# Complex integral using Cauchy's Theorem?

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?

-
Looks right to me. – anon Apr 23 '12 at 23:07
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. – rotskoff May 31 '12 at 18:54

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