Q: Evaluate $\displaystyle \int_{\gamma}\frac{1}{z^{2}-2z}dz$, where $\displaystyle \gamma$ is the circle of raduis 1 centered at 2 traveled once conterclockwise.

First thing I did was use partial fraction decomp to break up the integrand. I ended up with

$\displaystyle \int_{\gamma}\frac{1}{z^{2}-2z}dz=\frac{-1}{2}\int_{\gamma}\frac{1}{z}+\frac{1}{2}\int_{\ga mma}\frac{1}{z-2}$

$\displaystyle \gamma=\{z||z-2|=1\}$ traveled counter clockwise. So, I can let $\displaystyle x=cos(t)$ and $\displaystyle y=sin(t)$ for $\displaystyle 0\leq\\t<2\pi$. This gives the parametrisized equation

$\displaystyle \gamma(t)=(cos(t)-2)^{2}+sin(t)^{2}$.

From here I am not entirely sure what I should do. I tried breaking the integral up into is real and imaginary components and using greens thereom, but that seems like too much. I would like to use the fundamental thereom of contour integrals, but the integrand is not defined on at 2, so I don't think I can use it.

Doing it the way I just mentioned I ended up with 0 as my answer.

Thanks