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I'm confused because, in class, we've started working in the extended complex plane and just assuming all the properties we've worked with in the normal complex plane carry over. Here are the properties I'm confused with in particular:

  • Analyticity: I don't think this is equivalent to being holomorphic, because I can't see any way we can create a power series centered around infinity.
  • Holomorphicity: Maybe we can call a function holomorphic at infinity if it's differentiable for all complex numbers of a certain magnitude or greater, but I don't know how to make that rigorous, especially since a circle around infinity is normally thought of as a straight line.
  • Conformal maps: What would it even mean to have angles conserved for curves going through infinity.

If it helps, what got me thinking about this was when we were talking about the conformal automorphisms on the extended complex plane and our definition for an automorphism was an analytic, invertible map from a set to itself.

My question is how these properties are dealt with in the extended complex plane? Can we define them, and, if not, what do we use instead?

Jakobian
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Sandro
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1 Answers1

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Partial answer.

Often statements “at infinity” are, I think, converted to statements about $1/z$ at $z\to0$. For instance a function $f$ is holomorphic at infinity if the function $g(z)=f(1/z)$ has an extension to $z=0$, holomorphic in some neighbourhood of zero. In which case you can take the standard holomorphic-implies-analytic theorem to expand: $$g(z)=g(0)+\sum_{n\ge1}a_nz^n$$

And passing $z\mapsto1/z$:

$$f(z)=f(\infty)+\sum_{n\ge1}a_nz^{-n}$$

For $z$ in some neighbourhood of infinity. “$f(\infty)$” is just the value $g(0)$ which extends $g$ holomorphically. This is, I expect, your desired power series (I have never formally studied the extended complex plane myself, this is gathered from random digging I’ve done here and there).

FShrike
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  • Thank you! Do you know why we're allowed to make statements about the original function by considering the behavior of a function g(z)=f(1/z)? It makes sense intuitively, but I don't understand why it should be allowed – Sandro Oct 31 '22 at 07:12
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    @Sandro As far as I know, it’s by definition! But also topologically speaking, the adjoined point at infinity in the one point compactification is a true topological limit of $1/z$ as $z\to0$, and zero is a true topological limit of $1/z$ as $z\to\infty$. So the $z\mapsto1/z$ bijection preserves topological properties and it allows us to sensibly define things at infinity by instead defining them at zero. – FShrike Oct 31 '22 at 07:25
  • Okay, that seems really strange to me, but I think it's just because I don't have much experience with topology. I think I've managed to kind of understand how that gives us a way to define these properties as z goes to infinity in a way that is topologically invariant though, thank you! – Sandro Oct 31 '22 at 07:53