Cauchy's Residue Theorem

Theorem
Let $U$ be a simply connected open subset of the complex plane $\C$, and $a_1, a_2, \dots, a_n$ finitely many points of $U$.

Let $f:U \to \C$ be analytic in $U \setminus\left\{{a_1, a_2, \dots, a_n}\right\}$.

If $L = \partial U$ oriented counterclockwise, then:


 * $\displaystyle \oint_L f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( a_k )$

where $\operatorname{Res}$ is the residue of $ f$ at a point.

Proof
Let $\left\{{U_1,\dots,U_n }\right\}$ be a set of open sets such that $U_i \subset U$, $a_i \in U_i$, $a_i \notin U_j$ for $i \neq j$, and $U_i \cap U_i = \varnothing$ for $i \neq j$.

By the existence of Laurent series, around each $a_k$ there is an expansion:


 * $\displaystyle \sum_{j=-\infty}^\infty c_j (z-a_k)^j$

convergent in $f$ in $U_k$. Then


 * $\displaystyle \int_{\partial U_k} f(z)dz = \int_{\partial U_k} \sum_{j=-\infty}^\infty c_j (z-a_k)^j dz = \sum_{j=-\infty}^\infty \int_{\partial U_k} c_j (z-a_k)^j dz$

By Contour Integrals of Polynomials, this is just


 * $\displaystyle = \underbrace{\dots}_0 + c_{-2} \underbrace{\int_{\partial U_k}  (z-a_k)^{-2} dz}_0 + c_{-1} \underbrace{\int_{\partial U_k}   (z-a_k)^{-1} dz}_{2\pi i} + \underbrace{\int_{\partial U_k}  c_0  dz}_0 +  c_1 \underbrace{\int_{\partial U_k}  (z-a_k) dz}_0 + \underbrace{\dots}_{0}$


 * $\displaystyle = c_{-1} 2 \pi i = 2 \pi i \operatorname{Res}( a_k )$