Find Residue at z=1 for Essential Singularity Integral - Homework Help

[asi123]

Homework Statement

Hey guys.

So, I've got this integral:

http://img18.imageshack.us/img18/5742/scan0017.jpg

And I want to find the residue at z=1. I know it's an essential singularity point.

I tried to calculate the residue, but I'm completely not sure about my solution.
Can I please have some help?

Thanks a lot.

Homework Equations

The Attempt at a Solution

 

[Dick]

I don't see anything wrong with your solution. The residue is the coefficient of 1/(z-1), which comes from the exp(1/(z-1)) factor. Everything else is analytic at z=1.

 

[asi123]

I don't see anything wrong with your solution. The residue is the coefficient of 1/(z-1), which comes from the exp(1/(z-1)) factor. Everything else is analytic at z=1.

Yeah, well, the thing that I wasn't sure about is, that I took the residue of exp(1/(z-1)) and the residue of the rest of the function, and than multiply them.
So, are you saying that I can do that?

Thanks a lot.

 

[Dick]

No, you can't multiply two residues. But you didn't do that. The residue of the analytic part is zero. All you did was write the first order pole in the form f(z)/(z-1) and put 1 into f(z) getting f(1) as the residue.

 

[asi123]

No, you can't multiply two residues. But you didn't do that. The residue of the analytic part is zero. All you did was write the first order pole in the form f(z)/(z-1) and put 1 into f(z) getting f(1) as the residue.

Ok, just to make it clear, let's say the residue of exp(1/z-1) was 2, then the residue of f(z) would have been double the residue I found?

Thanks.

 

[Dick]

Sure. If you'd had exp(2/(z-1)) instead, the residue of that is 2. And the residue of f(z) would be twice what it was before.

 

[Dick]

Yeah, well, the thing that I wasn't sure about is, that I took the residue of exp(1/(z-1)) and the residue of the rest of the function, and than multiply them.
So, are you saying that I can do that?

Thanks a lot.

I think I was being pretty thoughtless when I agreed with your solution. Sorry. Thinking about it some more I realized I was wrong. If you write u=z-1, then exp(1/(z-1))=exp(1/u)=1+1/u+1/(u^2*2!)+... But now you have to expand (cos(u+1)-1)^3/((u+1)^7*((u+1)^2+1)) in a power series in u. It's analytic so you can write it as a0+a1*u+a2*u^2+... Multiply it by the exp series and collect ALL of 1/u terms. So you don't just get a0*1/u. You also get a1*u/(u^2*2!) and a2*u^2/u etc etc etc. Now I think I don't have any idea how to compute the residue. Unless you can somehow compute a contour integration around u=0. Which looks equally hard. Sorry to have been so dumb. You were right to object about 'multiplying residues'.