The order of a pole is an integer

Question

I try to solve a complex analysis question about isolated singularities, the question said:
Suppose that $f(z)$ has an isolated singularity at $z=z_0$, and that $\lim_{z\rightarrow z_0}{(z-z_0)^\alpha f(z)=M\neq0,\infty}$. Prove that $\alpha$ must be an integer.
I begin the proof by assuming that $\alpha \notin \mathbb{Z}$, and $f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^n$ and hence $(z-z_0)^\alpha f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^{n+\alpha}\rightarrow 0$ as $z\rightarrow z_0$ which a contradiction. Does my proof true? Thanks in advance.

Answer 1

No, it seems like you haven't considered the fact that $n + \alpha$ might be negative for some $n\in\mathbb{Z}$ such that $a_n \neq 0$. In this case, the limit $$\lim_{z\to z_0} \sum_{-\infty}^\infty a_n(z-z_0)^{n+\alpha}$$ may be nonzero. I think your approach is correct though -- proceed by the contrapositive, assuming that $\alpha\notin \mathbb{Z}$ and then showing that the limit cannot be finite and nonzero.