I’m trying to show that $z=0$ is a simple pole for the function $f(z)=\dfrac{\mathop{\text{Log}}(1+z)}{(e^z-1)^2}$.
I can see that $z=0$ is a pole of the function, but how do I show it? And that the order of this pole is one?
I was hoping that I could write the function as $f(z)=\dfrac{g(z)}{(z-0)^1}$ with the function $g(z)$ being analytic at $z=0$ and $g(0)≠0$ and suddenly not being sure this is even a viable approach.
Near $0$,$$\operatorname{Log}(z+1)=z-\frac{z^2}2+\frac{z^3}3-\cdots$$and$$(e^z-1)^2=z^2+z^3+\frac{7z^4}{12}+\cdots$$and therefore\begin{align}\frac{\operatorname{Log}(z+1)}{(e^z-1)^2}&=\frac{z-\frac{z^2}2+\frac{z^3}3-\cdots}{z^2+z^3+\frac{7z^4}{12}+\cdots}\\&=\frac1z\cdot\frac{1-\frac z{2}+\frac{z^2}3-\cdots}{1+z+\frac{7z^2}{12}+\cdots}\end{align}Since, near $0$, $\dfrac{1-\frac z{2}+\frac{z^2}3-\cdots}{1+z+\frac{7z^2}{12}+\cdots}$ is an analytic function and since it maps $0$ into $1$, it has a Taylor series of the type $1+a_1z+a_2z^2+\cdots$, and therefore, near $0$,$$\frac{\operatorname{Log}(z+1)}{(e^z-1)^2}=\frac1z+a_1+a_2z+\cdots$$
José Carlos Santos gave a good answer using series expansions. Here is another method based on the definition of poles. Recall: a function $f(z)$ has a pole of order $n$ at $z = z_0$ if and only if $\displaystyle\lim_{z\rightarrow z_0}(z-z_0)^mf(z)$ exists $\forall m \ge n$.
In the end, you will have to compute in your case : $$ \lim_{z\rightarrow0} \frac{z\ln(z+1)}{(e^z-1)^2} = \lim_{z\rightarrow0} \frac{\ln(z+1)+\frac{z}{z+1}}{2e^z(e^z-1)} = \lim_{z\rightarrow0} \frac{\frac{1}{z+1}+\frac{1}{(z+1)^2}}{2e^z(2e^z-1)} = 1 $$ using L'Hospital's rule twice. And you get the value of the residue at the same time.
