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Is the following simple argument correct? I'm looking for a simple, accessible argument.

  1. Assume a complex function can be represented by a Taylor series.

  2. Each term of the Taylor series is holomorphic because it is of the form $a(z-z_0)^n$.

  3. If each term is holomorphic, then so is the infinite sum of terms, the Taylor series.

  4. Therefore analytic complex functions are holomorphic.

If this is correct, is there a similarly accessible argument to show the reverse, that holomorphic functions are analytic?

Penelope
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    for me analytic on $\Bbb C$ == holomorphic by definition. It's just shorthand for a differentiable function defined on $\Bbb C$. – Henno Brandsma Jan 02 '22 at 22:54
  • @Henno Brandsma "holomorphic by definition". But the definition of holomorphy is different, it's about differentiability... – Jean Marie Jan 03 '22 at 00:08
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    @JeanMarie in my old textbook on complex function theory (Conway) we started with what a differentiable function on (open subsets of) $\Bbb C$ was. Later it was shown using Cauchy’s integral theorem and other tools that this was actually already a function with a power series expansion at every point of is domain and that such a function is infinitely many times differentiable etc. I found the book very clear, but in the approach it follows there is no need for a separate notion of analytic function and holomorphic was just shorthand for differentiable on all of $\Bbb C$. – Henno Brandsma Jan 03 '22 at 00:19

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Steps 1 and 2 are correct, but step 3 needs further justification. It's not obvious that an infinite sum of holomorphic functions is holomorphic. One typically shows this by first showing that the uniform limit of holomorphic functions is holomorphic (see here). One then applies the result to the partial sums of the infinite series, which are known to converge uniformly on compact subsets.

The reverse direction (holomorphic implies analytic) is much harder. It's probably the most important / deep result from complex analysis. The proof uses deep results like the Cauchy integral theorem, so I'm not aware of a simple proof like the one you're looking for. You might want to check out a proof in a complex analysis book, like Ahlfors or Churchill-Brown.

Adithya Chakravarthy
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    Thanks @chaad - on step 3 I had some caution so asked a specific question: https://math.stackexchange.com/questions/4347338/infinite-vs-finite-sum-of-holomorphic-functions-is-holomorphic – Penelope Jan 02 '22 at 23:12