# Complex Analysis questions

Laurent series

## Homework Equations

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

## 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... :/

## Homework Statement

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}}]##

## Homework 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})##

## 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...

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper

Laurent series

## Homework Equations

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

## 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... :/

## Homework Statement

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}}]##

## Homework 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})##

## 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...

## The Attempt at a Solution

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.

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.

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.

Dick