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    Residue at a pole of order $m$ as $\lim_{\tau\to a}\frac{d^{m-1}}{d\tau^{m-1}}\bigl((\tau-a)^m F(\tau)\bigr)$, and $\oint_{|z|=4}\frac{e^\tau}{(\tau^2+\pi^2)^2}\,d\tau$

    P-X7WUF

    Area
    Complex Analysis
    Topics
    Contour Integration, Poles, Residues
    Seen at
    UGA
    Years
    2016
    Status
    draft
    1. Let \(F\) be an analytic function inside and on a simple closed curve \(C\), except for a pole of order \(m\geq 1\) at \(z=a\) inside \(C\). Prove that

    \[\frac{1}{2 \pi i}\oint_{C} F(\tau) d\tau = \lim_{\tau\rightarrow a} \frac{d^{m-1}}{d\tau^{m-1}}\big((\tau-a)^m F(\tau))\big).\]

    1. Evaluate \[\oint_{C}\frac{e^{\tau}}{(\tau^2+\pi^2)^2}d\tau\] where \(C\) is the circle \(|z|=4\).
    Solution

    Part (a) is missing the factor \(1/(m-1)!\). The correct residue formula is \[ \frac{1}{2\pi i}\oint_C F(\tau)\,d\tau =\operatorname{Res}_{\tau=a}F =\frac1{(m-1)!}\lim_{\tau\to a} \frac{d^{m-1}}{d\tau^{m-1}} \bigl((\tau-a)^mF(\tau)\bigr). \] This follows immediately from the Laurent expansion: the derivative on the right equals \((m-1)!\) times the coefficient of \((\tau-a)^{-1}\).

    For part (b), the poles inside \(|z|=4\) are the double poles \(\pm i\pi\). Their residues are \[ \frac1{4\pi^2}+\frac{i}{4\pi^3}, \qquad \frac1{4\pi^2}-\frac{i}{4\pi^3}, \] respectively, so their sum is \(1/(2\pi^2)\). Therefore \[ \boxed{\oint_C{e^\tau\over(\tau^2+\pi^2)^2}\,d\tau={i\over\pi}.} \]

    The mathematical exercise is thereby resolved; its separate provenance issue remains, since this card does not occur in the official Fall 2016 UGA DOCX to which its collection currently points.

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