Residues

[monomocoso]

How doe I find the residues of the following function:

\(\displaystyle \frac{z^a}{1+z+z^2}\) ?

I found the poles at \(\displaystyle \frac{-1}{2} \pm \frac {i\sqrt{3}}{2}\)

but don't know how to get the residues.

 

[galactus]

First, factor the denominator and find the poles. I will write them in the equivalent polar form.

Doing so, results in \(\displaystyle z=e^{\frac{2\pi i}{3}}, \;\ e^{-\frac{2\pi i}{3}}\)

\(\displaystyle \displaystyle \lim_{z\to e^{\frac{2\pi i}{3}}}(z-e^{\frac{2\pi i}{3}})\cdot \frac{z^{a}}{(z-(e^{\frac{2\pi i}{3}}))(z-(e^{\frac{-2\pi i}{3}}))}\)

\(\displaystyle =\frac{e^{\frac{2\pi i a}{3}}}{\sqrt{3}i}=\frac{sin(\frac{2a\pi}{3})}{\sqrt{3}}-\frac{cos(\frac{2\pi a}{3})}{\sqrt{3}}i\)

Do the same to find the other residue.

\(\displaystyle \frac{e^{\frac{-2\pi i}{3}}}{-\sqrt{3}i}=\frac{sin(\frac{2a\pi}{3})}{\sqrt{3}}+\frac{cos(\frac{2\pi a}{3})}{\sqrt{3}}i\)
Do you have to integrate this as well?. If so, it is only valid if 'a' is between -1 and 1, exclusive.