TITLE: Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions?
QUESTION [42 upvotes]: The title pretty much says it all.  I am revisiting complex analysis for the first time since I "learned" some as an undergraduate.  I am trying to wrap my head around why it should be the case that a function which is differentiable once should be differentiable twice.  I know a proof (Use Cauchy integral formula and differentiate under the integral sign), but that proof doesn't do a whole lot to explain the magic.
Say you had never heard of complex numbers before, and someone told you that you had a function $f: \mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y) = (u(x,y), v(x,y))$, which locally looks like a rotation/expansion (i.e. it satisfies the Cauchy-Riemann equations everywhere).  Then why on Earth should this function be $C^\infty$?  I would love to see a picture, or just a proof that makes this feel less like a magic trick.  I hoped that "Visual Complex Analysis" would help me out here, but this seems to be the one theorem in the book which is not given a geometric motivation.

REPLY [2 votes]: This appears to be the earliest "topological" proof:
A TOPOLOGICAL  PROOF OF  THE CONTINUITY OF  THE 
DERIVATIVE OF A FUNCTION OF A COMPLEX 
VARIABLE 
BY ROBERT L. PLUNKETT 
1958
http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf
G.T. Whyburn himself (whose research was foundational to the result) gives his own proof, allegedly inspired by Connell and Porcelli:
Developments in topological analysis, 1961
http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf