# Riemann Mapping

If $$\Omega \subseteq {\mathbb{C}}$$ with $$\pi_1 \Omega = 1$$ then $$\Omega$$ is biholomorphic to $${\mathbb{D}}$$.

If $$\Omega$$ is simply connected, nonempty, and not all of $${\mathbb{C}}$$, then for every $$z_{0}\in \Omega$$ there exists a unique conformal map $$F:\Omega \to {\mathbb{D}}$$. Moreover, it can be arranged so that $$F(z_{0}) = 0$$ and $$F'(z_{0}) > 0$$.

Thus any two such sets $$\Omega_{1}, \Omega_{2}$$ are conformally equivalent.

Necessity of hypothesis:

• Not all of $${\mathbb{C}}$$: $$F: {\mathbb{C}}\to \Omega$$ with $$\Omega$$ bounded implies $$F$$ is constant by Liouville.
• Simply connected: since $$\pi_1 {\mathbb{D}}= 1$$, any closed curve in $$\Omega$$ is nullhomotopic by finding $$F$$ and composing $$F^{-1}$$ with a homotopy in $${\mathbb{D}}$$.

The basic idea:

• Consider the set $${\mathcal{F}}$$ of conformal $$h: \Omega\to {\mathbb{D}}$$ with $$h(z_0) = 0$$ and $$h'(z_0) > 0$$ which are not necessarily surjective.
• Show there is a maximal $$f\in {\mathcal{F}}$$, and show $$f$$ is surjective.

More details:

• Fix some $$z_0\in \Omega$$ and set $${\mathcal{F}}= \left\{{f\in \mathop{\mathrm{Hol}}(\Omega, {\mathbb{D}}) {~\mathrel{\Big\vert}~}f(z_0) = 0, f \text{ injective }}\right\}$$. A lemma will show $${\mathcal{F}}$$ is nonempty.

• Define the hyperbolic translations and compute \begin{align*} h_w(z) &\coloneqq{z-w \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \in \mathop{\mathrm{Hol}}({\mathbb{D}}) h'_w(0) &= 1 - {\operatorname{ab}}{w}^2 \\ h'_w(w) &= {1\over 1 - {\left\lvert {w} \right\rvert}^2} .\end{align*}

• Show that if $$f\in {\mathcal{F}}$$ is not surjective, one can find another member with a larger derivative of $$z_0$$, say $$g$$ with $${\left\lvert {g'(z_0)} \right\rvert} > {\left\lvert {f'(z_0)} \right\rvert}$$.

• Show that $${\mathcal{F}}$$ is uniformly bounded and thus normal by Montel.

• Define $$m\coloneqq\sup_{f\in {\mathcal{F}}} {\left\lvert {f'(z_0)} \right\rvert}$$, noting that $$0<m<\infty$$ since the maps are conformal and the Cauchy estimate supplies an upper bound: \begin{align*} {\left\lvert {f'(z_0)} \right\rvert} \leq \max_{{\left\lvert {z-z_0} \right\rvert} = R} { {\left\lvert {f(z)} \right\rvert} \over R} \leq {1\over R} .\end{align*}

• Produce a sequence $$\tilde {\mathcal{F}}= \left\{{f_k}\right\}$$ such that $${\left\lvert {f_k'(z_0)} \right\rvert} \overset{k\to\infty}\longrightarrow m$$.

• Apply Montel to $$\tilde {\mathcal{F}}$$ which are uniformly bounded by 1 to extract a subsequence that converges locally uniformly to some $$h$$.

• Show $$h\in {\mathcal{F}}$$ by showing $$h(z_0) = 0$$, is analytic, and non-constant. Apply Hurwitz’s theorem to conclude $$h$$ is injective as a uniform limit of injective functions.

• Show $$h$$ is surjective by showing $${\left\lvert {h'(z_0)} \right\rvert} \geq {\left\lvert {g'(z_0)} \right\rvert}$$ (contrapositive to earlier step).

A simply connected nonempty domain $$\Omega \subseteq {\mathbb{C}}$$ is conformally equivalent to exactly one of

• $${\mathbb{CP}}^1$$,
• $${\mathbb{C}}$$, or
• $${\mathbb{D}}$$.

Prove that if $$f:U\to V$$ is holomorphic and injective then $$f'(z)\neq 0$$ on $$U$$.

• Idea: Rouché to get multiple roots, and translate a bit to make sure they’re distinct.

• Toward a contradiction suppose $$f'(z_0)=0$$.

• Taylor expand and rearrange: \begin{align*} f(z) &= f(z_0) + f'(z_0)(z-z_0) + f''(z_0)(z-z_0)^2 + \cdots \\ \implies f(z) - f(z_0) &= \sum_{j\geq 1} f^{(j)}(z_0)(z-z_0)^j = a(z-z_0)^k + (z-z_0)^{k+1}H(z) ,\end{align*} where $$k\geq 2$$ is the first nonvanishing derivative and $$a\coloneqq f^{(k)}(z_0)$$.

• So we can write \begin{align*} f(z) - f(z_0) \coloneqq F_1(z) + G(z) && F_1(z) \coloneqq f^{(k)}(z_0)(z-z_0)^k ,\end{align*} where $$G$$ vanishes to order at least $$k+1$$ near $$z_0$$.

• For $${\left\lvert {z-z_0} \right\rvert}$$ small, note that $$\deg F_1 = k$$ and $$\deg G \geq k+1$$, so there is some neighborhood $$N$$ about $$z_0$$ where $${\left\lvert {F_1(z)} \right\rvert} > {\left\lvert {G(z)} \right\rvert}$$.

• Subtract off a small $$w$$: \begin{align*} f(z) - f(z_0) - w = F(z) + G(z) && F(z) \coloneqq a(z-z_0)^k - w .\end{align*}

Since $${\left\lvert {F_1} \right\rvert} > {\left\lvert {G} \right\rvert}$$, picking $$w$$ small enough preserves this inequality, so $${\left\lvert {F} \right\rvert} > {\left\lvert {G} \right\rvert}$$ on $$N$$.

• Apply Rouché: $${\sharp}Z(F) = {\sharp}Z(F+G) = {\sharp}Z(f(z) - f(z_0) - w)$$.

• But $$F(z) = a(z-z_0)^k - w$$ has exactly $$k$$ zeros, and $$k\geq 2$$. It only remains to check if they are distinct: $$F'(z) = f'(z)$$, and the claim is that we can choose a small enough neighborhood $$N'$$ of $$z_0$$ so that $$f'(z)\neq 0$$ for $$z\neq z_0$$, so $$F$$ and $$f$$ have distinct roots here. This would contradict injectivity.

• Toward a contradiction, suppose no such $$N'$$ exists. Form a sequence of shrinking neighborhoods $$N_j$$ about $$z_0$$.

• For each $$N_j$$, find a $$z_j$$ and $$w_j$$ such that $$f(z_j) - w_j = 0$$, so $$f(z_j) = w_j$$ with multiplicity at least 2 and $$f'(z_j) = 0$$.

• Choose them so that $${\left\lvert {w_{j+1}} \right\rvert} < {\left\lvert {w_j} \right\rvert}$$ so $$\left\{{w_j}\right\}\to 0$$. But then $$\left\{{z_j}\right\}\to z_0$$ with the $$z_j$$ distinct, making it a set with an accumulation point. By the identity principal, $$f' \equiv 0$$, again contradicting injectivity.

If $$f:U\to V$$ is holomorphic and injective, then $$f$$ is invertible on its image.

• Idea: show that the usual formula from Calculus will work by defining $$(f^{-1})'$$ directly.
• By the proposition, $$f'(z) \neq 0$$ for $$z\neq z_0$$ in $$U$$.
• Write $$g\coloneqq f^{-1}$$ on $$\operatorname{im}f$$, and without loss of generality, replace $$V$$ by $$\operatorname{im}f$$.
• For $$w, w_0\in V$$ with $$0 < {\left\lvert {w-w_0} \right\rvert}$$ small, write $$w = f(z)$$ and $$w_0 = f(z_0)$$ for $$z, z_o\in U$$.
• Check the formula for $$g'(w_0)$$ and use that $$f'(z_0)\neq 0$$: \begin{align*} \frac{g(w)-g\left(w_{0}\right)}{w-w_{0}}=\frac{1}{\frac{w-w_{0}}{g(w)-g\left(w_{0}\right)}}=\frac{1}{\frac{f(z)-f\left(z_{0}\right)}{z-z_{0}}} \overset{z\to z_0}\longrightarrow {1\over f'(z_0) } = {1\over f'(g(w_0))} .\end{align*}