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Calculate $\int_\gamma \frac{1}{z^2+2z+2}dz$

where $\gamma$ is the circle with centre 0 and radius 1.

I got the singularities of this as $(-1+i)$ and $(-1-i)$. And as the modulus of these is $\sqrt2$ they fall outside $\gamma$ so by Cauchy's Theorem this integral is 0. Does that look correct?

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    Yep. The quadratic formula gives $\frac{-2\pm \sqrt{-4}/2},$ which are the roots you've given. Thus, the function is holomorphic on the disc, so Cauchy's theorem says that the integral is 0, as the path is homologous to 0.2012-05-31

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Yes, it's definitively true. Your map is holomorphic "inside" and on the image of $\gamma$.