Application of Cauchy's residue theorem

[lyranger]

Really need help for this one. Cheers!

Homework Statement

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

Homework Equations

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.?

 

[dirk_mec1]

Please use the standard format of PF!

 

[vanhees71]

Yes, please use LaTeX.

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.

 

[lyranger]

sorry for not using LaTex, haven't used this forum in ages.

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

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

 

[vanhees71]

Looks good!