Analytic function with arbitrary, countable zero set

Question

Let $S\subset\mathbb{C}$ be a countable set with no accumulation points. Does there exist an analytic function $f(z)$ such that $f(z_i)=0$ for all $z_i\in S$ and $f(z)\neq 0$ for $z\neq S$?

Work so far: If we consider analytic to be equivalent to infinitely differentiable (at least for complex functions) then the analogous case has an affirmative answer for real functions. Basically, chain smooth transition functions together such that the construction is zero on the desired points non-zero everywhere else. For an example, see here for a function that is zero at $0$ and $3$ using the above construction. Since there are no accumulation points among the zeros, you are always assured that you can stretch and shrink these functions to your desired fit. The difficulty I am having is that this construction is not analytic, and does not extend to complex functions.

Answer 1

Below is Theorem 4.1 of Section 4 (Weierstrass infinite products) of Complex Analysis by Stein and Shakarchi:

Given any sequence $\{a_n\}$ of complex numbers with $|a_n|\to\infty$, there exists an entire function $f$ that vanishes at all $z=a_n$ and nowhere else. Any other such entire function is of the form $f(z)e^{g(z)}$ where $g$ is entire.

Since your $S$ is countable, you can write them as a sequence. Moreover, since $S$ has no accumulation points, in any finite disk there are only finitely many points of $S$. You can rearrange your prescribed zeros so that they form a sequence that tends to $\infty$. The theorem is thus applicable.