Show that \(\Aut_\CC(\CC) = \ts{ z \mapsto az+b\st a\in \unitsof{\CC}, b\in \CC }\).
Solution
Goal: Show that the holomorphic automorphisms of \(\CC\) are exactly the affine maps \(z \mapsto az + b\) with \(a \in \CC^{\times}\), \(b \in \CC\).
1. Every affine map \(z \mapsto az + b\), \(a \neq 0\), is a holomorphic automorphism of \(\CC\).
It is entire, and its inverse \(z \mapsto (z - b)/a\) is also entire; both are bijections of \(\CC\) onto itself.
2. Let \(f \in \Aut_\CC(\CC)\).Consider the singularity of \(f\) at \(\infty\).
\(f\) is entire, so it has a power series expansion \(f(z) = \sum_{n \ge 0} c_n z^n\) converging on all of \(\CC\).
The nature of the singularity at \(\infty\) is determined by the series.
3. The singularity at \(\infty\) is a pole, not an essential singularity.
If \(\infty\) were an essential singularity, then by the Casorati–Weierstrass theorem (or the open mapping theorem on neighborhoods of \(\infty\)), \(f\) would not be injective near \(\infty\) — in fact \(f(\CC)\) would be dense in \(\CC\) and \(f\) could not be injective on all of \(\CC\).
More elementarily: \(f\) is a bijection of \(\CC\), so as \(z \to \infty\) along any sequence, \(f(z) \to \infty\) (otherwise a sequence \(z_n \to \infty\) with \(f(z_n)\) bounded would have a convergent subsequence \(f(z_{n_k}) \to w\), and injectivity + open mapping would be violated — indeed \(f(z_{n_k}) \to w\) with \(z_{n_k} \to \infty\) is impossible for a biholomorphism of \(\CC\)). Hence \(\infty\) is a pole of \(f\).
4. A pole at \(\infty\) means \(f\) is a polynomial.
If \(\infty\) is a pole of order \(m\), then \(f(z) = \sum_{n=0}^m c_n z^n\) is a polynomial of degree \(m\): the power series expansion has only finitely many nonzero terms.
5. The polynomial \(f\) has degree exactly 1.
A polynomial of degree \(d \ge 2\) is not injective on \(\CC\) (the equation \(f(z) = w\) has \(d\) roots counting multiplicity for generic \(w\); by the fundamental theorem of algebra it is not one-to-one).
A polynomial of degree \(0\) is constant, not a bijection. Since \(f\) is a bijection, \(d = 1\).
6. Q.E.D.
1 and 5 show \(\Aut_\CC(\CC)\) consists exactly of the maps \(z \mapsto az + b\) with \(a \neq 0\).