If $f : D(0;1) \setminus \{ 0 \}$ holomorphic such that $f(1/n) = 0$ then either $f=0$ or $f$ has an essential singularity at $0$

Question

I'm trying to solve this question:

Let $f$ be a holomorphic function on $D(0; 1)\setminus\{0\}$ with the property that $f(1/n) = 0$ for every positive integer $n$. Show that $f$ is either identically zero, or that $f$ has an essential singularity at $0$. Give an example of the latter.

I'm thinking I need to use the Laurent expansion of the function to show the singularity is essential, but I'm unsure of how to do that since I don't know the function. I'm confused about how to show it could be identically zero as well. Any help is appreciated, thank you!

Answer 1

If $f$ does not have an essential singularity at $0$ then it has either a pole or a removable singularity. If it is a removable singularity then defining it to be $0$ at $0$ gives an analytic function in the open unit disk whose zeros have a limit point, so $f \equiv 0$. If it has a pole then necessarily $|f(z)| \to \infty$ as $ z \to 0$ but this is false because $f(\frac 1 n)=0$ for all $n$. An example where $f$ has an essential singularity is $f(z)=\sin (\frac {\pi} z)$.

Answer 2

Abridged proof. If $f$ does not have an essential singularity at zero and is not the function zero, Laurent development at zero gives an analytic function $f_1$ on the disc and an integer $m \geq 0$ such that $f(z) = \dfrac{1}{z^m} f_1(z),$ with $f_1(0) \neq 0.$ $f_1$ being analytic, cannot have cluster values for its set of zeroes, violating the hypothesis. Q.E.D.