# Complex Analysis questions

1. Oct 14, 2013

### d3nat

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

2. Relevant equations

$f(z) = sinh(z)$ around origin

3. The attempt at a solution

$sinh(z +\frac{1}{z}) = \sum_{-infty}^\infty A_nz^n$
where
$A_n = \frac{1}{2\pi i} \oint \frac{sinh(z'+\frac{1}{z'})}{z'^{n+1}} d'$

Let c = unit circle, $z'=e^{i \theta}$
$dz' = ie^{i\theta} d\theta$

using Euler relationships

$= \frac{1}{2\pi i} i \oint \frac{sinh(e^{i\theta}+e^{-i\theta}}{(e^{i\theta})^{n+1}} d\theta'$
Cancel out the $e^{i\theta}$ on top and bottom

$= \frac{1}{2\pi} \oint \frac{sinh(2cos\theta)}{e^{in\theta}} d\theta$

$= \frac{1}{2\pi} \int_0^{2\pi} sinh(2cos\theta) e^{-in\theta} d\theta$

$e^{-in\theta} = cos(n\theta)-isin(n\theta)$

Then I said that only the $cos(n\theta)$ part mattered because I stated the bounds of the integral were from $0 to 2\pi$

$= \frac{1}{2\pi} \int_0^{2\pi} sinh(2cos\theta) cos(n\theta) d\theta$

However, I've tried many ways of simplifying this and I can't get an answer. Any suggestions? I'm so stumped... :/

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

$\int_0^{2\pi} \frac\{sin^2{\theta)}{a+bcos(\theta)} d\theta = \frac{2\pi}{b^2} [a-(a^2-b^2)^{frac{1}{2}}]$

2. Relevant equations

$d\theta = \frac{-i dz}{z}$
$sin^2(\theta) = \frac{-1}{4}(z^2+\frac{1}{z^2}-2)$
$cos(\theta) = \frac{1}{2}(z+\frac{1}{z})$

3. The attempt at a solution

Subst. all in and assuming unit circle:

$= -i \oint \frac{dz}{z} \frac{\frac{-1}{4}(z^2+\frac{1}{z^2}-2)}{a+ \frac{b}{2}(z+\frac{1}{z})} dz$

rearranging

$= \frac{i}{2} \oint \frac{dz}{z} \frac{(z^2+\frac{1}{z^2}-2)}{bz^2+2az+b} dz$

Then I solved this using the quadratic equations, so I had two roots (for the denominator)

I know I have to use residue theory, but I've never had to do it with something in the numerator. Is it still the same method? I'm confused...

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

2. Relevant equations

3. The attempt at a solution

2. Oct 14, 2013

### Dick

I don't get where you are going here. sinh(z) has a Taylor series around z=0. It's (e^z-e^(-z))/2. There are no negative powers of z at all.

3. Oct 14, 2013

### d3nat

I mean, yah, I can solve it with a Taylor series, but I wasn't sure how to solve using a Laurent series, which is what the problem asks.

4. Oct 14, 2013

### Dick

A Laurent series is just a Taylor series possibly includes term of negative degree. sinh(z) doesn't have any. The Taylor series is the same as the Laurent series.

5. Oct 15, 2013

### d3nat

Ohhh. Well then, I feel dumb.
My professor made them out to be completely different, so I thought it was two different methods of solving.
Thanks!