# Thread: Complex integral

1. ## Complex integral

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

2. Partial fractions is good.

$\displaystyle \displaystyle\oint_{\gamma}\frac{1}{z^{2}-2z}\,dz=-\frac{1}{2}\oint_{\gamma}\frac{1}{z}\,dz+\frac{1}{ 2}\oint_{\gamma}\frac{1}{z-2}\,dz$
The first integral on the RHS is zero, because the integrand is analytic inside and on the entire contour. Just invoke the Cauchy-Goursat theorem.

For the second integral, there are multiple ways to do it. The easiest is residues, but if you haven't studied that yet, you could do a parametrization. However, I wouldn't pick your parametrization. The first thing I would do is a change of variables: $\displaystyle w=z-2.$ Let $\displaystyle \tilde{\gamma}$ be the unit circle centered at the origin. Then

$\displaystyle \displaystyle\frac{1}{2}\oint_{\gamma}\frac{1}{z-2}\,dz=\frac{1}{2}\oint_{\tilde{\gamma}}\frac{1}{w }\,dw.$

Can you continue from here?