# Determine convergence annulus of Laurent series without computing the Laurent series

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)?

• If you can assume/show that $\exp(2\pi\iota z)$ is an entire function with constant term 1, no zeroes and non-zero first derivative, then you can solve this, by using algebraic properties of analytic functions. – Kapil May 24 at 9:58

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 $$(1)$$.