Help with complex integral

  • #1

Homework Statement



Evaluate the integral along the path given:

integral(along a(t) of (b^2-1)/(b^2+1) db ) where a(t)=2*e^(it) , 0 <= t <= 2*pi


Homework Equations



none

The Attempt at a Solution



I am thinking of using the Residue Theorem.

I think there are poles at -i and +i.

But not sure what to do to show this explicitly.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Yes, the function
[tex]\frac{z^2- 1}{z^2+ 1}= \frac{z^2- 1}{(z+ i)(z- i)}[/tex]
has simple poles at z= i and z= -i.

That is true because
[tex]\lim_{z\to i}\frac{z^2- 1}{(z+i)(z-i)}[/tex]
does not exist while
[tex]\lim_{z\to i}(z- i)\frac{z^2- 1}{(z+i)(z-i)}= \lim_{z\to i|}\frac{z^2- 1}{z+ i}= i[/tex]
and similarly at z= -i.
 
  • #3
How can I use these facts to evaluate the integral?
 
  • #4
hunt_mat
Homework Helper
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You do know Cauchy's residue theorem right?
[tex]
\oint_{\gamma}f(z)dz=2\pi i\sum\textrm{Res}(f(z);\gamma )
[/tex]
Calculate the residues and you have your answer.
 
  • #5
so the integral would be zero in this case since the residues are -i and i?
 
  • #6
hunt_mat
Homework Helper
1,742
26
The contour is a circle of radius 2 centred on the origin, these poles certainly lie within that contour. Have you calculated the residues for the poles?
 
  • #7
No, not sure how to...
 
  • #8
hunt_mat
Homework Helper
1,742
26
To calculate the reside at i for example, compute:
[tex]
\textrm{Res}(f(x);i)=\lim_{z\rightarrow i}(z-i)f(z)
[/tex]
 
  • #9
Thanks....I think that helps!
 

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