Question

Let \( {H} = \{ z \in {C} : \operatorname{Im}(z)>0 \ \) and \( {D} = \{ z \in {C} : |z|<1 \} \). Then \[ \sup \{ |f'(0)| : f { is an analytic function from } {D} { to } {H} { and } f(0) = \frac{i}{2} \} \] is equal to:}

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( 1 \)
• D. \( 100 \)

Hint:

For extremal problems with analytic functions, use Schwarz-Pick theorem to find the conformal map achieving the supremum.

Answer 1

The Correct Option is C

Step 1: Schwarz-Pick theorem. For analytic functions \( f : {D} \to {H} \), the Schwarz-Pick theorem states that the supremum of \( |f'(0)| \) is determined by a conformal map. Step 2: Using a conformal map. The map \( f(z) = \frac{i(1 + z)}{1 - z} \) satisfies \( f(0) = \frac{i}{2} \) and achieves the supremum for \( |f'(0)| \). Step 3: Calculating \( |f'(0)| \). The derivative of \( f(z) \) is: \[ f'(z) = \frac{2i}{(1 - z)^2}, \quad f'(0) = 2i. \] Thus, \( |f'(0)| = 1 \). Step 4: Conclusion. The supremum is \( {(3)} 1 \).