Caunchy's Integral Formula w/ trigonometric pole

[bmanbs2]

Homework Statement

Solve the contour integral
\int_{\gamma}\frac{2+z}{1-cos\left(z\right)}dz
along a counter-clockwise circle of radius 1 centered at z=6

Homework Equations

Cauchy's integral formula
f\left(z\right)=\frac{1}{i*2\pi}\int_{\gamma}\frac{f\left(w\right)}{w-z}dw

The Attempt at a Solution

I know that the integrand has a pole at 2*pi inside the contour, but since this pole is inside a trig function, I don't know how to isolate it and find an f(z) to use in Cauchy's integral formula. Any tips?

 

[vela]

Try expanding the integrand as a Laurent series about z=2π.

 

[bmanbs2]

That's not really helping. I can calculate

[PLAIN]http://www.sharpermath.com/cgi/mimetex.cgi?\frac{1+z}{1-cos(z)}%20=%20\frac{1+z}{(z-2pi)[{\frac{(z-2\pi)}{2!}-\frac{(z-2\pi)^3}{4!}+\frac{(z-2\pi)^5}{6!}-+...]}

But I guess I don't see how this helps. Sorry we've never used Laurent series to solve path integrals in my class, just found the series themselves.

 

[vela]

That's a good start. Now write it like this:

\frac{1+z}{1-\cos z} = \frac{(1+2\pi)+(z-2\pi)}{(z-2\pi)^2\left[\frac{1}{2!}-\frac{(z-2\pi)^2}{4!}+\cdots\right]}

Everything's written in terms of (z-2π). You want to get a series in powers of (z-2π).

The reason you want to do this is because the contour integral is equal to 2\pi i times the residue at the pole at z=2π, and the residue is the coefficient of the (z-2\pi)^{-1} term. Note this means you don't actually have to calculate the whole series but just the one coefficient.

 

[bmanbs2]

OK I understand now, but is there an easier way to find that one coefficient other than lots and lots of algebraic manipulation?

 

[vela]

You can use whatever method you want to find the residue at the pole. I think using a Laurent series often ends up being the least tedious, but you have to learn how to work with series. The key is to only look at the terms that will give you the one you want. For instance, in this problem, you have

\frac{1+z}{1-\cos z} = \frac{(1+2\pi)}{(z-2\pi)^2\left[\frac{1}{2!}-\frac{(z-2\pi)^2}{4!}+\cdots\right]} + \frac{(z-2\pi)}{(z-2\pi)^2\left[\frac{1}{2!}-\frac{(z-2\pi)^2}{4!}+\cdots\right]}

The whole first term only has even powers of (z-2π), so it can not contribute to the residue. You can completely ignore it. The second term is proportional to (z-2π)^-1, so you only need the very first term of the series.

If you have a simple pole, it's sometimes straightforward to just multiply by (z-z₀) and take the limit. In this problem, you had a second-order pole, which means you'd have to differentiate, which means differentiating a quotient. I think it ends up being more algebra than figuring out the series.

 

[bmanbs2]

OK I understand what needs to be done and found the answer, so thank you so much. For posterity's sake however, I considered the term,

\frac{1}{(z-2\pi)\left[\frac{1}{2!}-\frac{(z-2\pi)^2}{4!}+\cdots\right]},

Found the closed-form of the series, found its reciprocal, and turned it back into a series. The first term of this series was the term I needed, and the term was 2.