Given: (b) \((0, 1, \infty) \mapsto (1, 1+i, 2)\) We are asked to find the explicit **fractional linear transformation** (also called Möbius transformation) that sends \(0 \mapsto 1\), \(1 \mapsto 1+i\), and \(\infty \mapsto 2\). A **fractional linear transformation** has the form: \[ f(z) = \frac{az + b}{cz + d} \] where \(a, b, c, d\) are complex numbers and \(ad-bc \neq 0\). --- ### **Step 1: Set up equations for the images** Let: - \(f(0) = 1\) - \(f(1) = 1+i\) - \(f(\infty) = 2\) #### **Plug in each value:** 1. For \(z=0\): \(f(0) = \frac{b}{d} = 1 \implies b = d\) 2. For \(z=1\): \(f(1) = \frac{a + b}{c + d} = 1 + i\) 3. For \(z=\infty\): \(f(\infty) = \frac{a}{c} = 2 \implies a = 2c\) --- ### **Step 2: Substitute and simplify** From above: - \(b = d\) - \(a = 2c\) Substitute into the second condition: \[ f(1) = \frac{a + b}{c + d} = 1 + i \] \[ \frac{2c + d}{c + d} = 1 + i \] Let \(d = 1\) (since fractional linear transformations are only up to scaling, we can pick one variable for convenience). Then \(b = d = 1\). So: \[ \frac{2c + 1}{c + 1} = 1 + i \] Now solve for \(c\): \[ 2c + 1 = (1 + i)(c + 1) \] \[ 2c + 1 = (1 + i)c + (1 + i) \] \[ 2c + 1 - (1 + i)c = 1 + i \] \[ 2c - (1 + i)c = i \] \[ c(2 - (1 + i)) = i \] \[ c(1 - i) = i \] \[ c = \frac{i}{1 - i} \] Multiply numerator and denominator by the conjugate \(1 + i\): \[ c = \frac{i(1 + i)}{(1 - i)(1 + i)} = \frac{i + i^2}{1^2 - i^2} = \frac{i - 1}{1 - (-1)} = \frac{i - 1}{2} \] --- ### **Step 3: Find all coefficients** - \(c = \frac{i - 1}{2}\) - \(a = 2c = i - 1\) - \(b = 1\) - \(d = 1\) So the transformation is: \[ f(z) = \frac{(i - 1)z + 1}{\frac{i - 1}{2}z + 1} \] --- ### **Step 4: Simplify the transformation** Multiply numerator and denominator by 2 to clear denominators: \[ f(z) = \frac{2(i - 1)z + 2}{(i - 1)z + 2} \] --- ### **Final Answer** \[ \boxed{ f(z) = \frac{2(i - 1)z + 2}{(i - 1)z + 2} } \] This Möbius transformation sends \(0 \mapsto 1\), \(1 \mapsto 1 + i\), and \(\infty \mapsto 2\).

Certainly! Here's a concise, two-line explanation focusing on the theory: A fractional linear transformation (or Möbius transformation) is uniquely determined by three points and their images; solving for its parameters involves setting up equations based on these correspondences and solving for the coefficients \(a, b, c, d\) in \(f(z) = \frac{az + b}{cz + d}\). Once the coefficients are found, the explicit form of the transformation is obtained, ensuring it maps the given points to their specified images.