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I am seeking a simple explanation for why complex function, if they are differentiable, must be infinitely differentiable.

I have looked at other questions on this site, and discussion elsewhere, and can't any explanation that doesn't require university level mathematics and terminology.

EDIT 1 - Several comments have suggested similar questions (eg), but sadly the answers there require university level mathematics and terminology.

Penelope
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    well, it's a very non-trivial result (btw note that the function is supposed to be differentiable on an entire open set; a single point is not enough), so you can't expect it to have a simple answer which can be completely dumbed down. The reason is Cauchy's integral formula: $f(z)=\int_{|w-a|=r}\frac{f(w)}{w-z},dw$. The ability to express the value of a complex-differentiable function via an integral over some curve gives rise to all sorts of wonderful results. – peek-a-boo Jun 02 '21 at 13:14
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    This answer explains why complex-differentiable functions can be expected to have nicer local properties, but of course it is very far from obvious why this simple heuristic would imply the amazing result that once-complex differentiable on an open set implies analytic (hence infinitely differentiable). – peek-a-boo Jun 02 '21 at 13:18
  • @peek-a-boo As an aside, MathOverflow has references to proofs that avoid integration, but it seems there is general agreement that they're harder. – Mark S. Jun 02 '21 at 13:22
  • @MarkS. thank you for the comment. I remember that comment made by Ahlfors when reading, but never bothered to look into it further and had forgotten about it till you mentioned it. – peek-a-boo Jun 02 '21 at 13:24
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    Does this answer your question? Why are differentiable complex functions infinitely differentiable? – Jean Marie Jun 02 '21 at 13:36
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    Also duplicate of https://math.stackexchange.com/q/3697024 – Jean Marie Jun 02 '21 at 13:37
  • Thanks everyone for suggestion the other links .. sadly the answers there aren't "simple enough" as requested by this question. – Penelope Jun 02 '21 at 18:01
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    If you're familiar with the fact discussed in most multivariable calculus courses that a vector field $\langle u, v \rangle$ on a simply connected (no holes) region in $\mathbb{R}^2$ with $v_x - u_y = 0$ is conservative, this fact implies that for $f$ holomorphic (with continuous derivatives) and $C$ as before, $\int_C f(z)dz = 0$. From which the Cauchy Integral formula for $f$ can be deduced. And this expression is infinitely differentiable. – Alex Nolte Jun 02 '21 at 20:07

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