# Polar Integrals

1. Apr 25, 2012

### Fixxxer125

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

Evaluate the integral

2. Relevant equations
I can substitute and thus end up with

3. The attempt at a solution
I then expand the denominator out and end up with 1/
However I then assume I need to factorise the top line of that fraction as this will be the denominator in my integral, so I want to factorise this and find where it equals zero and thus where the poles are. However I'm unsure how to do this and need help! Thanks

2. Apr 25, 2012

### clamtrox

Just multiply upstairs and downstairs by z, carry z^2 in downstair inside the brackets, and solve the 2nd order equation.

3. Apr 25, 2012

### Fixxxer125

sorry I don't really understand what you are saying!

4. Apr 25, 2012

### clamtrox

$$I = -i \oint_{|z|=1} \frac{dz}{z(5-3(\frac{z-z^{-1}}{2i}))^2} = -i \oint_{|z|=1} \frac{ z dz}{\left[5z-3(\frac{z^2-1}{2i})\right]^2}$$
and then solve
$$5z-3(\frac{z^2-1}{2i}) = 0$$
(and remember how many of each pole there are!)

5. Apr 26, 2012

### Fixxxer125

thanks! I'll give it a go