Laurent series and residue of $f(x)=\frac{1}{1+e^z}$

Question

I am having trouble trying to expand this function using Laurent series, and finding the residue$$f(x)=\frac{1}{1+e^z}$$ If I replace $e^z$ with its series I get $$f(x)=\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$ and the expansion is $$\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{n!}}+...$$ which is not useful in finding the residue at the singularity point(s) $z_0=(2k+1)\pi i$.

What would be a proper expansion, or is there another way to solve this?

edit:fixed summation formula layout

Answer 1

You can use the formula for finding residues of a pole $z_0$ of order $1$:

$$\text{Res}(f, z_0)=\lim \limits_{z\to z_0}\left[(z-z_0)f(z)\right].$$

The singularities here are the points $(2k+1)\pi i$, where $k$ ranges over the integers.

To use the above formula you need to first prove that these points are poles of order $1$, that is, you need to prove that these points are not removable discontinuities and you need to prove that $\lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)\in \mathbb C$. This limit actually is the same as in the above formula, so if you find it, you immediately find your residue.

One has $$\begin{align} \lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)&=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+(2k+1)\pi i}+1}\right)\\ &=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+\pi i}+1}\right)\\ &=\lim \limits_{w\to 0}\left(\dfrac{1}{e^{w+\pi i}}\right)\\ &=-1. \end{align}$$

Hence $\forall k\in \mathbb Z\left(\text{Res}(f,(2k+1)\pi i)=-1\right)$ which agrees with Wolfram Alpha.

In the above I used Cauchy's rule. If you want to avoid without having to resort to Laurent series, you can first prove that $\lim \limits_{w\to 0}\left(\dfrac{e^{w+\pi i}+1}w\right)=1$ and this you can do with the Taylor Series.

Answer 2

As Git explained, you need to calculate $$\lim_{z\to (2k+1)\pi i}\dfrac{z-(2k+1)\pi i}{e^z+1} \,.$$ The fastest way of doing this is to observe that $$\lim_{z\to (2k+1)\pi i}\dfrac{e^z+1}{z-(2k+1)\pi i} \,.$$ is just the definition of the derivative of $e^z$ at $z=(2k+1)\pi i$. Thus $$\lim_{z\to (2k+1)\pi i}\dfrac{e^z+1}{z-(2k+1)\pi i}=e^{(2k+1)\pi i}=-1$$