Principal part: $ \frac{1}{z(\exp(z)-1)^2} $ (Laurent series) I have to compute from the following, its principal part of the Laurent series around $z=0$.
$$ \frac{1}{z(\exp(z)-1)^2} $$
I have trouble with computing the series of $(\exp(z)-1)^2$. I wanted to take out the $\frac{1}{z}$ and then multiply it by the series of the exp term. 
Thanks already.

It is convenient to represent $\exp(z)-1$ using the big-O notation.
\begin{align*}
\exp(z)-1&=\sum_{n=1}^\infty\frac{z^n}{n!}\\
&=z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\tag{1}
\end{align*}

We obtain
  \begin{align*}
\frac{1}{z\left(\exp(z)-1\right)^2}&=\frac{1}{z\left(z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\right)^2}\tag{2}\\
&=\frac{1}{z\left(z^2+z^3+\frac{7}{12}z^4+\frac{1}{4}z^5+O(z^6)\right)}\tag{3}\\
&=\frac{1}{z^3}\cdot\frac{1}{1+z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)}\tag{4}\\
&=\frac{1}{z^3}\sum_{n=0}^\infty(-1)^n\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)^n\tag{5}\\
&=\frac{1}{z^3}\left(1-\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)\right.\\
&\qquad\qquad\left.+\left(z^2+\frac{7}{6}z^3+O(z^4)\right)-\left(z^3+O(z^4)\right)+O(z^4)\right)\tag{6}\\
&=\frac{1}{z^3}\left(1-z+\frac{5}{12}z^2-\frac{1}{12}z^3+O(z^4)\right)\tag{7}\\
&=\color{blue}{\frac{1}{z^3}-\frac{1}{z^2}+\frac{5}{12z}}-\frac{1}{12}+O(z)\tag{8}
\end{align*}
with the blue marked part in (8) 
  being the principal part of $$\frac{1}{z\left(\exp(z)-1\right)^2}$$ at $z=0$.

Comment:


*

*In (2) we use the representation of $\exp(z)-1$ from (1).

*In (3) we multiply out and collect all terms with power greater or equal $6$ in $O(z^6)$.

*In (4) we factor out $\frac{1}{z^3}$.

*In (5) we apply the geometric series expansion.

*In (6) we multiply out, write the terms for $n=0,1,2$ and $n=3$ and collect all other summands in $O(z^4)$.

*In (7) we collect the terms accordingly.
