5
$\begingroup$

This is from Conway's Complex Analysis:

Let $f$ be analytic in the plane except for isolated singularities at $a_1,\ldots,a_m$. The residue at infinity $\operatorname{Res}(f;\infty)$ is defined as the residue of $-\dfrac1{z^2}f\left(\dfrac 1z\right)$ at $z=0$.

Equivalently, $\operatorname{Res}(f,\infty)=-\dfrac1{2\pi i}\displaystyle\int_{|z|=R}f$ for sufficiently large $R$.

It is stated as an exercise that $\operatorname{Res}(f;\infty)=-\displaystyle\sum_{k=1}^m\operatorname{Res}(f;a_k)$. So I think that sufficiently large $R$ is an $R$ such that all $a_1,\ldots,a_m$ 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 $\dfrac1{z^2}f\left(\dfrac 1z\right)$ is related to that integral. Can anyone explain it to me?

Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $2\pi i$).

$\endgroup$
1
  • $\begingroup$ Succinct and complete! +1 - Mark $\endgroup$
    – Mark Viola
    Commented Jan 13, 2016 at 5:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .