Let \(\Omega \subset \mathbb{C}\) be an open and bounded domain and \(\phi: \Omega \to \Omega\) a holomorphic map. Prove that if there exists a point \(z_0 \in \Omega\) such that \(\phi(z_0) = z_0\) and \(\phi'(z_0) = 1\), then \(\phi(z) = z\) for all \(z \in \Omega\).
Solution
Translate so that \(z_0=0\). Thus \(0\in\Omega\), \(\phi(0)=0\), and \(\phi'(0)=1\). Suppose \(\phi\) is not the identity. Since \(\phi(z)-z\) is holomorphic and not identically zero, there is a least \(m\ge2\) such that \[ \phi(z)=z+az^m+O(z^{m+1}),\qquad a\ne0. \]
1. Let \(\phi^{\circ n}\) denote the \(n\)-fold iterate. Induction on \(n\) gives\[ \phi^{\circ n}(z)=z+na z^m+O(z^{m+1}). \]Indeed, composing \(z+na z^m+O(z^{m+1})\) with \(z+az^m+O(z^{m+1})\) changes the coefficient of \(z^m\) from \(na\) to \((n+1)a\).Hence\[ (\phi^{\circ n})^{(m)}(0)=m!na. \]
2. Because \(\Omega\) is bounded, choose \(M\) with \(|w|\le M\) for every \(w\in\Omega\). Also choose \(r>0\) with \(\overline{D_r(0)}\subset\Omega\). Every iterate maps \(\Omega\) into itself, so on \(D_r(0)\),\[ |\phi^{\circ n}(z)|\le M. \]Cauchy’s estimate gives\[ |(\phi^{\circ n})^{(m)}(0)|\le\frac{m!M}{r^m}. \]Thus\[ n|a|\le\frac{M}{r^m} \]for every \(n\), impossible because \(a\ne0\). Therefore \(\phi(z)=z\) near \(0\), and the identity theorem gives \(\phi(z)=z\) on the connected domain \(\Omega\).