# Help with complex integral

1. Jun 21, 2011

### Metric_Space

1. The problem statement, all variables and given/known data

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

2. Relevant equations

none

3. 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.

2. Jun 22, 2011

### HallsofIvy

Staff Emeritus
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.

3. Jun 22, 2011

### Metric_Space

How can I use these facts to evaluate the integral?

4. Jun 22, 2011

### hunt_mat

You do know Cauchy's residue theorem right?
$$\oint_{\gamma}f(z)dz=2\pi i\sum\textrm{Res}(f(z);\gamma )$$

5. Jun 22, 2011

### Metric_Space

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

6. Jun 22, 2011

### 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?

7. Jun 22, 2011

### Metric_Space

No, not sure how to...

8. Jun 22, 2011

### hunt_mat

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

9. Jun 22, 2011

### Metric_Space

Thanks....I think that helps!