10.2 Cauchy's Residue Theorem

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Cauchy's residue theorem tells us that - under certain broad circumstances - we can calculate contour integrals simply by calculating coefficients in Laurent series. Specifically, we will need the coefficient $a_{-1}$ which from now on has a special name.

Definition (Resdiue)

If $f:\mathsf{Ann}(b,r,R)\to \mathbb{C}$ is holomorphic then the coefficient $a_{-1}$ in its Laurent series centered at $b$ is called the residue of $f$ at $b$. Write $\mathsf{Res}(f,b)$ for the residue of $f$ at $b$.

The setup for Cauchy's residue theorem is a simply connected domain $D$ and a finite set of points $b_1,\dots ,b_n\in D$. Recall that $D$ is simply connected if $\mathsf{wind}(\mathrm{\Gamma },z)=0$ whenever $\mathrm{\Gamma }$ is a closed contour in $D$ and $z\in \mathbb{C}\setminus D$. Roughly speaking, simply connected means that $D$ does not have any holes in it.

Figure 1: The setting for Cauchy's residue theorem is a simply connected domain and marked points $b_1,\dots ,b_n$.

The connection between residues and contour integration comes from Laurent's theorem: it tells us that $$\mathsf{Res}(f,b)=a_{-1}={\displaystyle \frac{1}{2\pi i}}\underset{\gamma }{\int }f(z)\mathsf{d}z={\displaystyle \frac{1}{2\pi i}}\underset{0}{\overset{2\pi }{\int }}f(b+s{e}^{it})i{e}^{it}\mathsf{d}t$$ when $\gamma (t)=b+s{e}^{it}$ on $[0,2\pi ]$ for any $r<s<R$. Combining this with the generalized Cauchy theorem gives Cauchy's celebrated residue theorem.

Theorem (Cauchy's Residue Theorem)

Let $D\subset \mathbb{C}$ be a simply connected domain. Suppose $f:D\setminus \{b_1,\dots ,b_n\}\to \mathbb{C}$ is holomorphic with poles at each of the $b_i$. For any closed contour $\mathrm{\Gamma }$ in $D\setminus \{b_1,\dots ,b_n\}$ we have $$\underset{\mathrm{\Gamma }}{\int }f=2\pi i\sum _{j=1}^n\mathsf{wind}(\mathrm{\Gamma },b_j)\mathsf{Res}(f,b_j)$$

Proof:

The closed contour $\mathrm{\Gamma }$ has a winding number $\mathsf{wind}(\mathrm{\Gamma },b_j)$ around each of the points $b_j$. Whenever $\mathsf{wind}(\mathrm{\Gamma },b_j)$ is positive define ${\gamma }_j(t)=b_j+r_j{e}^{-it}$ on $[0,2\pi |\mathsf{wind}(\mathrm{\Gamma },b_j)|]$ with $r_j$ chosen so small that the circle centered at $b_j$ of radius $2r_j$ does not contain any other $b_k$. Similarly, whenever $\mathsf{wind}(\mathrm{\Gamma },b_j)$ is negative, define ${\gamma }_j(t)=b_j+r_j{e}^{it}$ on $[0,2\pi |\mathsf{wind}(\mathrm{\Gamma },b_j)|$ with $r_j$ again so small that no other $b_k$ is contained on or within ${\gamma }_j$. We have $$\mathsf{wind}(\mathrm{\Gamma },b_j)+\mathsf{wind}({\gamma }_1,b_j)+\cdots +\mathsf{wind}({\gamma }_j,b_j)+\cdots +\mathsf{wind}({\gamma }_n,b_j)=0$$ for all $1\le j\le n$. We can therefore apply the generalized Cauchy theorem to deduce that $$\underset{\mathrm{\Gamma }}{\int }f+\underset{{\gamma }_1}{\int }f+\cdots +\underset{{\gamma }_n}{\int }f=0$$ and calculate by hand that $$\begin{array}{rl}\underset{{\gamma }_j}{\int }f=& \pm \underset{0}{\overset{2\pi |\mathsf{wind}({\gamma }_j,b_j)|}{\int }}f(b_j+r_j{e}^{\pm it})i{e}^{\pm it}\mathsf{d}t\\ =& \mathsf{wind}({\gamma }_j,b_j)\underset{0}{\overset{2\pi }{\int }}f(b_j+r_j{e}^{\pm it})i{e}^{\pm it}\mathsf{d}t\\ =& -2\pi i\mathsf{wind}(\mathrm{\Gamma },b_j)\mathsf{Res}(f,b_j)\end{array}$$ for all $1\le j\le n$. $◻$

The applicability of the theorem goes hand in hand with our ability to calculate residues. We will cover the calculation of residues in the next section.