Residue Theorem

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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:

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

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

$\displaystyle \oint_{\partial \paren {U \setminus X} } \map f z \rd z = 0$

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:

\(\displaystyle \oint_{\partial U_k} \map f z \rd z\) \(=\) \(\displaystyle \oint_{\partial U_k} \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \oint_{\partial U_k} \paren {\sum_{j \mathop = -\infty}^{-2} c_j \paren {z - a_k}^j + \frac {c_{-1} } {z - a_k} + \sum_{j \mathop = 0}^\infty c_j \paren {z - a_k}^j} \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = -\infty}^{-2} \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z + \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z + \sum_{j \mathop = 0}^\infty \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z\) by Contour Integral of Power, terms where $j \ge 0$ and $j < -1$ vanish
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi i c_{-1}\) Contour Integral of Power
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi i \Res f {a_k}\) Definition of Residue

So:

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

$\blacksquare$