Let's break down the proof of the mapping \( f(z) = e^{i\theta} \frac{z - \beta}{z - \overline{\beta}} \): --- ### **Step 1: Mapping the Upper Half-Plane to the Unit Disk** **Explanation:** We want a bijective analytic (conformal) map from the upper half-plane \( U = \{z \in \mathbb{C} : \text{Im}(z) > 0\} \) to the unit disk \( D = \{z \in \mathbb{C} : |z| < 1\} \). A standard conformal map from \( U \) to \( D \) is: \[ \phi(z) = \frac{z - i}{z + i} \] - For \( z \in U \), \( \text{Im}(z) > 0 \), so \( |\phi(z)| < 1 \), hence \( \phi(z) \in D \). **Why this works:** - \( \phi(i) = 0 \), so \( i \) in \( U \) maps to the center of \( D \). - \( \phi \) is bijective and analytic. --- ### **Step 2: Generalizing to Any Bijective Analytic Map** **Explanation:** The most general form of a bijective analytic map (Möbius transformation) from \( U \) to \( D \) is: \[ f(z) = e^{i\theta} \frac{z - \beta}{z - \overline{\beta}} \] where \( \beta \) is a preimage of \( 0 \) under \( f \), i.e., \( \beta = f^{-1}(0) \) and \( \theta \in \mathbb{R} \) is a rotation parameter. **Why this works:** - \( f(z) \) maps \( U \) conformally onto \( D \). - \( f(\beta) = 0 \) and \( |f(z)| < 1 \) for \( z \in U \). - The \( e^{i\theta} \) factor represents rotation in \( D \). --- ## **Final Answer** The mapping \( f(z) = e^{i\theta} \frac{z - \beta}{z - \overline{\beta}} \) is a bijective analytic (conformal) map from the upper half-plane \( U \) to the unit disk \( D \), where \( \beta = f^{-1}(0) \) and \( \theta \in \mathbb{R} \). This generalizes the standard map by allowing for translation (choice of \( \beta \)) and rotation (\( \theta \)).