# Residue at infinity without singularity

I'm sorry if this question is a duplicate but I really have not found anything online that says this and I need a confirmation. My problem is the following: for a function on the complex plane to have a residue at a point there must be a singularity. This because the residue is defined to be the coefficient $$a_{-1}$$ of the Laurent series and therefore if the Laurent series has no negative terms there is no residue (or the residue is zero). However the situation on the Riemann sphere is different, that is to say it seems to me that the point at infinity might have a residue even without being a singularity.

Take for example $$f(z) = \frac{1}{z}$$. A simple calculation (using the definition) shows that the residue at infinity is -1. However, this function has a 0 at infinity and not a singularity. This can be seen both by taking the limit and by using the fact that a function $$f$$ has a pole at infinity iff $$f(\frac{1}{z})$$ has a pole at 0 and this is not the case. Moreover, the function has to have a nonzero residue at infinity because the sum of residues of a meromorphic function on the Rieman sphere must be zero. This is counterintuitive to me but I'm quite convinced this is how it is. Can someone confirm this to me (that the point at infinity is radically different from other points)? Moreover, if this is the case, is there any profound reason why the point at infinity behaves so differently?

• Possibly helpful: math.stackexchange.com/questions/629495/…. Jul 14, 2019 at 19:17
• Thank you! That is intresting. Still I'd be happy to find something that addresses the problem of residue without singularity explicitly! Jul 14, 2019 at 19:38

For a function $$f$$ analytic on $$\Bbb{C}$$ minus finitely many points, the residue at $$\infty$$ is what you need to obtain $$\sum_a Res(f(z),a) = 0$$.
For $$r$$ large enough $$f$$ is analytic for $$|z| \ge r$$ so that $$\frac{1}{2i\pi}\int_{|z| = r} f(z)dz = \sum_{a \ne \infty} Res(f(z),a)$$ and $$Res(f(z),\infty) \overset{def}=-\frac{1}{2i\pi}\int_{|z| = r} f(z)dz = \frac{1}{2i\pi}\int_{|s| = 1/r} f(1/s)d(1/s) \\= \frac{1}{2i\pi}\int_{|s| = 1/r} \frac{f(1/s)}{-s^2}ds = Res(\frac{f(1/s)}{-s^2},0)$$
Then for the compact Riemann surface viewpoint : the residues of $$f(z)$$ are chart dependent so they are ill-defined. To make it chart independent think to $$f(z)dz$$ as a 1-form on the Riemann sphere, holomorphic at all but finitely many points, we can define it in every chart it will become $$f(\phi(u))d\phi(u)$$ and the residue $$Res(f(z)dz,\phi(a))=\frac{1}{2i\pi}\int_{|u-a|=r} f(\phi(u))d\phi(u)$$ doesn't depend on the chosen chart, the sum of all residues is $$0$$.