# Math Help - Integrate - getting stuck

1. ## Integrate - getting stuck

I want to solve a problem.

$\displaystyle \int_{|z|=8} \frac{dz}{(1-e^z)}$

I thought I could get something decent by doing a Taylor expantion. But no!

How should i solve it?

2. Originally Posted by liquidFuzz
I want to solve a problem.

$\displaystyle \int_{|z|=8} \frac{dz}{(1-e^z)}$

I thought I could get something decent by doing a Taylor expantion. But no!

How should i solve it?

Calculate the residue of $\frac{1}{1-e^z}$ at $z=0$ ...

Tonio

3. Yes, I know that. But how do I do that?

4. 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 $1/(1-e^{z}) = 1/(1-(1+z+z^{2}/2+...))$ (write the Taylor's expansion for $f(z)=e^{z}$)

$= 1/(-z-z^{2}/2-...) = (-1/z)(1/(1+z/2+...))$

Now reduce this into a Laurent series expansion about z=0 and hence find the residue as coefficient of $1/z$

5. 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.
$\displaystyle - \frac{1}{z} +\frac{1}{2} - \frac{z}{12} + O(z^3)$

This is about it. I don't know what the next step is...

6. HUH!!! I googled residue lauernt series and got an interesting result.

Originally Posted by wikipedia
The coefficient a−1 of the Laurent expansion of such a function is called the residue of ƒ(z)
So my residue is -1, or?

7. Yep, in this case it is. The residue is the coefficient of the term $(z-z0)^{-1}$ in the Laurent Series expansion of a function about a point $z0$ In this case, it happens to be -1 as ur simplifications shows. here $z0=0$

8. Originally Posted by liquidFuzz
HUH!!! I googled residue lauernt series and got an interesting result.

So my residue is -1, or?

Er...yes. You didn't know that before attempting to solve such an integral?!?

Tonio

9. Originally Posted by tonio
Er...yes. You didn't know that before attempting to solve such an integral?!?

Tonio
Actually, there might be just a tiny bit more mathematics I don't know...

10. Originally Posted by tonio
Calculate the residue of $\frac{1}{1-e^z}$ at $z=0$ ...

Tonio
The residue at $z = 2 \pi i$ will also need to be calculated (since the closed contour $|z| = 8$ encloses both the isolated singular points $z = 0$ and $z = 2 \pi i$ of $\displaystyle \frac{1}{1 - e^z}$).