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

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

• What do you want exactly? The laurent series? The residue? Both? A closed formula for the laurent series or just a few terms? Jun 12, 2014 at 11:45
• I need to find the residue, and for that i thought I need a couple of terms from the expansion. The terms are not really necessary, if there is another way to find the residue. Jun 12, 2014 at 11:49
• This answer of mine should help you. The first part of the answer gives you a formula to find the residue. The second part hints at how you can find the laurent series to find the residue. Jun 12, 2014 at 11:50

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.

• Thank you for the help, but i still have one small issue, and this has been my problem from the start. How exactly do you prove that those singularity points are poles of order 1? Jun 12, 2014 at 12:35
• I addressed that issue in my answer "You need to prove that these points are not removable discontinuities and you need to prove that $\lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)\in \mathbb C$". I took care of the limit, you just have to prove that they aren't removable discontinuities. Jun 12, 2014 at 12:40

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 $$\lim_{z\to (2k+1)\pi i}\dfrac{e^z+1}{z-(2k+1)\pi i}=e^{(2k+1)\pi i}=-1$$

• Nice. I usually am on the look out for these tricks. I missed it. Jun 12, 2014 at 12:25