Let $U$ be the open set of all $z \in \mathbb C$ such that $-1< \text{Im } z <1$. Let $f: U \to \mathbb C$ be a bounded homomorphic function and let $A :=\sup_{z\in U}|f(z)| $. Then using Cauchy's integral formula for derivatives, I can easily show that $|f^{(n)}(x)| \le An! $ for all $n \ge 1, x\in \mathbb R$.
My question is: How to show that this bound is sharp? i.e. how to find a bounded homomorphic function $f: U \to \mathbb C$ such that $\sup_{x \in \mathbb R} |f^{(n)}(x)| = An! $ for all $n \ge 1$, where $A=\sup_{z\in U}|f(z)| $ ?