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Let $\Gamma$ be a circumference with center $3i$ and radius $5$ oriented counter-clockwise. Calculate: $$\displaystyle\oint_{\Gamma} \frac{z}{(z^2 -2z)(z^2 - 4z + 13)} dz$$

At a first glance I thought it was a line integral but then I realized the function $f(z)$ has only one unknown ($z$), while it should have 2 such that I can parameterized it. Any hint is appreciated.

Thanks,
rubik

P.S. Sorry for the english: the problem is a translation from another language and I don't know whether I used the correct terms.

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You need to use Cauchy Residue Theorem. In your case, the integrand has poles at $z = 0,2,2 \pm 3i$. The poles $0,2,2+3i$ (except $2-3i$) are all inside the circle over which you are integrating. Hence you need to compute the residue at all these poles and add them up and multiply by $2 \pi i$ to get the value of the integral.

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    Thank you! Poles are $f(z)$ singularities, aren't they?2012-06-24
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    @rubik Pole is a type of singularity. In your case, all the singularities are poles. http://en.wikipedia.org/wiki/Pole_(complex_analysis)2012-06-24
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    It is better to work out the residue at infinity.2012-06-24
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    @Marvis, The pole $\,z=2-3i\,$ is out the circle $\,\Gamma:|z-3i|\leq 5\,$2012-06-24
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    @DonAntonio Ah. Yes. Thanks for pointing it out.2012-06-24
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    @yzhao: What do you mean? Currently I calculate residues like this: $Res_{z=c}(f) = lim_{z->c}(z - c)f(z)$2012-06-25
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    Can someone please confirm that the result is $\dfrac{\pi i}{9}$? I reasoned as follows: $Res_{z=0}(f) = 0$, $Res_{z=2}(f) = \dfrac{1}{9}$, $Res_{z=2 + 3i}(f) = -\dfrac{1}{18}$ so result is $2\pi i \sum_{k=1}{3} Res_{k} = \dfrac{2\pi i}{18} = \dfrac{\pi i}{9}$2012-06-25
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    @rubik:A better way to compute this integral is to change the contour of integral.Consider a circle $C$ centered at origin with radius $R$.Let R tend to infinity.It is easy to work out that $I=\int_{C}f(z)dz $ is zero.But $I$ is the sum of your integral and the residue at $2-3i$,so you just need to compute one residue.The result is $\pi i/9$.2012-06-26
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    @yzhao: Thank you for explaining an alternative approach: very interesting!2012-06-26