Solution: Recall the Schwarz lemma We also have a generalized version, called Schwarz-Pick lemma.

Since the given function is a function from $\mathbb{D} $ to $\mathbb{H} $, we need to find an analytic function, say $g$, from $\mathbb{H} $ to $\mathbb{D} $ so that the composition $g\circ f$ is an analytic map from $\mathbb{D} $ to $\mathbb{D} $ and hence we can use any of the above results. Take \[ g(z) = \frac{z - \frac{\iota }{2}}{z + \frac{\iota }{2}} \text{ and } h(z) = \frac{z - \iota }{z + \iota }. \] Both of these functions are analytic and maps the upper half plane to the unit disk. Therefore, $g\circ f$ and $h \circ f$ are analytic functions on unit from unit disk to unit disk. We have \begin{align*} (g\circ f)'(z) & = g'(f(z)) \cdot f'(z) \\ & = \frac{\left( z + \frac{\iota }{2} \right) - \left( z - \frac{\iota }{2} \right) }{\left( z + \frac{\iota}{2} \right)^2 } \cdot f'(z) \\ & = \frac{\iota }{\left( z + \frac{\iota }{2} \right)^2 } \cdot f'(z) \\ \implies \left\vert (g\circ f)'(0) \right\vert & = \left\vert \frac{\iota }{\iota ^2 } \right\vert \left\vert f'(0) \right\vert = \vert f'(0) \vert . \end{align*} Since \[ g \circ f(0) = g\left( \frac{\iota}{2} \right) = 0, \] by Schwarz Lemma, we conclude that \begin{align*} \left\vert (g\circ f)'(0) \right\vert \leq 1 & \left\vert f'(0) \right\vert \leq 1. \end{align*} Thus, the supremum will be $1$.

We could have also used the function $h\circ f$. We have \begin{align*} (h \circ f)'(z) & = \frac{(z + \iota ) - (z - \iota )}{(z + \iota )^2}\cdot f'(z) \\ & = \frac{2\iota }{(z + \iota )^2} f'(z) \\ \implies \vert (h \circ f)'(0) \vert & = \left\vert \frac{2\iota }{(z + \iota )^2} \right\vert \cdot \left\vert f'(0) \right\vert \\ & = \frac{8}{9} \vert f'(0) \vert . \end{align*} Therefore using Schwarz-Pick lemma, we have \begin{align*} & \vert (h\circ f)'(0) \vert \leq \frac{1- \vert (h\circ f)(0) \vert^2 }{1} \\ \implies & \frac{8}{9}\cdot \vert f'(0) \vert \leq 1 - \frac{1}{9} \\ \implies & \vert f'(0) \vert \leq 1. \end{align*}