Conformal automorphism of complex numbers implies affine map

This article describes the computation of the conformal automorphism group of a domain or a Riemann surface
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Statement

Suppose $f:\mathbb{C}\to \mathbb{C}$ is a conformal automorphism: in other words, $f$ is an entire function from $\mathbb{C}$ to $\mathbb{C}$ with entire inverse. Then, $f$ is of the form:

$z\mapsto az+b$

Facts used

• Liouville's theorem: Any bounded and entire function must be constant
• Casorati-Weierstrass theorem: This states that near an essential singularity, the image of a holomorphic function is dense in $\mathbb{C}$
• Fundamental theorem of algebra

Proof

Proof using Casorati-Weierstrass

Suppose $f:\mathbb{C}\to \mathbb{C}$ is a conformal map. By composing $f$ with a translation, assume that $f(0)=0$. Then, consider the map $g:{\mathbb{C}}^{*}\to \mathbb{C}$ given by:

$g(z):=f(1/z)$

Now consider the singularity of $z$ at $0$. The following possibilities arise.

• The singularity is a removable singularity: In that case, it takes a finite complex value, and thus $g$ is bounded in a small neighborhood of $0$. Thus, $f$ is bounded outside a bounded subset, and since it is continuous, $f$ is globally bounded. Liouville's theorem (fact (1)) thus forces $f$ to be constant
• The singularity is an essential singularity: In that case, the Casorati-Weierstrass theorem tells us that the image under $g$ of a small neighborhood of $0$, is dense in $\mathbb{C}$. This clearly shows that $f$ isn't a homeomorphism (since it maps a non-dense set to a dense set) so it cannot be a conformal automorphism
• The singularity is a pole: Thus, we obtain:

$g(z)=z^{-n}h(z)$

where $h$ is a holomorphic. Plugging back $g$ in terms of $f$ we obtain:

$f(z)=z^{n}h(1/z)$

where $h$ is a holomorphic function. But we also have a global power series for $f$, which must match up with the above, so we see that $f$ is a polynomial of degree at most $n$. Thus, it remains to check which polynomial maps from $\mathbb{C}$ to $\mathbb{C}$ are conformal.

Suppose $f:\mathbb{C}\to \mathbb{C}$ is conformal and a polynomial map. Then, $f^{\text{'}}$ should vanish nowhere. But $f^{\text{'}}$ is also a polynomial map, and by the fundamental theorem of algebra, $f^{\text{'}}$ must be constant. Thus, $f$ must be a linear polynomial.