Residue of $\frac{1}{z(e^{z}-1)}$ I was trying to find the residue of   $\dfrac {1}{z (e^z - 1)}$. I have written the Taylor series for $e^z$ which is $1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}..$ Thus, for $e^z - 1$ I have series of the form $z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}..$. But now I am stuck as I have a problem dividing 1 with my series. 
 A: 
We obtain
  \begin{align*}
\color{blue}{\frac{1}{z\left(e^z-1\right)}}&=\frac{1}{z(z+\frac{1}{2}z^2+O(z^3))}\tag{1}\\
&=\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}\tag{2}\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{z^2}-\frac{1}{2z}+O(1)}
\end{align*}
  and we conclude the residue is $\color{blue}{-\frac{1}{2}}$.

Comment:


*

*In (1) we expand $e^z-1$ up to $z^3$.

*In (2) we factor out $z$.

*In (3) we do a geometric series expansion
\begin{align*}
\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}&=\frac{1}{z^2}\left(1-\left(\frac{1}{2}z+O(z^2)\right)-\left(\frac{1}{2}z+O(z^2)\right)^2+O(z^2)\right)\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)
\end{align*}
A: Another way to find the residue is using one of the well known residue formulas.
We have 


*

*$e^z-1$ has a zero of order 1 at $z=0$

*$\Rightarrow \frac{1}{z(e^z-1)}$ has a pole of order 2 at $z= 0$

*$\Rightarrow Res_0 \frac{1}{z(e^z-1)} = \lim_{z\rightarrow 0}\frac{d}{dz}\left(z^2\frac{1}{z(e^z-1)} \right)= \lim_{z\rightarrow 0}\frac{d}{dz}\frac{z}{e^z-1} = \lim_{z\rightarrow 0} \frac{e^z-1 - ze^z}{e^{2z}-2e^z+1}$

*Note that $\frac{z}{e^z-1}$ has a removable singularity at $z=0$ and the limit of its first derivative at $z=0$ exists. So, we can use L'Hospital to calculate the limit:
$$Res_0 \frac{1}{z(e^z-1)}=\lim_{z\rightarrow 0} \frac{e^z-1 - ze^z}{e^{2z}-2e^z+1} = \lim_{z\rightarrow 0} \frac{-ze^z}{2e^{2z}-2e^z}=\lim_{z\rightarrow 0} \frac{-z}{2e^{z}-2} =\lim_{z\rightarrow 0} \frac{-1}{2e^{z}}= -\frac{1}{2}$$

