Residue of $1/(e^z+1)$

Question

How do I find the residues of $\frac{1}{1+e^{z}}$. I have calculated that the root of $1+e^{z}=0$ and the answer is $z=i\pi(2k+1)$, but the problem is that I get stucked at $$Re(f,i\pi)=\lim_{z\rightarrow i\pi}(z-i\pi)\frac{1}{1+e^{z}}$$

Answer 1

Let $w=z-(2k+1)\pi i$. Then $$\lim_{z\to(2k+1)\pi i}\frac{z-(2k+1)\pi i}{1+e^z} =\lim_{w\to 0}\frac{w}{1+e^{w+(2k+1)\pi i}} =\lim_{w\to0}\frac{w}{1-e^w}=\lim_{w\to0}\frac{w}{-w-w^2/2-\cdots}=-1.$$

Answer 2

The limit is of type $\frac 00$. Using l'Hospital: $$\lim_{z\to i\pi}\frac{z-i\pi}{1+e^z}=\lim_{z\to i\pi}\frac{(z-i\pi)'}{(1+e^z)'}=\lim_{z\to i\pi}\frac1{e^z}=-1$$