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What makes the crucial difference between the reals and the complex numbers is that the complex numbers are algebraically closed. So while going through all the proofs that "being holomorphic implies being analytic" this must be the decisive step. Can you give me a sketchy plot how this enters any proofs?

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Yes, I'm not certain algebraic closure is really the crucial difference here. I would assume it has more to do with how "nice" a function has to be in order for the derivative to be consistent when we compute the limits in all possible directions, and this is more related to why analyticity "happens". –  pjs36 Jan 24 at 0:05
    
? Can you explain this more? –  Freeze_S Jan 24 at 0:15
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The second answer from the link given by @AymanHourieh explains this more fully. Essentially, when we have the quantity $f'(z) = \lim_{h \to 0} (f(z+h)-f(z))/h$, the complex number $h$ can approach $0$ from any direction, and must give a consistent answer no matter the approach. I can't speak to how to involve the holomorphic condition, I'm afraid. –  pjs36 Jan 24 at 0:24
    
As far as I know, holomorphic implies analytic comes from the Cauchy Integral Formula. Not sure if algebraic closure is the key property you are looking for. –  Braindead Jan 24 at 4:23

1 Answer 1

We have the important property for analytic functions

A function is analytic at a point if and only if it possess a power series expansion at that point.

The above statement, in general, is not true for infinitely differentiable real-valued functions.

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You forgot to add that this power series needs to converges to the function in an open neighborhood of the said point. –  Braindead Jan 24 at 3:48
    
Thanks but that is what is commonly understood under analytic and - I hope so - we all know what is its definition. But I'd like to know why differentiability implies analyticity when considering the complex numbers while when considering the reals it even fails for smooth functions to be analytic in general. –  Freeze_S Jan 24 at 4:19

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