The residue at infinity is given by:

$$\underset{z_0=\infty}{\operatorname{Res}}f(z)=\frac{1}{2\pi i}\int_{C_0} f(z)dz$$

Where $f$ is an analytic function except at finite number of singular points and $C_0$ is a closed countour so all singular points lie inside it.

It can be proven that the residue at infinity can be computed calculating the residue at zero.


The proof is just to expand $-\frac{1}{z^2}f\left(\frac{1}{z}\right)$ as a Laurent series and to see that the $1/z$ is the integral mentioned.

I can see that we change $f(z)$ to $f(1/z)$ so the variable tends to infinity.

But, is there any intutive reason of why we introduce the $-1/z^2$ factor?

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    $\begingroup$ Without specifying what $\;C\;$ is the first two lines make little sense... $\endgroup$ – DonAntonio Jan 6 '14 at 21:16
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    $\begingroup$ @DonAntonio Clarified. $\endgroup$ – jinawee Jan 6 '14 at 21:23
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    $\begingroup$ What is $d(1/z)$? $\endgroup$ – Andrés E. Caicedo Jan 6 '14 at 21:24
  • $\begingroup$ ...and if $\;\infty\;$ is a singular point, then what is $\;C\;$ ? $\endgroup$ – DonAntonio Jan 6 '14 at 21:24

The thing is that functions do not have residues, but rather differentials have residues. This is something which can be quite confusing in a first complex analysis class. The "residue of a function" is not invariant under a change of local parameter, but the residue of a differential is. For this reason, what is usually called the "residue at $0$ of $f(z)$" is actually the residue at $0$ of $f(z)dz$.

When you change the coordinate from $z$ to $w=1/z$, the differential $dz$ is transformed into $-dw/w^2$, which explains the change of sign and the extra factor. Thus,

$$f(z)dz = \frac{-1}{w^2} f(1/w) dw.$$

The "residue of $f$ at $\infty$" is the residue at $0$ of $\frac{-1}{w^2} f(1/w) dw$.

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    $\begingroup$ "The thing is that functions do not have residues, but rather differentials have residues." So does this explain why a function $\left(\text{e.g. }\frac{z+2}{z(z+1)}\right)$ can be analytic at $\infty$ yet still "have" a residue at $\infty$? $\endgroup$ – Shaun Jan 17 '14 at 16:30
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    $\begingroup$ @Shaun That's right! In fact, the differential $dz$ has a double pole at infinity! :) $\endgroup$ – Bruno Joyal Jan 17 '14 at 16:40
  • $\begingroup$ @BrunoJoyal consider $\sum_\limits{k=-\infty}^{\infty} c_k z^k$ this is equivalent to $\frac{1}{z^2}\left(\sum_\limits{k=-\infty}^{\infty} c_k \left(\frac{1}{z}\right)^k\right)$ so the $c_{-1}$ term is the same for $\frac{1}{z^2}f\left(\frac{1}{z}\right)$ at $z=0$ and $f(z)$ at $\infty$. How do we see that we need a minus sign? i.e the residue at infinity is the residue of $f(z)$ taken with negative sign. $\endgroup$ – Alexander Cska Sep 2 at 11:52

$-1/z^2$ comes from changing the variable from $z$ to $u=1/z$.

So $$ \int_{C_0} f(z)dz=\int_{C'} f(1/u)d(1/u)=-\int_{C'}\frac{-1}{u^2}f(1/u)du $$

where $C'$ is the trace of $C_0$ in the $u$ space. If all singularity lies within $C_0$ in the $z$ space, then every singularity (except for the one at $u=0$) of $-(1/u^2)f(1/u) $ would lie outside the $C'$ in the $u$ space.

Since $u=0$ is the only singular point of $-(1/u^2)f(1/u)$ inside $C'$, thus this integral gives the residue at $u=0$, or equivalently, $z=\infty$.

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