# Existence of holomorphic function with given conditions

I would like to answer this question:

Let $$\mathbb{E}:= \{z \in \mathbb{C}: |z| < 1\}$$. Does there exist a holomorphic function $$f: \mathbb{E} \to \mathbb{C}$$ with $$f(0) = 2i$$ and $$\lim_{|z| \to 1}|f(z)| = 1$$?

If we would replace $$\lim_{|z| \to 1}|f(z)| = 1$$ with the condition $$|f(z)| = 1$$ for all $$z \in \mathbb{E}$$, then I would know that the answer would be negative (identity theorem). Here I am unsure about whether or not the condition $$\lim_{|z| \to 1}|f(z)| = 1$$ changes something in regard to use the identity theorem.

I am thankful for some clarification.

Since $$\lim_{|z| \to 1}|f(z)| = 1$$, there is an $$r \in (0, 1)$$ such that $$|f(z)| \le 1.5$$ for $$r \le |z| < 1$$.
The maximum modulus principle then implies that $$|f(z)| \le 1.5$$ for $$|z| \le r$$, so that $$f(0) = 2i$$ is not possible.
Remark: The same argument shows that if $$f$$ is holomorphic in the unit disk with $$\lim_{|z| \to 1}|f(z)| = 1$$ then $$|f(z)| \le 1$$ for all $$z$$.
• You apply it to the closed disk with radius $r$. Jan 24 at 22:01