Show that \(\Aut(\CC) = \ts{ z \mapsto az+b\st a\in \unitsof{\CC}, b\in \CC }\).
Solution
\(\supseteq\): Clear, every affine function is bijective with inverse \(g(z) \da \inverseof{a}(z-b)\), and holomorphic since \[ g(f(z)) = z \implies g'(f(z)) f'(z) = 1 \implies g'(w) = {1\over f'(\inverseof{f}(w))} .\]
\(\subseteq\): Suppose \(f\) is entire and bijective. If \(f\) is bounded, then \(f(z) = b\) is constant by Liouville and we’re done, so suppose not. Then \(\abs{f(z)}\to \infty\) as \(\abs{z}\to \infty\), making \(z=\infty\) either an essential singularity or a pole, since it is isolated and not removable since \(f\) is unbounded in every small enough neighborhood of \(\infty\). If \(f(b) = 0\) without loss of generality we can translate to assume \(f(0) = 0\) by replacing \(f(z)\) with \(f(z) - b\). By bijectivity, \(z=0\) must be a zero of order 1, so \({1\over f(z)}\) has a pole of order 1 at \(z=0\). So define \(G(z) \da {z\over f(z)}\), which is now an entire function.
On a large enough disc, \(\abs{f(z)} > 1\) for all \(\abs{z} > R\), so \(\abs{G(z)} < 1\inverseof{R}\) on \(\abs{z} > R\). Since \(G\) is holomorphic and thus continuous on \(\abs{z}\leq R\), which is compact, \(G\) is bounded here too. Thus \(G\) is constant by Liouville, and \(G(z) = c \implies f(z) = \inverseof{c} z \da az\). Unwinding the initial translation that ensured \(f(0)=0\), we get \(f(z) = az + b\).