Notes on Complex Analysis
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§4.4.1 The Group of Holomorphic Automorphisms on ℂ

4.4.1 The Group of Holomorphic Automorphisms on ℂ

In complex analysis, three main sets of interest are 𝔻, ℂ, and ℂ̂. We will now find Aut(ℂ).

Theorem 4.4.1.1 (The Holomorphic Automorphism Group on ℂ).

∀𝑓∈Aut(ℂ), 𝑓 is linear and non-constant. In other words, ∃𝑎∈ℂ∖{0} and ∃𝑏∈ℂ such that

𝑓(𝑧)=𝑎𝑧+𝑏.
Proof.

First, assume that ∞ is not an essential singularity of 𝑓(𝑧), which we will prove later. Then ∞ must be a pole by trichotomy, as a removable singularity implies boundedness (Proposition 4.3.1Proposition 4.3.1Proposition 4.3.1). Therefore, 𝑓(𝑧) is a polynomial of degree 𝑚, where 𝑚∈ℕ.

Since 𝑓−1∈Aut(ℂ), it is true that (𝑓−1)′ is entire. Since

(𝑓−1)′=1𝑓′(𝑓−1),

it follows that 𝑓′ has no zeros in ℂ. By the Fundamental Theorem of Algebra (Theorem 3.3.1Theorem 3.3.1Theorem 3.3.1), if 𝑚>1, then 𝑓′ has a complex zero, which is a contradiction. Hence, 𝑓 must be linear, and all functions in Aut(ℂ) are in the form of 𝑎𝑧+𝑏, where 𝑎∈ℂ∖{0} and 𝑏∈ℂ are constants. In other words, any holomorphic automorphism on ℂ is a composition of a rotation, a dilation, and a translation.

We will now prove the primary assumption; the singularity at 𝑧=∞ cannot be an essential singularity of 𝑓(𝑧). Let 𝑤∈ℂ be arbitrary. Then by the Casorati–Weierstrass Theorem (Theorem 4.2.4Theorem 4.2.4Theorem 4.2.4), ∀𝜀>0 and ∀𝑅>0, ∃|𝑧|>𝑅 such that |𝑓(𝑧)−𝑤|<𝜀. Equivalently, ∀𝑅>0, ∃𝜁∈𝐷(𝑤,𝜀) such that |𝑓−1(𝜁)|>𝑅. Since 𝑓−1 is continuous on 𝐷(𝑤,𝜀)‾ by holomorphy, by Theorem 1.2.22Theorem 1.2.22Theorem 1.2.22, it is bounded, which is a contradiction. ⁠□