I need to classify the pole $z_0=0$ of the following function- $$f(z) = \frac{1}{1-e^z}$$ The only method that I know is to expand the function in its Laurent Series. I cannot use Bernoulli Numbers since I have not viewed in class. I began by expanding $e^z$ as its Taylor series, but I got a problem when I was constructing de funcion $f(z)$.
I got stucked in the following
$$ \frac{1}{1-e^z} = \frac{1}{1-\sum_{n=0}^{\infty} \frac{z^n}{n!}} $$
Also I tried to substitute de Taylor series of $e^z$ in the expansion os $Log(1+z)$ and the geometric series; nonetheless, I got a double series in each case.
This topic is very new form me. Any help is aprecciated. Thank you very much.
It is a simple pole. An easy way to check this is to see that $\lim_{z\to 0} \frac{z}{1-e^z}=\lim_{z\to 0}\frac{1}{-e^z}=-1$(which can be done by expanding the Taylor series of $e^z$ or you can just divide the numerator and denominator by z and use that $1-e^z$ is holomorphic at 0.)