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I am having a little confusion on the naming of functions in complex analysis. If $f$ is a holomorphic function on the complex plane and its domain is the complex plane then it's called an entire function.

If $g$ is holomorphic everywhere on the complex plane apart from its poles and its domain is the complex plane then it's a meromorphic function.

But... If $g$'s domain is extended to the Riemann sphere (and it doesn't have a essential singularity at infinity) is it a "meromorphic function with domain of Riemann sphere" or "holomorphic function with domain of Riemann sphere" or a "rational function"?

Thanks!

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1 Answer 1

Holomorphic functions are usually defined over an open set of the complex plane or Riemann sphere.

A holomorphic function defined over the whole complex plane is called entire. This is a definition.

A meromorphic function is one that is holomorphic except for isolated singularities, which must be poles. This is a definition.

A meromorphic function defined over the whole Riemann sphere must be a rational function. This is a theorem.

A holomorphic function defined over the whole Riemann sphere must be constant. This is a theorem.

The only entire functions that extend to meromorphic functions on the Riemann sphere are polynomials. (Thanks to Theo Buehler for pointing this out.)

See also http://en.wikipedia.org/wiki/Meromorphic#Meromorphic_functions_on_Riemann_surfaces.

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Ok, thanks. I just got confused by this wikipedia article; en.wikipedia.org/wiki/M%C3%B6bius_transformation#Definition . See it says that by extending the domain to the Riemann sphere the meromorphic function turns into a holomorphic function. –  Harry Barber Sep 13 '11 at 21:51
    
@Harry: The terms holomorphic and meromorphic can also be given more general definitions, for functions from one Riemann surface to another. And then it turns out that a "holomorphic function from the Riemann surface $S$ to the Riemann sphere" is (except for the constant function with value $\infty$) the same thing as "a meromorphic function from $S$ (minus poles) to the complex plane". –  Hans Lundmark Sep 14 '11 at 6:28

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