3.4 Residues

Definition-Proposition 3.22

Suppose \(f\) has an isolated singularity at \(z_0\in \mathbb {C}\). Then, there exists some \(r{\gt}0\) such that \(f\) is analytic on the punctured disk \(D^*(z_0, r) = \{ z\in \mathbb {C}: 0 {\lt} |z-z_0| {\lt} r\} \). Write \(f(z) = \sum _{i\in \mathbb {Z}} a_i (z-z_0)^i\) for the Laurent series of \(f\) on \(D^*(z_0, r)\). We call \(a_{-1}\) the residue of \(f\) at \(z_0\) and write \(\operatorname*{\mathrm{Res}}_{z=z_0} f := a_{-1}\).

Proof ▼

The assertion in the proposition about the existence of a punctured disk where \(f\) is analytic is obvious from the definition of isolated singularity.

Proposition 3.23

If \(f\) has a removable singularity at \(z_0\), then \(\operatorname*{\mathrm{Res}}_{z = z_0} f(z) = 0\).

Proof ▼

Obvious.

Examples 3.24

We have \(\operatorname*{\mathrm{Res}}_{z=0} 1/z = 1\). We have \(\operatorname*{\mathrm{Res}}_{z = z_0} 1/z = 0\) for all \(z_0 \neq 0\). We have \(\operatorname*{\mathrm{Res}}_{z = 0} 1/z^2 = 0\).

Proposition 3.25

Suppose \(f\in A(D^*(z_0,r))\), where \(z_0\in \mathbb {C}\) and \(r {\gt} 0\). Then

\begin{equation} \label{eres} \operatorname*{\mathrm{Res}}_{z=z_0} f(z) = \frac{1}{2\pi i} \int _C f(z)\, dz, \end{equation}

3.26

where \(C\) is any positively oriented simple closed contour around \(z_0\) that lies in \(D^*(z_0,r)\).

Proof ▼

In fact, \(a_{-1}\) is, by definition, the right-hand-side of 3.26.

Theorem 3.27 Cauchy’s Residue Theorem

Let \(U\subseteq \mathbb {C}\) be an open set and let \(S = \{ z_1,\ldots , z_n\} \subseteq U\) be a finite set of \(n\) points points in \(U\). Let \(C\) be a simple, closed, counterclockwise contour in \(U\) such that the points \(S\) lies inside of \(C\). Then

\[ \oint _C f(z)\, dz = 2\pi i\sum _{k=1}^n \operatorname*{\mathrm{Res}}_{z=z_k} f(z). \]

Sketch ▼

Draw disjoint closed disks \(D_i\) around each of the points \(z_i\) inside of \(C\), and let \(C_i\) denote the circle at the boundary of \(D_i\) oriented counterclockwise. Then, as in the proof of Cauchy-Goursat, you can break the region \(R\) bounded by \(C\) into two regions \(R_1\) and \(R_2\) on which \(f\) is analytic and the disks \(D_i\). Since \(f\) is analytic on the regions \(R_i\), we get that

\[ \oint _C f(z)\, dz = \sum _{k=1}^n \oint _{C_i} f(z)\, dz = 2\pi i\sum _{k=1}^n \operatorname*{\mathrm{Res}}_{z=z_k} f(z). \]

Example 3.28

Let’s compute \(\int _C \frac{2z-1}{z(z+1)}\, dz\) where \(C\) is the contour given by \(|z| = 2\). Working in partial fractions, we see that

\[ \frac{2z-1}{z(z+1)} = \frac{3}{z+1} - \frac{1}{z}. \]

So \(\operatorname*{\mathrm{Res}}_{z=0} \frac{2z-1}{z(z+1)} = -1\), while \(\operatorname*{\mathrm{Res}}_{z=-1} \frac{2z-1}{z(z+1)} = 3.\) From this, it follows that

\[ \int _C \frac{2z-1}{z(z+1)}\, dz = 2\pi i(3 - 1) = 4\pi i. \]