Complex Integral Along a Path with Residue Theorem

[Metric_Space]

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.

 

[HallsofIvy]

Yes, the function
\frac{z^2- 1}{z^2+ 1}= \frac{z^2- 1}{(z+ i)(z- i)}
has simple poles at z= i and z= -i.

That is true because
\lim_{z\to i}\frac{z^2- 1}{(z+i)(z-i)}
does not exist while
\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
and similarly at z= -i.

 

[Metric_Space]

How can I use these facts to evaluate the integral?

 

[hunt_mat]

You do know Cauchy's residue theorem right?
<br /> \oint_{\gamma}f(z)dz=2\pi i\sum\textrm{Res}(f(z);\gamma )<br />
Calculate the residues and you have your answer.

 

[Metric_Space]

so the integral would be zero in this case since the residues are -i and i?

 

[hunt_mat]

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?

 

[Metric_Space]

No, not sure how to...

 

[hunt_mat]

To calculate the reside at i for example, compute:
<br /> \textrm{Res}(f(x);i)=\lim_{z\rightarrow i}(z-i)f(z)<br />

 

[Metric_Space]

Thanks...I think that helps!