Residue of $\frac{1}{z(e^{z}-1)}$

Question

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.

Answer 1

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*}

Answer 2

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: $$Res_0 \frac{1}{z(e^z-1)}=\lim_{z\rightarrow 0} \frac{e^z-1 - ze^z}{e^{2z}-2e^z+1} = \lim_{z\rightarrow 0} \frac{-ze^z}{2e^{2z}-2e^z}=\lim_{z\rightarrow 0} \frac{-z}{2e^{z}-2} =\lim_{z\rightarrow 0} \frac{-1}{2e^{z}}= -\frac{1}{2}$$