Can the zeros of an analytic function have an accumulation point outside the set of definition?

Question

$f\in \mathrm H (\Omega)$, the region $\Omega \subsetneq \mathbb C$. Can the zeros of $f$ have an accumulation point on $\mathbb {\hat C}\backslash \Omega$?

Answer 1

The answer is yes! A function that is analytic in domain $\Omega$ can have an accumulation point of zeros outside $\Omega$. Consider the function $\sin(1/z)$. This function is not analytic at $0$, but since $\sin(\pi n) = 0$ for all $n \in \mathbb{Z}$, $0$ is an accumulation point of the zeros of $\sin(1/z)$. But $\sin(1/z)$ is analytic in a lot of places, for example around $z = 1/\pi$, so functions that are somewhere analytic can have accumulation points of their zeros outside the domains in which they are analytic.

(Since this holds for $\mathbb{C}$, it of course holds for $\hat{\mathbb{C}}$).

Answer 2

$\textbf{Proposition}$. Let $\Omega$ be a dominion (open and connected set). Then, for each $a \in \{z\in \mathbb{C}: f(z)=0\}$ exists $\delta > 0$ such that $D(a,\delta) ⊂ \Omega$ and $f(z) 6\neq 0$ for all $z \in D(a,\delta) \backslash \{a\}$. For that reason, $Z(f):=\{z\in \mathbb{C}: f(z)=0\}$ has no accumulations points in $\Omega$.

$\textbf{Proof}$. Consider $a \in Z(f)$ and let $m$ be the order of the zero of $f$ in $a$, so then exists a function $g$ such that

$$f(z)=(z-a)^mg(z) \qquad \text{and} \qquad g(a)\neq 0.$$

As $g$ is a continuous function in $a$, exists $\delta>0$ such that $D(a,\delta) \subset \Omega$ and $g(z) \neq 0$ for all $z ∈ D(a,\delta)$. For $z ∈ D(a,\delta) \backslash \{a\}$ we have that $(z−a)^m \neq 0$ y $g(z)\neq 0$, so $f(z) = (z−a)^m g(z) \neq 0$. For all $a \in Z(f)$ we got $\delta > 0$ such that $D(a,\delta)\cap Z(f) = \{a\}$, so every point of $Z(f)$ are isolated, that is to say, $Z(f)∩Z(f)' = \emptyset$ . However, $Z(f)$ is a closed subset of $\Omega$, because $f$ is continuous, so $Z(f)' ∩\Omega ⊂ Z(f)$, and $Z(f)' ∩\Omega = \emptyset$ .