How can Laurent series be applied to complex analysis problems?

[d3nat]

Homework Statement

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...

 

[Dick]

Homework Statement

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...

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.

 

[d3nat]

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]

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.

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.

 

[d3nat]

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.

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!