Let \(U\) be a simply connected bounded domain in \(\CC\) and let \(P\in U\). Prove that for every \(t\in\RR\) there is a unique holomorphic function \(f:U\to U\) such that \(f(P)=P\) and \(f'(P)=e^{it}\).
Solution
1. A normalized Riemann map constructs the required function.
The domain \(U\) is nonempty since it contains \(P\), and is a proper subset of \(\CC\) since it is bounded. By the Riemann mapping theorem, there is a biholomorphism \(\phi:U\to D\), where \(D\) is the unit disk. We normalize it by \(\phi(P)=0\). The normalization is obtained, if necessary, by composing a Riemann map with a disk automorphism sending the image of \(P\) to zero.
For fixed \(t\in\RR\), put \(\lambda=e^{it}\) and define \[ f_t(z)=\phi^{-1}\bigl(\lambda\phi(z)\bigr). \] Rotation by \(\lambda\) maps \(D\) bijectively to itself, so \(f_t\) is a holomorphic automorphism of \(U\) and \(f_t(P)=P\). Differentiating \(\phi^{-1}\circ\phi\) at \(P\) gives \((\phi^{-1})'(0)\phi'(P)=1\). Therefore \[ f_t'(P)=(\phi^{-1})'(0)\lambda\phi'(P)=\lambda=e^{it}. \] This proves existence.
2. Every holomorphic self-map with the prescribed data equals \(f_t\).
Let \(f:U\to U\) be any holomorphic map satisfying those data. Set \(G=\phi\circ f\circ\phi^{-1}:D\to D\). Then \(G(0)=0\), and the chain rule gives \[ G'(0)=\phi'(P)f'(P)(\phi^{-1})'(0)=e^{it}. \] The Schwarz lemma yields \(\abs{G(w)}\leq\abs{w}\). Thus \(H(w)=G(w)/w\) extends holomorphically across zero with \(H(0)=G'(0)=e^{it}\) and \(\abs{H}\leq1\) on \(D\). It attains modulus one at zero, so the maximum modulus principle makes \(H\) the constant \(e^{it}\). Hence \(G(w)=e^{it}w\). Undoing the conjugation gives \(f(z)=\phi^{-1}(e^{it}\phi(z))=f_t(z)\), proving uniqueness.
3. Q.E.D.
Step 1 proves existence, and step 2 proves uniqueness.