1
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Kapil May 24 at 9:58
1
$\begingroup$

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 $(1)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see, thank you! $\endgroup$ – user792123 May 24 at 10:47
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos May 24 at 11:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.