Classify the singularities (removable, pole and essential)

Question

I have to classify the singularities (removable, pole and essential) of $\frac{1}{e^z-1}$.

I know that $e^z-1=0 \iff e^z=1 \iff z = 2\pi k i$ for each $k \in \mathbb{Z}$. Is there an easy to find out all type of singularity without passing through the Laurent series directly? (Please explain in details.)

Thanks!

Answer 1

For each $k\in\mathbb{Z}$, the expression $$ \frac{z-2\pi ik}{e^z-1} $$ is the reciprocal of the quotient which appears in the definition of the derivative of $e^z$ at $z=2\pi ik$. Since this derivative is equal to $1$, it follows that $$\lim_{z\to 2\pi ik}\frac{z-2\pi ik}{e^z-1}=1 $$ hence $\frac{1}{e^z-1}$ has a simple pole at $2\pi ik$.

Answer 2

The series expansion for $e^z - 1$ is $$ e^z - 1 = \sum_{j=1}^\infty \frac{z^j}{j!} $$ so it has a simple zero (divide by $z$ to see this) at every $z = 2\pi i k$ by the periodicity of $\exp$. Hence the reciprocal $\frac{1}{e^z - 1}$ has a simple pole at every $z = 2\pi i k$.