Currently completing a practice test and have come unstuck at the following question:
For the function $\frac{1}{e^{2z}-e^z-2}$ state the location and order of each pole and find the corresponding residue.
Thus far I have Identified the 2 simple poles as being $Log(-1)$ and $Log(2)$ - however in the answers I have been provided, my $Log(-1)$ pole has been subsequently changed for $-i\pi$?? I am unsure why or how this has occurred? I'm sure its something simple I am missing, any help would be greatly appreciated.
Note that\begin{align}e^{2z}-e^2-2=0&\iff(e^z)^2-e^z-2=0\\&\iff e^z=-1\vee e^z=2\\&\iff z=\pi i+2k\pi i\text{ for some }k\in\mathbb{Z}\vee z=\log(2)+2k\pi i\text{ for some }k\in\mathbb{Z}.\end{align}It's not a good idea to talk about logarithms in this context, unless it is the logarithm of a real number greater than $0$.
In this case, each pole is a simple pole. For instance, if $z_0=\pi i+2k\pi i$ for some integer $k$, then$$\operatorname{res}_{z=z_0}\left(\frac1{e^{2z}-e^z-2}\right)=\frac1{2e^{z_0}-e^{z_0}}=\frac1{-2-(-1)}=-1.$$