Take the holomorphic function: $$\mathbb{C} \setminus \{2k \pi i \ \mid \ k \in \mathbb{Z} \} \ni z \mapsto \frac{1}{e^z - 1} \in \mathbb{C}. $$ How can we determine the annulus of convergence of the Laurent series around the point $z = 0$ of the above function, without actually computing the Laurent series?
I don't really know how to start. Of course, for $z = 0$, the above function is not defined, so the annulus of convergence will be $\{z \in \mathbb{C} \ \mid \ 0 < |z| < R, \}$ for some $R \in \mathbb{R}_+$. Also, the function is again not defined in $2\pi i$ (by periodicity of the exponential). So I believe that the annulus of convergence will be $$\{z \in \mathbb{C} \ \mid \ 0 < |z| < |2\pi i| = 2\pi \}. $$ Is this correct? Also, how would I formally prove this (besides saying that a bigger radius of convergence would include the point $2\pi i$, in which the function is not defined)?
The Laurent series $\sum_{n=-\infty}^\infty a_nz^n$ of $f$ must converge on any annulus contained in its domain. So, it must converge in$$\{z\in\Bbb C\mid0<|z|<2\pi\}.\tag1$$But, if it converged on a larger annulus, the limit$$\lim_{z\to2\pi i}\sum_{n=-\infty}^\infty a_nz^n\tag2$$would exist (in $\Bbb C$). But $(2)$ is equal to $\lim_{z\to2\pi i}f(z)$, which does not exist (again, in $\Bbb C$).
So, the answer is $(1)$.