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:
To solve this problem, we need to find the supremum of \( |f(4i)| \) over all bijective analytic functions \( f: U \to D \) such that \( f(i) = 0 \). Here, \( U = \{z \in \mathbb{C}: \operatorname{Im}(z) > 0\} \) is the upper half of the complex plane, and \( D = \{z \in \mathbb{C}: |z| < 1\} \) is the open unit disk in the complex plane.
For such transformations, the biholomorphic mapping between the upper half-plane and the unit disk, which is also bijective and analytic, is given by the Cayley transform. One such transform is: \(f(z) = \frac{z - i}{z + i}\)This function maps the upper half-plane to the unit disk with \( f(i) = 0 \).
Now, we need to check where \( f(4i) \) maps under this transformation:
Substituting \( z = 4i \) into \( f(z) = \frac{z-i}{z+i} \), we calculate: \(f(4i) = \frac{4i - i}{4i + i} = \frac{3i}{5i} = \frac{3}{5}\)
Since this mapping is conformal, the function \( f \) does indeed map \((4i)\) to \(\frac{3}{5}\) in the unit disk. By the properties and definition of such bijective and analytic mappings, this is the supremum of \( |f(4i)| \).
Hence, the supremum value of \( |f(4i)| \) is \(\frac{3}{5}\).
Therefore, the correct answer is: \(\frac{3}{5}\)