Method of residues

  • #1

Homework Statement



evaluate using the method of residues [itex]\int^{\pi}_{-\pi}[/itex] [itex]\frac{d\theta}{1+sin^2\theta}[/itex] (=[itex]\pi\sqrt{2})[/itex]

Homework Equations





The Attempt at a Solution

[itex] sin^2 \theta = \frac{1}{2i}(z^2 - \frac{1}{z^2}) [/itex] so our integral becomes [itex]\int \frac{2iz^2}{z^4+2iz^2-1}\frac{dz}{iz} [/itex]

but before i continue on from here im not sure what the story with the limits. Normally i've just seen limits of 0 to 2[itex]\pi[/itex] so should iI put a [itex]\frac{1}{2}[/itex] in from of the integral?
 

Answers and Replies

  • #2
sorry got that wrong [itex] sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2 [/itex] should be [itex] \int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}[/itex] but still unsure about my limits....
 
  • #3
329
0
As theta varies from -pi to pi, how much of the unit circle does it sweep out?
 
  • #4
I think its Half a sweep, thats why i think i put a half in front of the integral
 
  • #5
329
0
Are you sure?
 
  • #6
a whole sweep is 2[itex]\pi[/itex], half a sweep is o to [itex]\pi[/itex] so is [itex]\pi[/itex] to -[itex]\pi[/itex] half a sweep...backwards? so i put -[itex]\frac{1}{2}[/itex] in front of it?
 
  • #8
329
0
What is pi - (-pi)?
 
  • #9
uart
Science Advisor
2,776
9
sorry got that wrong [itex] sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2 [/itex] should be [itex] \int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}[/itex] but still unsure about my limits....
^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.
 
  • #11
^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.
[itex]\int \frac{-4z^2}{z^4-2z^2+2}\frac{dz}{iz} [/itex]

thanks a million
 
  • #12
**** thats not right either
 
  • #13
Now i'm getting [itex] \int \frac{-4z^2}{z^4-6z^2+2}\frac{dz}{iz} [/itex] which doesnt work out that easy. So im guessing I got it wrong again. im going to use formula and see what it throws up anywho.thanks guys. Do either of you have a clue what my other post (jordans lemma) is about? i havent a clue...
 
  • #14
329
0
I think you need to double-check your algebra, one more time.
 
  • #15
yes your right, it should be [itex] \int \frac{-4z^2}{z^4-6z^2+1}\frac{dz}{iz} [/itex]
 

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