Discussion
Let \(\mathbb{H} = \{z \in \mathbb{C} : \text{Im}(z) \gt 0\}\) and \(\mathbb{D} = \{z \in \mathbb{C} : |z| \lt 1\}\). Then
\(\sup\left\{|f'(0)| : f\) is an analytic function from \(\mathbb{D}\) to \(\mathbb{H}\) and \(f(0) = \frac{i}{2}\right\}\)
is equal to
Answer: C
The supremum is 1.
Compose with the Cayley-type map \(\varphi(w) = \frac{w - i/2}{w + i/2}\), which sends \(\mathbb{H}\) onto \(\mathbb{D}\) and \(i/2\) to 0. Then \(\varphi\circ f\) maps \(\mathbb{D}\) into \(\mathbb{D}\) and fixes 0, so by the Schwarz lemma \(|(\varphi\circ f)'(0)| \le 1\).
\[ \begin{aligned} \varphi'(w) &= \frac{i}{(w + i/2)^{2}} \\ |\varphi'(i/2)| &= \frac{1}{|i|^{2}} = 1 \end{aligned} \]
So \(|f'(0)| \le 1\), with equality for \(f = \varphi^{-1}\) (times a rotation), which is analytic from \(\mathbb{D}\) onto \(\mathbb{H}\) with \(f(0) = i/2\). In general \(|f'(0)| \le 2\,\text{Im} f(0)\). The values 1/4 and 1/2 are smaller than what \(\varphi^{-1}\) attains, and 100 exceeds the bound.
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