9.6: Residue at ∞

The residue at $\infty$ is a clever device that can sometimes allow us to replace the computation of many residues with the computation of a single residue.

Suppose that $f$ is analytic in $C$ except for a finite number of singularities. Let $C$ be a positively oriented curve that is large enough to contain all the singularities.

Figure $\PageIndex{1}$: All the poles of $f$ are inside $C$. (CC BY-NC; Ümit Kaya)

Definition: Residue

We define the residue of $f$ at infinity by

$$\text{Res} (f, \infty) = -\dfrac{1}{2\pi i} \int_C f(z)\ dz. \nonumber$$

We should first explain the idea here. The interior of a simple closed curve is everything to left as you traverse the curve. The curve $C$ is oriented counterclockwise, so its interior contains all the poles of $f$. The residue theorem says the integral over $C$ is determined by the residues of these poles.

On the other hand, the interior of the curve $-C$ is everything outside of $C$. There are no poles of $f$ in that region. If we want the residue theorem to hold (which we do –it’s that important) then the only option is to have a residue at $\infty$ and define it as we did.

The definition of the residue at infinity assumes all the poles of $f$ are inside $C$. Therefore the residue theorem implies

$$\text{Res} (f, \infty) = -\sum \text{ the residues of } f. \nonumber$$

To make this useful we need a way to compute the residue directly. This is given by the following theorem.

Theorem $\PageIndex{1}$

If $f$ is analytic in $C$ except for a finite number of singularities then

$$\text{Res} (f, \infty) = -\text{Res} \left(\dfrac{1}{w^2} f(1/w), 0\right). \nonumber$$

Proof

The proof is just a change of variables: $w = 1/z$.

Figure $\PageIndex{1}$: Changing variables. (CC BY-NC; Ümit Kaya)

Change of variable: $w = 1/z$

First note that $z = 1/w$ and

$$dz = -(1/w^2)\ dw. \nonumber$$

Next, note that the map $w = 1/z$ carries the positively oriented $z$-circle of radius $R$ to the negatively oriented $w$-circle of radius $1/R$. (To see the orientation, follow the circled points 1, 2, 3, 4 on $C$ in the $z$-plane as they are mapped to points on $\tilde{C}$ in the $w$-plane.) Thus,

$$\text{Res} (f, \infty) = -\dfrac{1}{2\pi i} \int_C f(z)\ dz = \dfrac{1}{2\pi i} \int_{\tilde{C}} f(1/w) \dfrac{1}{w^2}\ dw \nonumber$$

Finally, note that $z = 1/w$ maps all the poles inside the circle $C$ to points outside the circle $\tilde{C}$. So the only possible pole of $(1/w^2) f(1/w)$ that is inside $\tilde{C}$ is at $w = 0$. Now, since $\tilde{C}$ is oriented clockwise, the residue theorem says

$$\dfrac{1}{2\pi i} \int_{\tilde{C}} f(1/w) \dfrac{1}{w^2}\ dw = -\text{Res}(\dfrac{1}{w^2} f(1/w), 0) \nonumber$$

Comparing this with the equation just above finishes the proof.

Example $\PageIndex{1}$

Let

$$f(z) = \dfrac{5z - 2}{z(z - 1)}. \nonumber$$

Earlier we computed

$$\int_{|z| = 2} f(z)\ dz = 10 \pi i \nonumber$$

by computing residues at $z = 0$ and $z = 1$. Recompute this integral by computing a single residue at infinity.

Solution

$$\dfrac{1}{w^2} f(1/w) = \dfrac{1}{w^2} \dfrac{5/w - 2}{(1/w)(1/w - 1)} = \dfrac{5 - 2w}{w(1 - w)}. \nonumber$$

We easily compute that

$$\text{Res} (f, \infty) = -\text{Res} (\dfrac{1}{w^2} f(1/w), 0) = -5. \nonumber$$

Since $|z| = 2$ contains all the singularities of $f$ we have

$$\int_{|z| = 2} f(z)\ dz = -2\pi i \text{Res} (f, \infty) = 10 \pi i. \nonumber$$

This is the same answer we got before!