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I solved this problem in complex integrals.

Is my answer a correct ?

Here $z$ is a complex value:

$$ C:|z-1|=1 \ \ \ \ \ \mbox{integral path} $$

$$ \int_C\ \frac{2z^2-5z+1}{z-1}\ dz $$

My answer

$$ z=1+e^{i\theta} \ \ \ \ \frac{dz}{d\theta}=ie^{i\theta} $$

$$ \int_{0}^{2\pi}\ \frac{-e^{i\theta}+2e^{2i\theta}-2}{e^{i\theta}} \cdot\ ie^{i\theta} d\theta $$

$$ =\left[ -e^{i\theta}+ e^{2i\theta} -2i\theta \right]^{2\pi}_0=-4\pi i $$

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  • 2
    $\begingroup$ The approach seems correct. $\endgroup$ – glebovg Oct 20 '12 at 2:00
  • 1
    $\begingroup$ The answer is correct. $\endgroup$ – Mhenni Benghorbal Oct 20 '12 at 2:03
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If you are allowed to use the Cauchy's integral formula, then $$\int_{C} \frac{2 z^2 - 5x + 1}{z-1} dz = {2\pi i} \big( 2 z^2 - 5z + 1)_{z=1} = -4 \pi i,$$ showing that you did a great job.

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