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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.

$$\underset{z_0=\infty}{\operatorname{Res}}f(z)=\underset{z_0=0}{\operatorname{Res}}\frac{-1}{z^2}f\left(\frac{1}{z}\right)$$

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
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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
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$-1/z^2$ 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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