Let \(Y\) be a simply-connected region contained inside the disk \(D = \{z : |z| < 1\}\). Assume \(0 \in Y \neq D\). Without appealing to the Riemann mapping theorem itself, prove that there exists an \(r\) with \(1 > r > 0\) and a univalent conformal map \(f: Y \to D(r) = \{z : |z| < r\}\) with \(f(0) = 0\) and \(f'(0) = 1\).
Hint: You may use part of the proof of the Riemann mapping theorem.
Solution
Choose \(a\in\mathbb D\setminus Y\). Since \(0\in Y\), we have \(a\ne0\). Let \[ \phi_a(z)=\frac{z-a}{1-\overline a z}, \] an automorphism of \(\mathbb D\) sending \(a\) to \(0\). The function \(\phi_a\) has no zeros on \(Y\). Since \(Y\) is simply connected, it has a holomorphic square root there: choose \(q\) with \[ q(z)^2=\phi_a(z). \] Because \(\phi_a\) is injective, so is \(q\). Also \(|q(z)|<1\) on \(Y\).
Put \(b=q(0)\) and let \[ \psi_b(w)=\frac{w-b}{1-\overline b w}. \] Then \[ F=\psi_b\circ q:Y\to\mathbb D \] is univalent and satisfies \(F(0)=0\). Since \(b^2=\phi_a(0)=-a\), \(|b|^2=|a|\). Furthermore \[ |q'(0)|=\frac{|\phi_a'(0)|}{2|b|} =\frac{1-|a|^2}{2\sqrt{|a|}}, \] and \[ |\psi_b'(b)|=\frac1{1-|b|^2}=\frac1{1-|a|}. \] Thus \[ |F'(0)| =\frac{1+|a|}{2\sqrt{|a|}}>1, \] the strict inequality following from \((1-|a|)^2>0\).
Multiply \(F\) by a unimodular constant so that \(F'(0)=\lambda>1\) is positive real, and define \[ f(z)=\frac{F(z)}\lambda, \qquad r=\frac1\lambda. \] Then \(0<r<1\), \(f\) is univalent and conformal, \[ f:Y\to D(r),\qquad f(0)=0,\qquad f'(0)=1, \] as required.