Conformal mapping is the key object of studying in geometric function theory of one complex variable. In this area, the most important and fundamental result is the Riemann mapping theorem, which states that any proper simply connected domain D on $\mathbb{C}$ is conformal to the unit disk Δ. The Riemann mapping theorem is originally discovered by Riemann and a complete proof was first given by Koebe. Since then, the explicit construction of the Riemann mapping become a basic problem in complex analysis and other related area such as fluid mechanics and computational conformal geometry.

For polygon areas, Riemann mappings have some special forms, known as the Schwarz-Christoffel formula^[1]. In Ref.[2], Thurston gave a geometric approach to the construction of Riemann mappings. Thurston's method was further developed in Ref.[3]. Based on Bishop's methods^[4], Cheng^[5] studied explicit construction of Riemann mapping, by considering a condition called mappability.

In this note, we give a different explicit construction of iterative approximation of Riemann mappings. Our method is motivated by Koebe's proof of the Riemann mapping theorem, as presented in Ahlfors' classic Ref.[1]. The construction in the present note can be easily realized as a computer program, except the choices of the points w_k (see section 2 for details) which will be further studied in forthcoming works.

1 Proof of Riemann mapping theorem

In this section, we present a proof of the Riemann mapping theorem which is due to Koebe (see Ahlfors^[1]). Inspired by the idea in this proof, we will give an explicit iterative construction of conformal mappings from simply connected domains onto the unit disc.

Theorem 1.1  Let D⊆$\mathbb{C}$ be a simply connected domain, for any z₀∈D, there exists a unique biholomorphic mapping f:D→Δ such that f(z₀)=0, f'(z₀)>0.

Proof  The proof is taken from Ref.[1]. The analytic property encoded by the simply connectedness that will be used in the proof is the existence of square root of any holomorphic function on D without zeros. The uniqueness part follows easily from Schwarz Lemma. We give a proof of the existence of f, which is divided into several steps as follows.

Step 1. After a translation, we may assume 0∉D. Since D is simply connected, $\sqrt{z}$ has two single-valued branches on D, which are denoted by σ₊ and σ_－. It is clear that

$ {\sigma _ + }(D) \cap {\sigma _ - }(D) = \varnothing . $

In fact, if the formula does not hold, then there exist z, w∈D such that σ₊(z)=σ_－(w), and then z=σ₊²(z)=σ_－²(w)=w. This implies σ₊(z)=σ_－(w)=－σ₊(z) and z=0. We get a contradiction since 0∉D.

It is also clear that σ₊ and σ_－ are injective, hence we have σ₊(D)$\cong $D.

For a point a∈σ_－(D), we can find r>0 such that $\overline{\Delta \left(a, r \right)}\subseteq {{\sigma }_{-}}(D)$. Consider the function $\tau (z)=\frac{1}{z-a}$ on σ₊(D), we have |τ(z)| < $\frac{1}{r}$ for all z∈σ₊(D). Hence τ$\circ $σ₊ is injective and its image τ$\circ $σ₊(D)⊆Δ(0, $\frac{1}{r}$) is a bounded domain. Thus D is biholomorphic to a bounded domain.

Step 2. We consider the set: $\mathcal{F}$={g∈$\mathcal{O}$(D)|g:D→Δ is injective, g(z₀)=0}. From Step 1 we know that $\mathcal{F}$ is not empty. Let λ=sup{|g'(z₀)| |g∈$\mathcal{F}$}. From the Cauchy inequality we have λ < +∞. Take a sequence g_j∈$\mathcal{F}$, such that $\underset{j\to +\infty }{\mathop{\text{lim}}}\, $|g'_j(a)|=λ. By Montel's theorem, we may assume {g_j} converge to a holomorphic function f∈$\mathcal{O}$(D) on every compact subset K of D. We have

$ \left| {{f^\prime }\left( {{z_0}} \right)} \right| = \mathop {\lim }\limits_{j \to + \infty } \left| {g_j^\prime \left( {{z_0}} \right)} \right| = \lambda > 0. $

Hence f is injective by the Hurwitz's lemma. Note that f(D)⊆Δ, we can get f∈$\mathcal{F}$.

Step 3. The last step is to show that f is surjective. We argue by contradiction. If f is not subjective, there exists a point w₀∈Δ\f(D). Set

$ {\psi _{{w_0}}}(z) = \frac{{z - {w_0}}}{{1 - \overline {{w_0}}z }}, $

then ψ_w0∈Aut(Δ) and 0≠ψ_w0$\circ $f(D). Let h=$\sqrt{z} $ be a branch of the square root of z on ψ_w0$\circ $f(D). Then

$ \tilde f = {\psi _{h\left( { - {w_0}} \right)}} \circ h \circ {\psi _{{w_0}}} \in \widetilde {\cal F}. $

A straightforward calculation shows:

$ \left| {{{\tilde f}^\prime }\left( {{z_0}} \right)} \right| = \frac{{1 + \left| {{w_0}} \right|}}{{2\sqrt {\mid {w_0}} \mid }}\left| {{f^\prime }\left( {{z_0}} \right)} \right| > \left| {{f^\prime }\left( {{z_0}} \right)} \right| = \lambda . $

This leads to a contradiction by the definition of λ.

After a gyration, we have f'(z₀)>0.

The mapping in the above theorem is called a Riemann mapping.

2 Iterative construction of Riemann mapping

Motived by the proof of existence of Riemann mappings presented in the previous section, we give an iterative construction of Riemann mapping on a bounded domain D. More precisely, given a bounded domain D⊆$\mathbb{C}$ and a∈D, we construct a sequence of holomorphic injective mappings {f_j:D→Δ} with f_j(a)=0. Then we will prove that {f_j} converges to the Riemann mapping. The construction divides into several steps as follows.

Step 1. We imbed D into the unit disk Δ by complex affine mapping. There exists r>0 such that f₁(z)=r(z－a) maps D into Δ with f₁(z)=0 and f'₁(a)>0. If D is a disc, then we get our results because f is clearly surjective for suitable r; if D is not a disc, we continue our construction.

Step 2. Choose a point w₁∈Δ\f₁(D) such that $|{{w}_{1}}| < \frac{1+d(0, \partial {{f}_{1}}(D))}{2}$. We set

$ {f_2}(z) = {\alpha _1}{\psi _{h\left( { - {w_1}} \right)}} \circ h \circ {\psi _{{w_1}}} \circ {f_1}(z), $

where ${{\psi }_{{{w}_{1}}}}(z)=\frac{z-{{w}_{1}}}{1-\overline{{{w}_{1}}z}}, h(z)=\sqrt{z}$, and α₁∈S¹ is a constant chosen so that f'₂(a)>0. Then f₂ is a holomorphic injective mapping on D such that f₂(a)=0 and f'₂(a)>0.The computation shows that

$ \left| {f_2^\prime (a)} \right| = \left| {f_1^\prime (a)} \right|\frac{{1 + \left| {{w_1}} \right|}}{{2\sqrt {\left| {{w_1}} \right|} }}. $

Note that f₂ is not surjective since $-\sqrt{-{{w}_{1}}}\notin {{f}_{2}}(D)$.

Step 3. We construct f_j+1 inductively, by repeating the method in Step 2. Assume that we have constructed f_j. Choose w_j∈Δ\f_j(D) such that $|{{w}_{j}}| < \frac{1+d(0, \partial f(D))}{2}.$ We set

$ {f_{j + 1}}(z) = {\alpha _j}{\psi _{h\left( { - {w_j}} \right)}} \circ h \circ {\psi _{{w_j}}} \circ {f_j}(z), $

where ${{\psi }_{{{w}_{j}}}}(z)=\frac{z-{{w}_{j}}}{1-\overline{{{w}_{j}}z}}, h=\sqrt{z}$, and α₁∈S¹ is a constant chosen so that f'_j+1(a)>0. Then f_j+1 is a holomorphic injective on D such that f_j+1(a)=0 and f'_j+1(a)>0. A computation shows that

$ \left| {f_{j + 1}^\prime (a)} \right| = \left| {f_j^\prime (a)} \right|\frac{{1 + \left| {{w_j}} \right|}}{{2\sqrt {\left| {{w_j}} \right|} }}. $

From the above steps we get a sequence of holomorphic injective mappings {f_j:D→Δ} with

$ {f_j}(a) = 0,f_j^\prime (a) > 0. $

Theorem 2.1  Let {f_j} be the sequence constructed above. Then {f_j} converges to a biholomorhpic mapping g from D to the unit disk Δ such that g(a)=0 and g'(a)>0, uniformly on compact subsets of D.

Proof  By the uniqueness of the Riemann mapping f:D→Δ with f(a)=0 and f'(a)>0, it suffices to show that any subsequence of {f_j} has a subsequence that converges to the Riemann mapping f uniformly on compact sets of Δ.

By the Cauchy's inequality, there exists b>0 such that |f'_j(a)|≤b∈$\mathbb{R}$ for all j. Note that

$ \left| {f_{j + 1}^\prime (a)} \right| = \left| {f_1^\prime (a)} \right|\prod\limits_{k = 1}^j {\frac{{1 + \left| {{w_k}} \right|}}{{2\sqrt {\left| {{w_k}} \right|} }}} , $

we get $\frac{1+|{{w}_{j}}|}{2\sqrt{\left| {{w}_{j}} \right|}}\to 1$, and hence

$ \left| {{w_j}} \right| \to 1. $

Recall that we have

$ \left| {{w_j}} \right| < \frac{{1 + d\left( {0,\partial {f_j}(D)} \right)}}{2} \le 1 $

thus we can obtain that

$ d\left( {0,\partial {f_j}(D)} \right) \to 1. $

By Montel's theorem, any subsequence of {f_j} has a subsequence, say {f_jk}, that converges to a holomorphic function $\tilde{f}$ on D. By construction, it is clear that $\tilde{f}$(a)=0 and $\tilde{f}$'(a)>0. By Hurwitz's lemma, $\tilde{f}$ is injective and hence $\tilde{f}$(D)⊆Δ. To prove that $\tilde{f}$=f, it suffices to show that $\tilde{f}$:D→Δ is surjective.

Set g_j=f_j^－1. By Montel's theorem again, we may assume {g_jk} converges to a holomorphic function g on Δ uniformly on compact sets of Δ. Note that $g'\left(0 \right)=\underset{k\to +\infty }{\mathop{\text{lim}}}\, |{{{g}'}_{{{j}_{k}}}}\left(0 \right)|=\underset{k\to +\infty }{\mathop{\text{lim}}}\, \frac{1}{|f{{'}_{{{j}_{k}}}}\left(a \right)|}=\frac{1}{f'\left(a \right)}>0, g$ is nonconstant. For any w∈Δ, let z=g(w). For any small neighborhood U of w in Δ, by Rouché's theorem and the fact that d(0, ∂f_jk(D))→1, there is w_k∈U such that g_jk(w_k)=z, that is f_jk(z)=w_k, for k large enough. We have w_k→w as k→+∞. Hence

$ \tilde f(z) = \mathop {\lim }\limits_{h \to + \infty } {f_{{j_k}}}(z) = \mathop {\lim }\limits_{h \to + \infty } {w_k} = w. $

Since w∈Δ is arbitrary, $\tilde{f}$ is surjective.