# Contour integral, taylor and residue theory question

1. Aug 17, 2009

### Ian_Brooks

1. The problem statement, all variables and given/known data

http://img243.imageshack.us/img243/4339/69855059.jpg [Broken]
I can't seem to get far. It makes use of the Exponentional Taylor Series:

2. Relevant equations

http://img31.imageshack.us/img31/6163/37267605.jpg [Broken]

3. The attempt at a solution
taylor series expansions for cos and 1/z i assume - and stick it into the residual theorem, but i need to get it into the laurent series form first so that i can find out b1. I think it's like pole order 4 or 5 so it's going to be a pain

Last edited by a moderator: May 4, 2017
2. Aug 18, 2009

### Ian_Brooks

my attempt at a solution
is this on the right track?
http://img219.imageshack.us/img219/2425/image123t.jpg [Broken]

Last edited by a moderator: May 4, 2017
3. Aug 18, 2009

### Cyosis

It has been a while but it does look correct to me. You can do it a little bit easier though if you know the series expansion of the cosine.

$$z^5 \cos(\frac{1}{z})=z^5 \sum_{k=0}^{\infty} (-1)^k \frac{\left(\frac{1}{z}\right)^{2k}}{(2k)!}=...+\frac{-1}{6!z}+...$$

Last edited: Aug 18, 2009
4. Aug 18, 2009

### HallsofIvy

Why is the Taylor's series for ez a "relevant" equation? Use the Taylor's series for cos(z), replace z by 1/z so that you get a power series in negative powers of z (a "Laurent series") and then multiply by $z^5$ so you have a power series with highest power z5 and decreasing. It should be easy to see that, integrating term by term, every term gives 0 except the z-1 term.