= Solution
The <identity theorem> states: if $V\subseteq\mathbb C$ is a nonempty <connected> <open set> and $p,q$ are <holomorphic functions> on $V$, and their equality set has a <limit point> in $V$, then
$$
\boxed{p=q\quad\text{throughout }V}.
$$
The interior location of the <limit point> and <connectedness> are essential hypotheses.
For the proof put $h=p-q$. At any $z_0\in V$, its convergent <Taylor series> either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\geq0$. In the latter case,
$$
h(z)=(z-z_0)^m k(z),\qquad k(z_0)\ne0,
$$
where $k$ is <holomorphic> near $z_0$. By <continuity>, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior <limit point> therefore forces the neighbourhood alternative.
Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne\varnothing$, and $E$ is an <open set>. It is also a <closed set> relative to $V$: a point of its <closure> outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same <Taylor series> argument would place it in $E$. Since $V$ is <connected>, $E=V$. Thus $h=0$ on $V$, proving the theorem.
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