# Alternative definitions of residue at infinity

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.

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 $2\pi i$).