Simple argument that analytic complex functions are holomorphic? [duplicate]
Is the following simple argument correct? I'm looking for a simple, accessible argument.
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Assume a complex function can be represented by a Taylor series.
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Each term of the Taylor series is holomorphic because it is of the form $a(z-z_0)^n$.
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If each term is holomorphic, then so is the infinite sum of terms, the Taylor series.
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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?
Solution 1:
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.