Cauchy-Goursat 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 $$U - \left\{{a_1, a_2, \dots, a_n}\right\} \ $$.

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


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

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


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

convergent in $$f \ $$ in $$U_k \ $$. Then


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

By Contour Integrals of Polynomials, this is just


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


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