exsits two holomorphic and injective functions \(f:U\rightarrow V\) and \(g: V\rightarrow U\) with \(g(f(z))=z\) and \(f(g(w))=w\) for all \(z\in U\) and \(w\in V\).

Actually some books define a conformal map \(f:U\rightarrow V\) if \(f'(z)\neq 0\) for all \(z\in U\). This is a weaker condition, since \(f(z)=z^2\) is not injective but satisfies \(f(z)\neq 0\) for all \(z\in \mathbb{C}-\{0\}\). See the following example for further description.

Example. A holomorphic function \(f: U\rightarrow V\) is a local bijection on \(U\) if for all \(z\in U\), there exists a disc centered at \(z\) such that \(f:D\rightarrow f(D)\) is bijective (actually \(f\) is injective on \(D\)).

Prove \(f\) is a local bijection on \(U\) if and only if \(f'(z)\neq 0\) for all \(z\in U\).

The same logic as we show in the proof of Holomorphism of the inverse of a conformal mapping.

A holomorphic function satisfying the above condition preserve angles. Check the following example.

Example. A holomorphic function \(f\) defined near \(z_0\) is said to preserve angles at \(z_0\) if for any smooth curves \(\gamma\) and \(\eta\) intersecting at \(z_0\), the angle formed between \(\gamma\) and \(\eta\) at \(z_0\) equals the angle formed between \(f\circ \gamma\) and \(f\circ \eta\) at \(f(z_0)\).

Prove a complex-valued function \(f:\Omega \rightarrow \mathbb{C}\) is holomorphic and \(f'(z_0)\neq 0\) if and only if \(f\) preserves angles.

The Upper half-plane & disc

We give a demonstration of the conformal equivalence between the unit disc and the upper plane.

Denote the upper half-plane \(\mathbb{H}\) to be

\[ \mathbb{H}=\{z\in \mathbb{C}:\text{Im}(z)>0\} \]

Conformal function between disc & upper half-plane

Let

\[ F(z)=\frac{i-z}{i+z},\quad G(w)=i\frac{1-w}{1+w} \]

Then \(F: \mathbb{H}\rightarrow \mathbb{D}\) is a conformal map with inverse \(G: \mathbb{D}\rightarrow \mathbb{H}\).

  • Show that both are holomorphic in their respective domains. Easy to show that \(|F(z)|<1\) so the range of \(F\) is \(\mathbb{D}\). To show that \(G\) maps \(\mathbb{D}\) into \(\mathbb{H}\), let \(w=u+iv\), and calculate the imaginary part
\[ \begin{align*} G(w)&=i\frac{1-u-iv}{1+u+iv}\\ &=i\frac{(1-u-iv)(1+u-iv)}{|1+w|^2}\\ &=\frac{2v+i(1-u^2-v^2)}{|1+w|^2}. \end{align*} \]

so \(\text{Im}(G(w))>0\). And

\[ F(G(w))=\frac{i-i\frac{1-w}{1+w}}{i+i\frac{1-w}{1+w}}=\frac{2w}{2}=w,\quad \forall w\in \mathbb{D}, \]
\[ G(F(z))=i\frac{1-\frac{i-z}{i+z}}{1+\frac{i-z}{i+z}}=i\frac{2z}{2i}=z,\quad \forall z\in \mathbb{H}. \]

and we are done by Corollary: method of proving conformal equivalence of two open sets.

\(\square\)

Mappings of the form

\[ z\mapsto \frac{az+b}{cz+d} \]

where \(a,b,c,d\) are complex-valued number, and \(bc\neq ad\), are called fractional linear transformations.

Schwarz Lemma

Schwarz Lemma

Assume \(f:\mathbb{D} \rightarrow \mathbb{D}\) and \(f(0)=0\), then

(i) \(|f(z)|\leq |z|\) holds for all \(z\in \mathbb{D}\),

(ii) If there exists \(z_0\in \mathbb{D}\) such that \(f(z_0)=z_0\), then \(f\) is a rotation.

(iii) \(|f'(0)|\leq 1\), and if the equation holds, then \(f\) is a rotation.

(i) Expand \(f\) using power series

\[ f(z)=a_1z + \cdots \]

\(f(z)/z\) has removable singularity at \(0\), so

\[ \left|\frac{f(z)}{z}\right|\leq \frac{1}{r},\quad |z|=r<1, \]

By Maximum modulus principle, the above holds for \(|z|\leq r\). Let \(r\rightarrow 1\) and we have the result.

(ii) By assuption, \(f(z)/z\) reaches the maximum modulus inside the circle, so \(f(z)/z\) is a constant, i.e. \(f(z)=cz\). Substituting \((z_0, f(z_0))\) and taking the absolute value, we have \(|c|=1\), so \(c=e^{i\theta}\), that is, \(f\) is a rotation.

(iii) Let \(g(z)=f(z)/z\), then \(|g(z)|\leq 1\) for all \(z\in \mathbb{D}\). Since \(f'(0)=\lim\limits_{z\rightarrow 0}\frac{f(z)-f(0)}{z-0}=\lim\limits_{z\rightarrow 0}\frac{f(z)}{z}=g(0)\), so \(|f'(0)|=|g(0)|\leq 1\).

Moreover, if \(|f'(0)|=1\), then still by maximum modulus principle, \(g\) is a constant, and is also a rotation.

\(\square\)

A conformal map from an open set \(\Omega\) to itself is called Automorphism. The set of all automorphisim of \(\Omega\) is called \(Aut(\Omega)\), and carries the structure of group. Recall definition of group in Definition of group.

Here the group operation is the composition of map "\(\circ\)", the identity is the map \(z\mapsto z\), and the inverses are simply the inverse function. For inverse and composition, we have

\[ (f\circ g)^{-1}=g^{-1}\circ f^{-1}. \]

Automorphisms of the disc

In homework, we have known that the \(\psi\) function

\[ \psi_\alpha (z)=\frac{\alpha -z}{1-\overline{\alpha}z},\quad \alpha \in \mathbb{D} \]

is a bijection from \(\mathbb{D}\) to itself, thus is an automorphism of \(\mathbb{D}\). Interestingly, \(\psi_\alpha\) interchanges \(0\) and \(\alpha\) by

\[ \psi_\alpha(0)=\alpha,\quad \psi_\alpha(\alpha)=0. \]

Expression of all automorphism of disc

If \(f\) is an automorphism of the disc, then there exist \(\theta\in \mathbb{R}\) and \(\alpha \in \mathbb{D}\), such that

\[ f(z)=e^{i\theta}\frac{\alpha -z}{1-\overline{\alpha }z}. \]

Using Schwarz Lemma.

By property of automorphism, there exists a unique \(\alpha\) such that \(f(\alpha)=0\). So by operation of group, define another automorphism \(g=f\circ \psi_\alpha\), which satisfies \(g(0)=0\), and \(g^{-1}(0)=0\) i.e. normalization. Applying Schwarz Lemma to \(g\) and \(g^{-1}\), we have

\[ |g(z)|\leq |z|, |g^{-1}(w)|\leq |w|,\quad \forall z,w\in \mathbb{D} \]

which gives \(|g(z)|\leq|z|=|g^{-1}(w)|\leq |w|=|g(z)|\). So also by Schwarz Lemma, \(g\) is a rotation, i.e. \(g=e^{i\theta}z\). Replacing \(z\) by \(\psi_\alpha(z)\) and we have \(g\circ \psi_\alpha =f\circ\psi_\alpha\circ\psi_\alpha=f\), which gives the result.

\(\square\)

Corollary: automorphisms that fix the origin

The only automorphism of the unit disc that fix the origin is rotation.

Setting \(\alpha=0\) in Expression of all automorphism of disc.

We could see that the group of automorphism acts transitively, in the sense that given \(\alpha, \beta\in \mathbb{D}\), there is an automorphism \(\psi\) mapping \(\alpha\) to \(\beta\) by \(\psi=\psi_\beta\circ \psi_\alpha\).

Automorphism of the upper half-plane

By using \(F:\mathbb{H}\rightarrow \mathbb{D}\) in Conformal function between disc & upper half-plane and automorphism of the disc, we could determine the automorphism of the upper half-plane \(\text{Aut}(\mathbb{H})\).

Construction of automorphism of the upper plane

Consider map

\[ \Gamma: \text{Aut}(\mathbb{D})\rightarrow \text{Aut}(\mathbb{H}) \]

defined by

\[ \Gamma(\varphi) = F^{-1}\circ \varphi\circ F, \quad \varphi\in \text{Aut}(\mathbb{D}). \]

Actually \(\Gamma\) is an isomorphism between \(\text{Aut}(\mathbb{D})\) and \(\text{Aut}(\mathbb{H})\), that is,

\[ \begin{align*} \Gamma(\varphi_1\circ \varphi_2)&=F^{-1}\circ \varphi_1\circ \varphi_2\circ F\\ &=F^{-1}\circ \varphi_1\circ F\circ F^{-1}\circ \varphi_2\circ F\\ &=\Gamma(\varphi_1)\circ \Gamma(\varphi_2) \end{align*} \]

and \(\phi\) itself is a bijection.

Now we give a description of automorphism of the upper plane.

Denote \(SL_2(\mathbb{R})\) as the group of all \(2\times 2\) matrices with real entries and determinant \(1\), namely

\[ SL_2(\mathbb{R}) = \left\{ M=\left[\begin{array}{cc} a& b\\ c&d\end{array}\right]:\quad a,b,c,d\in \mathbb{R}, \quad ad-bc=1\right\} \]

which is called the special linear group.

Given a matrix \(M\in SL_2(\mathbb{R})\), define the mapping \(f_M\) by

\[ f_M = \frac{az+b}{cz+d}. \]

Expression of automorphism of the upper plane

Every automorphism of \(\mathbb{H}\) takes the form \(f_M\) for some \(M\in SL_2(\mathbb{R})\). Conversely, every map of his form is an automorphism on \(\mathbb{H}\).

  • Show that \(\forall M\in SL_2(\mathbb{R})\), \(f_M\) maps to from \(\mathbb{H}\) to itself.

  • Show that \(f_M \circ f_{M'}=f_{MM'}\).

  • Show that using elements in \(SL_2(\mathbb{R})\), we could map any \(z\in \mathbb{H}\) to \(i\).

  • Show that

\[ M_\theta=\left[\begin{array}{cc} \cos \theta & -\sin \theta\\ \sin\theta & \cos \theta \end{array}\right] \]

belongs to \(SL_2(\mathbb{R})\), \(F\circ f_{M_\theta}\circ F^{-1}\) corresponds to the rotation of angle \(-2\theta\) in the disc.

  • Complete the proof.

Suppose that \(f\) is a automorphism of \(\mathbb{H}\), then there exists a unique \(\beta\) such that \(f(\beta)=i\), and consider a matrix \(N\in SL_2(\mathbb{R})\) such that \(f_N(i)=\beta\), then \(g=f\circ f_N\) satisfies \(g(i)=i\), therefore \(F\circ g\circ F^{-1}\) is a automorphism of the unit disc that fixes origin. So \(F\circ g\circ F^{-1}\) is a rotation, by step 4 there exists \(\theta\) such that

\[ F\circ f_{M_\theta}\circ F^{-1}=F\circ g\circ F^{-1}. \]

Hence \(g = f_{M_\theta}\), so \(f=g\circ f^{-1}_{N}=f_{M_\theta N^{-1}}\), which is of the desired form.

\(\square\)

Riemann Mapping Theorem

The basic problem lies in determining the condition for existence of \(F:\Omega\rightarrow \mathbb{D}\). Here we give some observations.

Firstly, \(\Omega\neq \mathbb{C}\). Since, otherwise \(F\) is entire and bounded (by definition), then by Liouville's theorem \(F\) is constant, thus is not conformal. Then, since \(\mathbb{D}\) is connected, then we must also impose the requirement that \(\Omega\) is connected. The following example gives another condition.

Example. Suppose \(U\) and \(V\) are conformally equivalent. Prove that if \(U\) is simply connected, then so is \(V\).

For brevity, we shall call \(\Omega\) proper, if it is non-empty and not the whole of \(\mathbb{C}\).

Riemann Mapping Theorem

Suppose \(\Omega\) is proper, and simply connected. If \(z_0\in \Omega\), then there exists a unique conformal map \(F:\Omega \rightarrow \mathbb{D}\) such that

\[ F(z_0)=0,\quad F'(z_0)>0. \]

Check relative compect set description. Here we only need uniform boundedness and equicontinuity. That is,

Suppose a famliy \(\mathcal{F}\) of holomorphic functions on an open set \(\Omega\). If it is uniformly bounded on every compact subset of \(\Omega\), then it is equicontinuous and thus normal(every sequence of \(\mathcal{F}\) has a subsequence that converges uniformly on every compect subset of \(\Omega\).)

Similar to our proof in Ascoli-Arzelà Theorem.

We give a definition of exhaustion. Assume \(\Omega\) is an open set. A sequence \(\{K_l\}_{l=1}^\infty\) of compact subsets of \(\Omega\) is called an exhaustion of \(\Omega\), if

(i) \(K_l\) is contained in the interior of \(K_{l+1}\), for each \(l=1,2,\cdots\)

(ii) Every compact subset of \(\Omega\) is contained in some \(K_l\) for some \(l\). In particular \(\Omega = \bigcup_{l=1}^\infty K_l\).

Now we give a lemma for open subset \(\Omega\) of the complex plane, that is, Any open set \(\Omega\) in the complex plane has an exhaustion.

The proof is by construction. If \(\Omega\) is bounded, we let \(K_l\) denote the set of all points in \(\Omega\) at distance \(\geq 1/l\) from the boundary of \(\Omega\). If not, the above construction would cause \(K_l\) to be non-compact, so we add for each \(l\), \(K_l'=K_l\cap \{z:|z|\leq l\}\).

If \(\Omega\) is a connected open subset of \(\mathbb{C}\), and \(\{f_n\}\) a sequence of injective holomorphic functions on \(\Omega\) that converges uniformly on every compact subset of \(\Omega\) to a holomorphic function \(f\), then \(f\) is either injective or constant.

The proof is by contradiction. If the convergent function \(f\) is not injective, so there exists two complex number \(z_1,z_2\) such that \(f(z_1)=f(z_2)\). Now we define a sequence of functions \(g_n(z)=f_n(z)-f_n(z_1)\), which has no other zeros besides \(z_1\), since \(\{f_n\}\) is injective. By condition, \(\{g_n\}\) converges uniformly on every compact subset of \(\Omega\) to \(g(z)=f(z)-f(z_1)\). If \(g(z)\) is not identically zero, then \(z_2\) is another isolated zero. Using Augument Principle,

\[ \frac{1}{2\pi i}\int_\gamma \frac{g'(\zeta)}{g(\zeta)}d\zeta = 1 \]

for \(\gamma\) is a small circle centered at \(z_2\) such that \(g\) does not vanish on \(\gamma\) and its interior. Therefore, \(1/g_n\) converges uniformly to \(1/g\) on \(\gamma\). Since \(g'_n\) converges uniformly to \(g'\) on \(\gamma\) we have

\[ \frac{1}{2\pi i}\int_\gamma \frac{g'_n(\zeta)}{g_n(\zeta)}d\zeta \rightarrow \frac{1}{2\pi i}\int_\gamma \frac{g'(\zeta)}{g(\zeta)}d\zeta \]

However the leeft hand side equals \(0\) since \(g_n\) contains no zero inside \(\gamma\).

  • Consider all injective holomorphic functions \(F:\Omega\rightarrow \mathbb{D}\) with \(f(z_0)=0\), and find an \(f\) from these functions whose image fills out all of \(\mathbb{D}\), by making \(f'(z_0)\) as large as possible.

  • Extract this function as a limit from a sequence of functions.

  • Prove Uniqueness. Let \(F,G\) satisfies the above conditions. Then \(H=F\circ G^{-1}\) is an automorphism of \(\mathbb{D}\) that fix the origin. Therefore by Corollary: automorphisms that fix the origin, \(H(z)=e^{-i\theta}z\). Since \(H'(0)>0\) we must have \(e^{i\theta}=1\), which gives \(F=G\).

  • Suppose \(\Omega\) is a proper simply connected open subset of \(\mathbb{C}\). We claim that \(\Omega\) is conformally equivalent to a open subset of \(\mathbb{D}\) that contains the origin.

Indeed, choose a complex number \(\alpha \in \mathbb{C}-\Omega\), then \(z-\alpha\) does not vanish in \(\Omega\), so define

\[ f(z)=\log (z-\alpha) \]

so \(e^{f(z)}=z-\alpha\), which is apparently injective. So we claim that choose \(w\in \Omega\),

\[ f(z)\neq f(w)+2\pi i,\quad \forall z\in \Omega. \]

Since it is injective. So there exists a small circle centered at \(f(w)+2\pi i\) of radius \(r\) such that \(S_r(f(w)+2\pi i)\) does not contain \(f(\Omega)\). Otherwise choose a sequence \(\{z_n\}\) such that \(f(z_n)\rightarrow f(w)+2\pi i\), then exponentiating it we have \(z_n \rightarrow w\), so we have \(f(z_n)\rightarrow w\).

Define

\[ F(z)=\frac{1}{f(z)-(f(w)+2\pi i)},\quad z\in \Omega, \]

which is a conformal function that maps \(\Omega\) to \(F(\Omega)\), a subset of \(\mathbb{D}\). Since \(f\) is injective, so is \(F\). \(F\) is uniformly bounded on \(\Omega\), so we could translate and rescale \(F\) in order to get the result function \(F'\).

  • By step one, we assume \(\Omega\) is an open subset of \(\mathbb{D}\) with \(0\in \Omega\). Consider the family \(\mathcal{F}\) of all injective holomorphic functions on \(\Omega\) that map into \(\mathbb{D}\) and fix the origin, i.e.
\[ \mathcal{F}=\{f: f\in H(\Omega)\text{ and injective}, \quad f(0)=0 \}. \]

and extract function out the desired function. Note the set is non-empty since \(z\mapsto z\) is contained. By Cauchy's integral formula, we have

\[ |f'(0)|=\frac{1}{2\pi}\left|\int_\gamma \frac{f(\zeta)}{\zeta^2}d\zeta\right|\leq \frac{1}{r} \]

for some circle \(\gamma\) centered at the origin with radius \(r\), so \(|f'(0)|\) is uniformly bounded for all \(f\in \mathcal{F}\).

Define

\[ s=\sup_{f\in \mathcal{F}}|f'(0)| \]

Choose a sequence of \(\{f_n\}\in \mathcal{F}\) such that \(|f_n'(0)|\rightarrow s\). \(f_n\) is uniformly bounded on compact subset of \(\mathbb{D}\), so by Montel's Theorem, we have a subsequence \(f_{n_k}\) converges uniformly to \(f\) on the sompact subset. Since \(|f_n'(0)|\geq 1\) (\(z\mapsto z\) gives the minimum value of \(s\)), so \(|f'(0)|\geq 1\), and \(f\) is non-constant. By Lemma 2 for limit function of injective holomorphic functions, \(f\) is injective.

Apparantly \(f(0)=0\). And \(|f(z)|\leq 1\) since it is the limit function, and by maximum modulus principle, \(|f(z)|<1\). So \(f\in \mathcal{F}\), then \(|f_n'(0)|=s\).

  • Prove the limit function \(f\) is a conformal function from \(\Omega\) to \(\mathbb{D}\), i.e. prove \(f\) is surjective. We prove by contradiction.

If \(f\) is not surjective, then there exsits \(\alpha \in \mathbb{D}\) such that \(f(z)\neq \alpha\) for all \(z\in \Omega\). Choose \(\psi_\alpha\), an automorphism of \(\mathbb{D}\), such that \(\psi_\alpha\circ f\) does not contain the origin. Use square root function \(g(z)=e^{\frac{1}{2}\log z}\) to define

\[ F=\psi_{g(\alpha)}\circ g\circ \psi_\alpha \circ f, \]

which is injective and maps into \(\mathbb{D}\), since all the components are injective and map into \(\mathbb{D}\), and satisfies \(F(0)=0\). Let \(h(z)=z^2\) to be the inverse of \(g\), then

\[ f=\psi_\alpha^{-1}\circ h\circ \psi_{g(\alpha)}^{-1}\circ F:=\Phi\circ F \]

\(\Phi\) also maps \(\mathbb{D}\) into \(\mathbb{D}\), \(\Phi(0)=0\). But \(\Phi\) is not injective since \(h\) is not, so it is not a rotation, then by Schwarz Lemma, \(|\Phi'(0)|<1\). So

\[ |f'(0)|=|\Phi'(F(0))|\cdot |F'(0)|<|F'(0)| \]

which contradicts! (\(|f'(0)|=s\) matters!)

Then let \(\beta\) to be the augument of \(f'(0)\), and let \(f_{rot}=e^{-i\beta}f\) and we have \(f_{rot}'(0)>0\).

Note that simple-connectivity occurs in using the logarithm.

Corollary of RMT

Any proper simply connected open subsets in \(\mathbb{C}\) is conformally equivalent.

Conformal mappings onto Polygons

Example. Conformal mapping \(f(z)=z^\alpha\) from \(\mathbb{H}\) to sector, i.e.

\[ \{z: 0< \arg z<\alpha \pi\},\quad \alpha\in (0,2). \]

Write it as

\[ z^\alpha=f(z)=\int_0^z f'(\zeta)d\zeta=\alpha \int_0^z \zeta^{-\beta}d\zeta. \]

where \(\alpha+\beta=1\), and the integral is taken along any path in the upper half-plane.

Example. Conformal mapping

\[ f(z)=\int_0^z \frac{d\zeta}{(1-\zeta^2)^{1/2}},\quad z\in \mathbb{H}. \]

where the integral is taken from \(0\) to \(z\) along any path in the closed upper half-plane. We choose the branch for square root of \((1-\zeta^2)\) that makes it holomorphic in the upper half-plane and positive when \(\zeta\in (-1,1)\). This is actually the inverse of \(\sin z\).

Example. Consider Elliptic integrals (varients of this kind occur in the calculation of arc-length of an ellipse)

\[ f(z)=\int_0^z \frac{d\zeta}{[(1-\zeta^2)(1-k^2\zeta^2)]^{1/2}},\quad z\in \mathbb{H}, \]

where \(k\in (0,1)\) and the branch is chosen to make it positive when \(\zeta\in (-1,1)\).

Using the above examples, we could define the Schwarz-Christoffel integral and prove it maps the real line to a polygonal line.

General Schwarz-Christoffel integral

Define Schwarz-Christoffel integral as

\[ \begin{align} S(z)=\int_0^z \frac{d\zeta}{(\zeta-A_1)^{\beta_1}\cdots(\zeta-A_n)^{\beta_n}},\label{SC-integral} \end{align} \]

where \(A_1<\cdots<A_n\) are \(n\) distinct points on the real axis. The exponents \(\beta_k\) will be assumed to satisfy the conditions \(\beta_k<1\) for each \(k\) and \(1<\sum_{k=1}^n \beta_k\).

The integrand in \(\ref{SC-integral}\) is defined as follows. \((z-A_k)^{\beta_k}\) is that branch which is positive when \(z=x\) is real and \(x>A_k\). That is,

\[ (z-A_k)^{-\beta_k}= \begin{cases}(x-A_k)^{-\beta_k},\quad &x>A_k\\ (A_k - x)^{-\beta_k}e^{-i\pi \beta_k} ,\quad &x<A_k\end{cases} \]

\(\beta_k<1\) allows \((x-A_k)^{-\beta_k}\) is integral near \(A_k\), so the integral \(S(x)\) is continuous on the whole real axis. The simple connectivity of \(S'(z)\) in \(\mathbb{C}-\bigcup_{k=1}^n\{A_k+iy: y\leq 0\}\) allows it to have primitive and \(S(z)\) is holomorphic in the region. The continuity of \(S\) implies the integral path can be taken along any path in \(\mathbb{C}-\bigcup_{k=1}^n\{A_k+iy: y< 0\}\).

And estimation of growth order gives

\[ \left|\prod_{k=1}^n\frac{1}{(z-A_k)^{-\beta_k}}\right|\leq c |z|^{-\sum_{k=1}^n \beta_k} \]

so the condition \(\sum \beta_k>1\) allows the integral to converge at infinity, which is also independent of argument \(\theta\). We call the limit \(a_\infty\). Define \(a_k=S(A_k)\) for each \(k\).

Image of Schwarz-Christoffel integral

For the above integral \(\ref{SC-integral}\), we have

(i) If \(\sum \beta_k =2\), then the image is the polygonal region \(P\) with polygon \(\mathfrak{p}\) whose vertices are \(a_1,\cdots,a_n\), and \(a_\infty\) lies on \([a_n, a_1]\). Moreover, the interior angle at \(a_k\) is \(\pi \alpha_k\) with \(\alpha_k=\pi(1-\beta_k)\).

(ii) If \(\sum \beta_k \in (1,2)\), then the image is the polygonal region \(P\) with polygon \(\mathfrak{p}\) whose vertices are \(a_1,\cdots,a_n,a_\infty\). The interior angle at \(a_k\) for \(k=1,\cdots,n\) are the same as above, and the interior angle at \(a_\infty\) is \(\pi \alpha_\infty\) with \(\alpha_\infty=2-\sum \beta_k\).

We prove for \(\sum_k \beta_k=2\). For \(x\in [A_k,A_{k+1}]\), \(k=1,\cdots,k-1\), we consider the argument of \(S'(x)\), that is,

\[ \arg S'(x)=\arg \left(\prod_{j>k} (x-A_k)^{-\beta_k}\right)=\arg e^{-i \pi\sum \beta_k}=-\pi \sum \beta_k. \]

So the image of \(S\) on \([A_k,A_{k+1}]\) is the linear line \([a_k, a_{k+1}]\) with argument \(-\pi \sum \beta_k\). As for \(x>A_n\), the above argument equals \(-2\pi\), so \([A_n,\infty]\) is mapped into a linear line parallel to \(x\) axis. Similar for \(x\in [-\infty, A_1]\).

Same logic for \(\sum \beta_k<2\).

\(\square\)

Boundary behavior

We consider a polygonal region \(P\), namely a bounded, simply connected open set whose boundary is a polygonal line \(\mathfrak{p}\). We refer to this \(\mathfrak{p}\) as a polygon. To study conformal maps from \(\mathbb{H}\) to \(P\), we first consider the conformal maps from \(\mathbb{D}\) to \(P\) and their boundary behavior.

Theorem of the boundary behavior of conformal map from the unit disc to polygon

If \(F:\mathbb{D}\rightarrow P\) is a conformal mapping, then \(F\) extends to a continuous bijection from the closure \(\overline{\mathbb{D}}\) of the disc to the closure \(\overline{P}\) of the polygon. In particular, \(F\) gives rise to a bijection from the boundary of \(\mathbb{D}\) to the boundary of \(P\).

To prove this, we have to make use of the following lemma.

Lemma 1:

Suppose \(z_0\) is on the unit circle, for all small \(r\in (0,1/2)\), define \(C_r\) be the circle centered at \(z_0\) with radius \(r\). Given two points \(z_r\) and \(z_r'\) on \(C_r\) and contained in the unit disc, let \(\rho(r)=|f(z_r)-f(z_r')|\), then there exsits a sequence of \(\{r_n\}\) of radii tending to zero, and \(\lim\limits_{n\rightarrow \infty}\rho(r_n)=0\).

Show by contradiction.

If not, then there exsit \(c>0\) and \(R\in (0,1/2)\), such that \(\rho(r)>c\) for all \(r\in (0,R]\). Choose a \(r\), and have a corresponding \(z_r\) and \(z_r'\) on the circle \(C_r\) and contained in the unit disc. Define \(\alpha\subset C_r\) to be the arc connecting \(z_r\) and \(z_r'\), and parameterize it as \(z=z_0+re^{i\theta}\), with \(\theta\in (\theta_1(r), \theta_2(r))\), we have

\[ f(z_r)-f(z_r')=\int_\alpha f'(\zeta)d\zeta=\int_{\theta_1{r}}^{\theta_2(r)} |f'(z_0+re^{i\theta})|re^{i\theta}d\theta. \]

Take an absolute value and by Cauchy-Schwarz inequation

\[ \begin{align*} \rho(r)&\leq \int_{\theta_1{r}}^{\theta_2(r)} |f'(z)|rd\theta\\ & \leq \left(\int_{\theta_1{r}}^{\theta_2(r)} |f'(z)|^2 rd\theta\right)^{1/2}\left(\int_{\theta_1{r}}^{\theta_2(r)} rd\theta\right)^{1/2}\\ \frac{\rho(r)^2}{r} &\leq \left(\int_{\theta_1{r}}^{\theta_2(r)} |f'(z)|^2 d\theta\right)\left(\int_{\theta_1{r}}^{\theta_2(r)} d\theta\right)\\ &\leq 2\pi \int_{\theta_1{r}}^{\theta_2(r)} |f'(z)|^2 d\theta\\ \int_0^R \frac{\rho(r)^2}{r} dr &\leq 2\pi \int_0^Rdr\int_{\theta_1{r}}^{\theta_2(r)} |f'(z)|^2 d\theta\\ &\leq 2\pi \int_0^Rdr\int_0^{2\pi} |f'(z)|^2 d\theta=\iint_{f(D_r)}dxdy. \end{align*} \]

The left hand side is larger than \(c^2\int_0^R \frac{1}{r}dr=\infty\), while the right hand side is a finite number (actually the area of \(f(D_r)\)), which contradicts!

\(\square\)

Lemma 2: Existence of value on the boundary of unit disc

Suppose \(z_0\) is on the unit circle, and \(\{z_n\}_{n\geq 1}\subset \mathbb{D}\) tend to \(z_0\), then \(\{F(z_n)\}\) has a limit on the boundary of \(P\).

Show by contradiction.

If not, then there exist \(\{z_n\}\) and \(\{z_n'\}\) that both converges to \(z_0\) but \(\{F(z_n)\}\) and \(\{F(z_n')\}\) converges to two distinct values, denoted by \(\zeta\) and \(\zeta'\) respectively. Since \(F\) is conformal, then \(\zeta\) and \(\zeta'\) lie on the boundary of \(P\). Choose two two small isolated disc \(D\) and \(D'\) with a distance \(d>0\) that centered at \(\zeta\) and \(\zeta'\). For \(n\) large enough, \(F(z_n)\in D\) and \(F(z_n')\in D'\). There must exist two smooth curve \(\gamma\) which connect all \(F(z_n)\) and ends at \(\zeta\), and similar to curve \(\gamma'\).

Let \(\lambda=F^{-1}(\gamma)\) and \(\lambda'=F^{-1}(\gamma')\), both of which are smooth and contain infinitely many points of \(\{z_n\}\) and \(\{z_n'\}\) respectively. For small enough \(r\), let \(z_r=\lambda \cap C_r\), \(z_r'=\lambda'\cap C_r\), so by lemma 1, we have \(\rho(r)\rightarrow 0\) as \(r\rightarrow 0\), which contradicts \(|F(z_n)-F(z_n')|>d\) for large \(n\).

\(\square\)

Now we have prove the existence of

\[ \lim_{z\rightarrow z_0}F(z) \]

for \(z_0\) on the unit circle. Then we define \(F(z_0)\) to be the above limit value.

Lemma 3

The conformal mapping \(F\) extends to a continuous function from \(\overline{\mathbb{D}}\) to \(\overline{P}\).

We only consider the continuity of \(F\) from \(\overline{\mathbb{D}}\) to \(\overline{P}\). By continuity of \(F\) on \(\mathbb{D}\), \(\forall z_0\) on the unit circle, \(\forall \epsilon>0\), \(\exists \delta>0\), such that \(|F(z_0)-F(z)|<\varepsilon/2\) whenever \(z\in \mathbb{D}\cap D_\delta(z_0)\). So for \(z\) on the boundary of the disc, we also choose \(z\in \partial \mathbb{D}\cap D_\delta(z_0)\), then there must exist \(w\in \mathbb{D}\cap D_\delta(z_0)\cap D_\delta(z)\), such that \(|F(z)-F(w)|<\varepsilon/2\) and \(|F(z_0)-F(w)|<\varepsilon\). Then by triangle inequation we complete the proof.

Now we give the proof of Theorem of the boundary behavior of conformal map from the unit disc to polygon.

We could apply the same logic to the inverse \(G\) of \(F\), to show that \(G\) could extend to a continuous bijection from \(\overline{P}\) to \(\overline{\mathbb{D}}\). Then we only need to show that the extension of \(F\) and \(G\) are the inverse of each other. Given \(z_0\in \partial \mathbb{D}\), there exsit \(\{z_k\}\subset \mathbb{D}\), such that \(z_k\rightarrow z_0\), and we have \(G(F(z_k))=z_k\), and then take the limit we have \(G(f(z_0))=z_0\). The same logic for \(F(G(w_k))\rightarrow F(G(w_0))=w_0\).

\(\square\)

The mapping formula

Now we turn to the actual formula for a conformal map from \(\mathbb{H}\) to \(P\). The existence for \(F\) is guaranteed by Riemann Mapping Theorem. For \(f:\mathbb{D}\rightarrow \mathbb{H}\), \(f(w)=i(1-w)/(1+w)\), the point at \(-1\) on the unit circle corresponds to \(\infty\) on the upper half-plane. Here known info is the given \(P\) with its boundary \(\mathfrak{p}\), whose vertices ordered consecutively \(a_1,\cdots, a_n\), with \(n\geq 3\).

In Boundary behavior, we have discuss the boundary behavior of a conformal mapping from \(\mathbb{H}\rightarrow P\). We first consider \(P\) polygon \(\mathfrak{p}\), of which none of the vertices correspond to the point at infinity. Therefore, there exist real number \(A_1,\cdots, A_n\) such that \(F(A_k)=a_k\). Since

Formula

There exist complex number \(c_1\) and \(c_2\) such that for a known conformal map \(F:\mathbb{H}\rightarrow P\) could be given by

\[ F(z)=c_1S(z)+c_2, \]

where \(S(z)\) is the Schwarz-Christoffel integral formula \(\ref{SC-integral}\).

Using Schwarz reflection principle to extend the upper half-plane to the two-way strip. We aim to formulate a linear line on real axis.

For two segment \([A_{k-1}, A_{k}]\) and \([A_{k}, A_{k+1}]\) on the real axis, their image of \(F\) is two line with vertices \(a_{k-1}, a_{k}\) and \(a_{k+1}\) with inner angle \(\alpha_k\) at \(a_k\). Use

\[ h_k(z)=(F(z)-a_k)^{1/\alpha_k}. \]

Then this function maps \([A_{k-1}, A_{k}]\) and \([A_{k}, A_{k+1}]\) to a single line \(L_k\). By reflection, \(h_k\) is holomorphic in the whole strip \(\text{Re}(A_{k-1})<\text{Re}(z)<\text{Re}(A_k+1)\). We claim first \(h_k'\) does not vanish on the whole strip. Since \(F'(z)\neq 0\) on the upper strip, then by

\[ h_k'(z)=1/\alpha_k (F(z)-a_k)^{1/\alpha_k-1}F'(z), \]

\(h'(z)\neq 0\) holds for all \(z\) in the upper strip. Since \(F\) is injective up to the real axis, so is \(h_k\), which guarantees \(h'(x)\neq 0\) on real line \([A_{k-1}, A_{k+1}]\).

By taking derivative, we have \(F=h_k^{\alpha_k}+a_k\), \(F'=\alpha_k h_k^{\alpha_k-1}h'_k\), \(F'' =\alpha_k h_k^{\alpha_k-1}h''_k+\alpha_k (\alpha_k-1) h_k^{\alpha_k-2}h'^2_k\).

so

\[ \frac{F''(z)}{F'(z)}=\frac{h''_k}{h'_k} + \frac{(\alpha_k-1)h'_k}{h_k} \]

we rewrite it as (single pole at \(A_k\) and use argument principle)

\[ \frac{F''(z)}{F'(z)}=\frac{\alpha_k-1}{z-A_k} + E_k(z) \]

where \(E_k(z)\) is a holomorphic function on the corresponding strip. So

\[ \frac{F''(z)}{F'(z)} + \sum_{k=1}^n\frac{1-\alpha_k}{z-A_k} \]

is an entire function, then by Liouville's theorem, we have

\[ \frac{F''(z)}{F'(z)} = - \sum_{k=1}^n\frac{\beta_k}{z-A_k} \]

Denote \(Q(z)=c(z-A_1)^{-\beta_1}\cdots (z-A_n)^{-\beta_n}\), then \(\frac{Q'(z)}{Q(z)}\) also equal the right hand side of the above equation. So

\[ \frac{d}{dz}\left(\frac{F'(z)}{Q(z)}\right)=0 \]

And an integration gives the result.

\(\square\)