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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.

    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

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


    The proof is just a change of variables: \(w = 1/z\).

    9.6 hidden.svg
    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 \(\PageIndex{1}\)


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

    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.


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

    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.