This is a question from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming exam.
Problem:
Suppose that $f$ is holomorphic on $\mathbb{C}\setminus\{0\}$ and that $f(n)=(-1)^n$ for each positive integer $n$. Prove that $\inf_{z\neq 0} |f(z)|=0$.
Thoughts so far:
If $f(n)=(-1)^n$ for each positive integer $n$, then it seems like it is some kind of periodic function. Also, if it's holomorphic, then it's smooth and continuous. In real numbers, this means it would attain every value in between $-1$ and $1$, which includes zero, thus $\inf_{z\neq 0}|f(z)| = 0$ clearly holds.
I somehow don't think it quite works this way in the complex plane?
My other thought was that, since $f$ is holomorphic everywhere except 0, I would need to start with the existence of a Laurent series $$ f(z) = \sum_{-\infty}^\infty a_k z^k, $$
but I'm not sure where to go from here.
If not, then there would exist a $\delta>0$, such that $$ \inf_{z\in\mathbb C\setminus\{0\}} |f(z)|=\delta. $$ This would in turn imply that $g(z)=1/f(z)$ is holomorphic in $\mathbb C\setminus\{0\}$, not constant, since $g(n)=(-1)^n$, and bounded since $$ |g(z)|\le \frac{1}{\delta}. $$ But, as $g$ is bounded near 0, then it is analytically extended at $z=0$, and hence it is extended to a entire analytic function, which is non-constant and bounded. Contradiction.