Definition : A isolated singularity is a pole of ordre $m$ if $f(z)= \sum_{k=-m}^{\infty} a_k (z-z_0)^k$, $a_m \not= 0$
I have to classify the singularity (removable, pole and essential) of $\frac{1}{e^z-1}$. I know that $e^z-1=0 \iff e^z=1 \iff z = 2\pi k i = z_k$ for each $k \in \mathbb{Z}$. In using what I found, the solution of the book explain that as the derivative of $e^z$ doesn't vanish at $z_k$, so each $z_k$ are a simple pole, but it is not clear at all how the definition is used (??). Someone explain that $\lim_{z\to i2n\pi}\frac{z-i2n\pi}{e^z-e^{i2n\pi}}=\left. \frac{1}{\frac {de^z}{dz}}\right|_{z=i2n\pi}=1$, but how is it related to the definition?
Is anyone could explain to me the solution of the book? (Please explain in details) It is maybe a silly question, but I am only 13 years old, and sometime I need help to unblock on a subject.
Some preliminaries: If $f$ is a complex function with $f(z_0) = 0$, then $f$ has a simple zero at $z_0$ if there exists a holomorphic function $g$ within a neighbourhood of $z_0$ such that $g(z_0) \neq 0$, and $f(z) = (z-z_0)g(z)$.
If $f$ has a simple zero at $z_0$, then $1/f$ has a simple pole at $z_0$.
Expanding the series for $e^z - 1$, we find $$ e^z - 1 = -1 + \sum_{j=0}^\infty \frac{z^j}{j!} = \sum_{j=1}^\infty \frac{z^j}{j!} $$ Hence $e^z - 1$ has a simple zero at $0$, and simple zeroes at $2\pi i k$ by the periodicity of $e^z$. To see this, you can divide the series by $z$.
Hence $\frac{1}{e^z - 1}$ has a simple pole at $2\pi i k$.