I'm a bit rusty on my complex integration, I'm trying to solve $\int(z+1)/(2z^3-3z^2-2z)dz$ over the unit circle.
By partial fractions I have $(z+1)/(2z^3-3z^2-2z)$ = $3/(10(z-2)) - 1/2z +2/5(1+2z)$
Now, for the first fraction I got residue 3/10, for the second clearly -1/2, but for the third I'm having trouble. The solution to this is supposed to be -3/10.
Any help is appreciated!
As Daniel Fischer pointed out, the residue of the first fraction inside the unit disc is zero, since it is analytic for $|z|<2$. The other two fractions have residues $-1/2$ and $1/5$ respectively, which add up to $-3/10$.