# 9.6: Residue at ∞

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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.

## 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: $$w = 1/z$$.

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!

This page titled 9.6: Residue at ∞ is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.