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Thread: Complex integral

  1. #1
    Senior Member Danneedshelp's Avatar
    Apr 2009

    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.

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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    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?
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