Cauchy's Residue Theorem

Theorem
Let $U$ be a simply connected open subset of the complex plane $\C$.

Let $a_1, a_2, \dots, a_n$ be finitely many points of $U$.

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

Let $L = \partial U$ be oriented counterclockwise.

Then:
 * $\displaystyle \oint_L \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$

Proof
Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.

Let $U_i \cap U_j = \varnothing$ for all $i \ne j$.

By Existence of Laurent Series, around each $a_k$ there is an expansion:


 * $\displaystyle \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$

convergent in $U_k$.

Write:


 * $\displaystyle X = \bigcup_{i \mathop = 1}^n U_i$

Then, by Contour Integral of Concatenation of Contours:


 * $\displaystyle \oint_L \map f z \rd z = \oint_{L \setminus X} \map f z \rd z + \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$

As all poles of $f$ in $L$ are contained in $X$, $f$ is holomorphic on $L \setminus X$, and so by the Cauchy-Goursat Theorem:


 * $\displaystyle \oint_{L \setminus X} \map f z \rd z = 0$

Giving:


 * $\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$

We have:

So:


 * $\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$