Calculate $\int_{C_1(0)} \frac{1}{z^2+2z+2} dz$ where $C_1(0)$ is the unit circle.

I'm going to apply the theorem:

Let $f(z)$ be analytic in a domain $\Omega$ and let $\Gamma$ be a Jordan curve inside $\Omega$ whose interior is contained in $\Omega$ so that $f(z)$ is analytic on and inside $\Gamma$. Then $\int_{\Gamma} f(z) dz=0$.

I proved that $f(z)$ is analytic everywhere except the points $-i-1$ and $i-1$ (it's a quotient of analytic functions).

Question: I'm trying to find a domain $\Omega$ that contains the Jordan curve $\Gamma$ but doesn't contain the points $-i-1$ and $i-1$. Would the open disk $D_R(0)$ where $1<R<\sqrt{2}$ work? It contains $\Gamma$ and its interior and $f(z)$ is analytic in it since it doesn't contain $-i-1$ and $i-1$ (because $\left|-1 \pm i - 0 \right|^2=2>R^2$.


1 Answer 1


Yes, everything is fine. $\frac{1}{z^2 + 2z + 2}$ has no poles inside the unit circle so the integral is zero.


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