## Problem Restatement Given the transformation: \[ w = \frac{az + b}{cz + d}, \quad \Delta = ad - bc \neq 0 \] Show that this map can be inverted to find a unique (single-valued) \( z \) as a function of \( w \) everywhere. --- ## **Step 1: Solve for \( z \) in terms of \( w \)** Start from the given transformation: \[ w = \frac{az + b}{cz + d} \] Multiply both sides by \( cz + d \) to clear the denominator: \[ w(cz + d) = az + b \] \[ wcz + wd = az + b \] Bring all terms involving \( z \) to one side: \[ wcz - az = b - wd \] \[ z(wc - a) = b - wd \] \[ z = \frac{b - wd}{wc - a} \] **Explanation:** By algebraic manipulation, we have isolated \( z \) in terms of \( w \) (as long as \( wc - a \neq 0 \)). This gives us an explicit formula for \( z \) as a function of \( w \). --- ## **Step 2: Check for Uniqueness and Single-Valuedness** The denominator, \( wc - a \), can be zero for some specific value of \( w \), but since the determinant \( \Delta = ad - bc \neq 0 \), the transformation is **one-to-one (invertible) everywhere except possibly at the point where \( wc - a = 0 \)**. Solve \( wc - a = 0 \) for \( w \): \[ wc - a = 0 \implies w = \frac{a}{c} \] At this value, the original transformation maps \( z = \infty \) to \( w = \frac{a}{c} \), which is consistent with the Möbius transformation's behavior at infinity. Everywhere else, the transformation is invertible. **Explanation:** Except at \( w = \frac{a}{c} \), where the formula breaks down (corresponding to \( z = \infty \)), the expression for \( z \) in terms of \( w \) is well-defined and single-valued. The nonzero determinant ensures the transformation is invertible everywhere else. --- ## **Final Answer** **Yes, the map can be inverted to find a unique (single-valued) \( z \) as a function of \( w \) everywhere, except possibly at the point corresponding to \( z = \infty \) (i.e., \( w = \frac{a}{c} \)).** The inverse is: \[ z = \frac{b - wd}{wc - a} \]