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This is from Conway's Complex Analysys:

Let $f$ be analytic in the plane except for isolated singularities at $a_1,...,a_m$. The residue at infinity $Res(f;\infty)$ is defined as the residue of $-\dfrac{1}{z^2}f\left(\dfrac 1z\right)$ at $z=0$.

Equivalently, $Res(f,\infty)=-\dfrac{1}{2\pi i}\displaystyle\int_{|z|=R}f$ for sufficiently large $R$.

It is stated as an exercise that $Res(f;\infty)=-\displaystyle\sum_{k=1}^mRes(f;a_k)$. So I think that sufficiently large $R$ is an $R$ such that all $a_1,...,a_m$ are inside of the circle $|z|=R$ so we can apply the residue theorem to $\displaystyle\int_{|z|=R}f$.

What I would to know is why those two definitions are equivalent. I don't see how the function $\dfrac{1}{z^2}f\left(\dfrac 1z\right)$ is related to that integral. Can anyone explain to me?

Thank you.

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Make the change of variables $w=1/z$ in that integral. You then get $$\int_{|z|=R} f(z)dz=-\int_{|w|=1/R}f(1/w)\frac{dw}{w^2},$$ the $-1/w^2$ coming from $dz=-dw/w^2$. When $R$ is large enough so that $g(w)=-f(1/w)/w^2$ is analytic on $|w|\leq 1/R$ except at $w=0$, that integral computes the residue of $g$ at $0$ (up to a factor of $2\pi i$).

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  • $\begingroup$ Succinct and complete! +1 - Mark $\endgroup$ – Mark Viola Jan 13 '16 at 5:02

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