# Application of Cauchy's residue theorem

1. Aug 21, 2013

### lyranger

Really need help for this one. Cheers!

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

Question: calculate function z/(1-cos z) integrated in ac ounterclockwise circular contour given by |z-2pi|= 1

2. Relevant equations

3. The attempt at a solution

Clearly the pole in the given contour is 2pi. But the problem is: if it's a simple pole, then apply formula

we have Residue=lim (z-2pi) * z/(1-cos z) where z->2pi. This limit does not exist.

So I reckon 2pi might be a higher order pole but this actually makes no sense and even if it's true,

there is a really nasty differentiation.

Any thoughts.?

Last edited: Aug 21, 2013
2. Aug 21, 2013

### dirk_mec1

Please use the standard format of PF!

3. Aug 21, 2013

### vanhees71

What you need is a Laurent expansion around the pole at $z=2 \pi$. For your function
$$f(z)=\frac{z}{1-\cos z}$$
you have
$$1-\cos z=\frac{(z-2 \pi)^2}{2}+\mathcal{O}[(z-2 \pi)^4].$$
You need the single-order pole contribution. You can get this by either using the standard formula
$$\text{res}_{z \rightarrow 2 \pi} f(z)=\lim)_{z \rightarrow 2 \pi} \frac{\mathrm{d}}{\mathrm{d} z} (z-2 \pi)^2 f(z),$$
which is a bit cumbersome here, or you use the series expansion of $1-\cos z$ as given above and use $z=(z-2 \pi)+2 \pi$ in the numerator.

4. Aug 21, 2013

### lyranger

sorry for not using LaTex, havent used this forum in ages.

And thanks so much for the help! very smart move!!

Correct me if im wrong, the answer is 4i*pi?

Last edited: Aug 21, 2013
5. Aug 22, 2013

Looks good!