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 \(\mathop{\mathrm{Aut}}_{\mathbb{C}}(\Omega)\).
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 \(\mathop{\mathrm{Aut}}({\mathbb{CP}}^1) \cong \operatorname{PGL}_2({\mathbb{C}})\), 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-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z} .\end{align*} Some nice properties:
- \(\psi_a(a) = 0\) and \(\psi_a(0) = a\)
- \(\psi_a\) has a simple pole at \(1/\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu\) 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.
- \(\psi_a'(z) = {{\left\lvert {a} \right\rvert}^2 - 1\over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z)^2}\)
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 -\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu 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.