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Complex Analysis questions

  • Thread starter d3nat
  • Start date
  • #1
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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...

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

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...

Homework Statement





Homework Equations





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.
 
  • #3
102
0
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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #5
102
0
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!
 

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