- #1

TheCanadian

- 367

- 13

## Homework Statement

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Find and classify the isolated singularities of the following:

$$ f(z) = \frac {1}{e^z - 1}$$

## Homework Equations

## The Attempt at a Solution

I have the solution for the positions of the singularities, which is: ## z = 2n\pi i## (for ##n = 0, \pm 1, \pm 2, ...##) and this makes sense. But the solution also says these are simple poles. This part I do not quite understand. For example, if I take the derivative of ##f(z)##, then ##f'(z) = -(e^z -1)^{-2} e^z## where ##f'(2n\pi i)## is also ##\infty##. Perhaps my understanding of simple poles is wrong, but I don't see any singularities of order 1 when considering the Laurent series of ##f(z)##.

Any ideas on why ## z = 2n\pi i## is considered a simple pole here?