Question

Let \( U = \{z \in \mathbb{C}: \operatorname{Im}(z) > 0\} \) and \( D = \{z \in \mathbb{C}: |z| < 1\} \), where \( \operatorname{Im}(z) \) denotes the imaginary part of \( z \).
Let \( S \) be the set of all bijective analytic functions \( f: U \to D \) such that \( f(i) = 0 \).
Then, the value of \( \sup_{f \in S} |f(4i)| \) is:

• A. 0
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{3}{5} \)

Hint:

The Schwarz-Pick theorem provides a useful bound on the magnitude of analytic functions mapping between the upper half-plane and the unit disk. Apply it to find the supremum.

Answer 1

The Correct Option is D

Step 1: Recognizing the Problem Setup
We are given that \( U \) is the upper half-plane, and \( D \) is the unit disk in the complex plane. We are asked to find the supremum of \( |f(4i)| \) for functions \( f \) that are bijective analytic maps from \( U \) to \( D \) and satisfy \( f(i) = 0 \).

Step 2: Applying the Schwarz-Pick Theorem
The Schwarz-Pick theorem gives an upper bound for the magnitude of a function in terms of the distance from the point to the boundary of the domain. For functions mapping the upper half-plane to the unit disk, the theorem implies that:
\[ |f(z)| \leq \frac{|z - i|}{|z + i|} \] for any \( z \in U \). We can apply this theorem to \( f(4i) \).

Step 3: Calculating \( |f(4i)| \)
Using the Schwarz-Pick theorem and the specific value \( z = 4i \), we compute:
\[ |f(4i)| = \frac{|4i - i|}{|4i + i|} = \frac{3}{5} \]

Final Answer
\[ \boxed{D} \quad \frac{3}{5} \]