# Two Approaches to Calculate the Residue at Infinity

I am a little bit confused about the residue at infinity.

Generally, we have two approaches to compute $Res(f,∞)$, by residues at other singular points or by laruent series.

Consider $f(z)=\frac{1}{z^3-z^5}=\frac{1}{z^3(1+z)(1-z)}$, whose singular points include $0,±1,$and $∞$.

First approach:

We can easily obtain $Res(f,0)=1,Res(f,1)=-\frac{1}{2}$ and $Res(f,-1)=\frac{1}{2}$.

Therefore, $Res(f,∞)=-[Res(f,0)+Res(f,1)+Res(f,-1)]=-1.$

Second approach:

Suppose the Laurent series of $f(z)$ at $z=∞$ is $f(z)=\sum\limits_{n=-∞}^{+∞}a_nz^n$, then $Res(f,∞)=-a_{-1}$.

Let $w=\frac{1}{z}$. $f(\frac{1}{w})=\frac{w^5}{w^2-1}=-w^5\sum\limits_{n=0}^{+∞}w^{2n}$.

Therefore, $f(z)=-\frac{1}{z^5}\sum\limits_{n=0}^{+∞}\frac{1}{z^{2n}}$.

$a_{-1}=0$

There must be something wrong. But I can't find out it. Any help? Many thanks.

Strictly speaking, functions $f(z)$ don't have residues, differentials $f(z)\,dz$ do.
Take $f(z)=1/(z^3-z^5)$ and consider the residue of $f(z)\,dz$ at $z=\infty$. Let $w=1/z$, so that $z=1/w$. We need to substitute for $dz$ too. Indeed, $dz=-dw/w^2$. Therefore $$f(z)\,dz=-\frac{dw}{w^2(w^{-3}-w^{-5})}=\frac{w^3\,dw}{1-w^2}$$ so the residue at $w=0$ (that is at $z=\infty$) is zero.
This does square with the sum of the residues of a rational differential over the Riemann sphere being zero, as the residue at $z=-1$ is actually $-\frac12$.