 Solution

The identity theorem for holomorphic functions says that a holomorphic function on a connected open set which has distinct zeros accumulating at a point of the set is identically zero. Equivalently, two holomorphic functions agreeing on such a set of points agree everywhere in that connected open set.

To prove it, expand at an accumulation point $a$. If the first nonzero Taylor coefficient had index $m$, the function would factor as $(z-a{)}^{m}h(z)$ with $h(a)\ne 0$. Continuity would make $h$ nonzero near $a$, so no other zeros could approach $a$, a contradiction. Thus all coefficients vanish and the function vanishes on a neighborhood of $a$. Let $E$ be the set of points having such a zero neighborhood. It is nonempty and open. If a point in the domain is a limit point of $E$, it is either already in $E$ or is an accumulation point of distinct zeros, and the same Taylor argument puts it in $E$. Thus $E$ is relatively closed as well. Connectedness forces $E$ to be the whole domain. Apply this result to the difference of two functions for the equivalent formulation.