# Elegant complex analysis

lark
Complex analysis has a lot of nice theorems that real analysis doesn't have: if you can take the complex derivative once, you can take it $$\infty$$ many times. Maximum modulus theorem; inside the radius of convergence the Taylor series of a function converges to the function.
So what I wonder is, is the elegance of complex analysis related to the fact that the complex #'s are algebraically complete?
Complex analysis can be used to show that the complex #'s are algebraically complete. So one could ask if that same proof works over other fields that are both metrically complete and algebraically complete. To try and get a handle on the question.
Laura

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## Answers and Replies

Studiot
How would you cope with the fact that complex numbers don't possess an ordering relation?

Tinyboss
Analyticity is a very strong condition to put on a function, and there are correspondingly strong results that can be shown.

lark
Analyticity is a very strong condition to put on a function, and there are correspondingly strong results that can be shown.

being real analytic does not give the strong results that being complex analytic does. So I'm wondering if that's related to the algebraic completeness of complex no's.

elibj123
The structure of the complex numbers which poses the conditions of analyticity (the Cauchy-Riemann equations) and its two-dimensional nature, is the cause of the special behaviour analytic functions exhibit.

Real analyticty is not as strong and somewhat different in nature from complex analyticity, and many ideas don't even exist in the one dimensional world.

lark
The structure of the complex numbers which poses the conditions of analyticity (the Cauchy-Riemann equations) and its two-dimensional nature, is the cause of the special behaviour analytic functions exhibit.

The Cauchy-Riemann equations are a consequence of the particular polynomial $$x^2+1=0$$ that $$i$$ satisfies over R. If you were extending analysis in a field F to some other field G of finite index over F, where $$G=F[a]$$ and $$a$$ satisfies some other polynomial over F, then you'd have a different set of equations.

The http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf" [Broken] says that if an algebraically closed field G has a subfield F of finite index > 1, then G = F, and G is of characteristic 0.

So if you are doing analysis where the algebraically closed field is of finite index over the Cauchy-complete field, then you'd get the Cauchy-Riemann equations!

By http://en.wikipedia.org/wiki/Ostrowski%27s_theorem" [Broken], a field complete with respect to an Archimedean absolute value is (algebraically and topologically) isomorphic to either the real numbers or the complex numbers.

So what you're saying may be equivalent to my guess that "the elegance of complex analysis is a consequence of complex numbers being algebraically closed".

I don't know what happens in p-adic analysis - if you were doing analysis in the algebraic closure of the p-adics, which is of infinite index over the p-adics, whether there would be an analog of the Cauchy-Riemann equations.

Laura

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lark
The structure of the complex numbers which poses the conditions of analyticity (the Cauchy-Riemann equations) and its two-dimensional nature, is the cause of the special behaviour analytic functions exhibit.

I don't think that's true. $$Q_p$$ is a quadratic extension of the p-adics, and it has Cauchy-Riemann equations, but it doesn't have the nice theorems like Cauchy integral theorem, residue theorem, maximum modulus.

But $$\Omega_p$$, which is algebraically closed and contains the p-adics $$Q_p$$, does have such theorems.

So that suggests those theorems have something to do with being algebraically closed. The reference I looked up actually used the fact that $$\Omega_p$$ is alg. closed to prove a residue theorem.

And I don't know that $$\Omega_p$$ is two-dimensional in any sense.

Laura