# Theory and Background: Conformal Maps

Conformal Mapping Dictionary:

Parts I,II ,III ,IV, and V

## Conformal Map Facts

It’s a theorem that holomorphic and $$f'\neq 0$$ implies conformal. Write $$f(z+{\varepsilon}) = f(z) + {\varepsilon}f'(z) + { \mathsf{O}}({\varepsilon}^2)$$, then \begin{align*} \operatorname{Arg}(f(z+{\varepsilon}) - f(z)) \approx \operatorname{Arg}({\varepsilon}f'(z)) = \operatorname{Arg}({\varepsilon}) + \operatorname{Arg}(f'(z))\to \operatorname{Arg}(f'(z)) ,\end{align*} so all tangent vectors near $$z_0$$ are rotated by approximately the same angle $$f'(z_0)$$, preserving their relative angles.

A map $$f$$ is conformal on $$\Omega$$ iff $$f$$ is complex-differentiable, $$f'(z)\neq 0$$ for $$z\in \Omega$$, and $$f$$ preserves signed angles (so $$f$$ is orientation-preserving).

A bijective conformal map $$f:U\to V$$ biholomorphism, and we say $$U$$ and $$V$$ are biholomorphic.

To check if a map is conformal at $$p$$, it suffices to check that $$f'(p)\neq 0$$.

Conformal implies holomorphic, and a bijective conformal map has conformal inverse automatically. Importantly, bijective holomorphic maps always have holomorphic inverses. Self-biholomorphisms of a domain $$\Omega$$ form a group $$\mathop{\mathrm{Aut}}_{\mathbf{C}}(\Omega)$$.

The bijectivity condition can be weakened: an injective holomorphic map satisfies $$f'(z) \neq 0$$ and $$f ^{-1}$$ is well-defined on its range and holomorphic.

## The Cross-Ratio Construction

Define generalized cross-ratios as \begin{align*} (z_1, z_2, z_3, z_4) &\coloneqq{z_1 - z_3\over z_1-z_4}{z_2 - z_4 \over z_2 - z_3} \\ R(z) \coloneqq(z, a,b,c) &\coloneqq{z - b\over z-c}{a - c\over a - c} \\ .\end{align*}

Given any three points, $$R(z)$$ sends \begin{align*} a &\to 1 \\ b &\to 0 \\ c &\to \infty .\end{align*}

One can use this to produce a map sending any three points to any other three points: \begin{align*} T(z) \coloneqq (w; w_1, w_2, w_3)^{-1} \circ (z; z_1,z_2, z_3) .\end{align*}

If any of the $$z_i$$ are $$\infty$$, the convention is to remove the corresponding terms where they appear:

## Linear Fractional/Mobius Transformations

A map of the following form is a linear fractional transformation ( or a Mobius transformation): \begin{align*} T(z) = {az + b \over cz + d} ,\end{align*} where the denominator is assumed to not be a multiple of the numerator. These have inverses given by \begin{align*} T^{-1}(w) = {dw-b \over -cw + a} \end{align*} and derivatives given by \begin{align*} T'(z) = {ad-bc \over (cz+d)^2} ,\end{align*} so this is always a conformal map when $$ad-bc\neq 0$$.

Using that $$\mathop{\mathrm{Aut}}({\mathbf{CP}}^1) \cong \operatorname{PGL}_2({\mathbf{C}})$$, there is a nice matrix representation if you act on projective coordinates: \begin{align*} { \begin{bmatrix} {a} & {b} \\ {c} & {d} \end{bmatrix} } \cdot {\left[ {z: 1} \right]}^t = {\left[ { {az+b \over cz + d }: 1} \right]} = {\left[ {f(z): 1} \right]} .\end{align*} This yields a quick way of finding $$f^{-1}$$: invert the matrix and ignore the determinant that shows up since it just scales every entry: \begin{align*} {az + b\over cz+ d} \leadsto { \begin{bmatrix} {a} & {b} \\ {c} & {d} \end{bmatrix} } ^{-1}= { \begin{bmatrix} {d} & {-b} \\ {-c} & {a} \end{bmatrix} } \leadsto {dw-b \over -cw + a} .\end{align*}

An LFT that fixes three points is the identity.

## Blaschke Factors

A very useful variant that shows up in applications of the Schwarz’ lemma: \begin{align*} \psi_a \coloneqq{a-z \over 1-\overline{a} z} .\end{align*} Some nice properties:

• $$\psi_a(a) = 0$$ and $$\psi_a(0) = a$$
• $$\psi_a$$ has a simple pole at $$1/\overline{a}$$ and a simple zero at $$z=a$$.
• $$\psi_a({{\partial}}{\mathbb{D}}) = {{\partial}}{\mathbb{D}}$$, i.e. $${\left\lvert {\psi_a(z)} \right\rvert} = {\left\lvert {z} \right\rvert}$$ when $${\left\lvert {z} \right\rvert} = 1$$.
• $$\mathop{\mathrm{Aut}}({\mathbb{D}}) = \left\{{ e^{i\theta} \psi_{a_k} {~\mathrel{\Big\vert}~}a_k\in {\mathbb{D}}}\right\}$$, i.e. these form the factors of automorphisms of the disc after including rotations.
• $$\psi_a'(z) = {{\left\lvert {a} \right\rvert}^2 - 1\over (1-\overline{a} z)^2}$$

## Exercise

Write a Mobius transformation $$f(z) = {az+b\over cz + d}$$ as a cross-ratio.

\begin{align*} T(z) = \qty{ z; {d-b\over a-c}, -{b\over a}, -{d\over c}} .\end{align*}

Let $$f$$ be meromorphic on $${\mathbb{D}}$$ with no poles on $${{\partial}}{\mathbb{D}}$$. Show that there exists a meromorphic $$g$$ with no poles in $${\mathbb{D}}$$ such that $${\left\lvert {f(z)} \right\rvert} = {\left\lvert {g(z)} \right\rvert}$$ when $${\left\lvert {z} \right\rvert} = 1$$.

Write $$\left\{{a_1,\cdots, a_n}\right\}$$ for all of the poles of $$f$$, indexed with multiplicity, and define \begin{align*} g(z) \coloneqq\prod_{1\leq k\leq n} \psi_{a_k}(z) f(z) \coloneqq\qty{ \prod_{1\leq k \leq n}{z-a_k\over 1 -\overline{a_k} z}} f(z) .\end{align*} Then $$g$$ has no poles, and since $${\left\lvert { \psi_{a_k} } \right\rvert} = 1$$ on $${{\partial}}{\mathbb{D}}$$, this works.

Show that if $$f$$ is injective, then $$f'$$ is nowhere vanishing and thus $$f$$ is conformal.

Contrapositive: if $$f'(z_0)=0$$ at some point, then $$f$$ fails injectivity in a neighborhood of $$z_0$$. Without loss of generality, we can assume $$f(0) = 0$$ after changes of coordinates in the domain/codomain, since this won’t affect injectivity or vanishing of derivatives. If $$z_0$$ is a zero of $$f$$ of some order $$n$$, we can write $$f(z) = z^n g(z)$$ with $$g(0) \neq 0$$ and $$g$$ nonvanishing in some neighborhood of zero. Choosing a branch of $$z\mapsto z^{1\over n}$$, write $$g(z) = (h(z))^n$$ for some $$h$$, then \begin{align*} f(z) = z^n h(z)^n = (zh(z))^n .\end{align*} Write $$H(z) \coloneqq zh(z)$$, then $$H'(z) = h(z) + zh'(z)$$, so $$H'(0)\neq 0$$. By the inverse function theorem, $$H$$ is invertible on a small neighborhood of $$0$$, making $$f$$ $$n$$-to-one for some $$n$$. Writing $$f(z) = \sum_{k\geq 0} c_k z^k$$, we have $$c_0 = 0$$ since $$f(0) = 0$$ and $$c_1 = 0$$ if $$f'(0) = 0$$, making $$n\geq 2$$, so $$f$$ fails injectivity in this neighborhood.