I want to solve a problem.
$\displaystyle \displaystyle \int_{|z|=8} \frac{dz}{(1-e^z)}$
I thought I could get something decent by doing a Taylor expantion. But no!
How should i solve it?
To help you a bit in finding the residue at z=0, since our aim is to compute the residue at the singularity z=0,
consider $\displaystyle 1/(1-e^{z}) = 1/(1-(1+z+z^{2}/2+...))$ (write the Taylor's expansion for $\displaystyle f(z)=e^{z}$)
$\displaystyle = 1/(-z-z^{2}/2-...) = (-1/z)(1/(1+z/2+...))$
Now reduce this into a Laurent series expansion about z=0 and hence find the residue as coefficient of $\displaystyle 1/z$
I'm sorry if I was vague when I described my problem. I was on a train tapping the message on my phone.
bandedkrait
As I stated in my first post, I actually did the expansion, but I'm not sure of how to interpret the result. When I played around with expansions I used the similarity to geometric series to get the next step from where you stopped and got this.
$\displaystyle \displaystyle - \frac{1}{z} +\frac{1}{2} - \frac{z}{12} + O(z^3)$
This is about it. I don't know what the next step is...
Yep, in this case it is. The residue is the coefficient of the term $\displaystyle (z-z0)^{-1}$ in the Laurent Series expansion of a function about a point $\displaystyle z0$ In this case, it happens to be -1 as ur simplifications shows. here $\displaystyle z0=0$
The residue at $\displaystyle z = 2 \pi i$ will also need to be calculated (since the closed contour $\displaystyle |z| = 8$ encloses both the isolated singular points $\displaystyle z = 0$ and $\displaystyle z = 2 \pi i$ of $\displaystyle \displaystyle \frac{1}{1 - e^z}$).