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The Cauchy Riemann equations in effect say that a function $f(z) = u(z)+iv(z)$ can be approximated as roughly a scaled rotation

$$f(c+h) \approx f(c) + f'(c)h = f(c) + \begin{bmatrix}u_x & -v_x\\v_x & u_x\end{bmatrix} \begin{bmatrix}h_x \\ h_y \end{bmatrix}$$

Informally, if $f$ has this kind of approximation at every point in a domain then it admits a power series representation at each point.

Intuitively why should having the scaled rotation approximation give a full power series representation? Suppose at some point $c$ the Cauchy Riemann equations are satisfied but there does not exist a power series representation at $c$, then presumably there should be a geometric point of view to see that this is ridiculous and leads to $f$ not satisfying the Cauchy Riemann equations elsewhere.

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I am not sufficiently prepared to answer you now but I can point you to a book where you will find a very convincing treatment of your question: it is Visual Complex Analysis by Tristan Needham. –  Giuseppe Negro Dec 17 '12 at 4:08
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2 Answers

It is quite possible for the Cauchy-Riemann equations to be satisfied at a particular point, or even along a curve, without the function being analytic in a neighbourhood of the point. You only get analyticity if the equations are satisfied everywhere in an open set.

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For example, any non-holomorphic function satisfies the C-R equations at every point where is Jacobian is zero. –  Mariano Suárez-Alvarez Dec 17 '12 at 4:13
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Another question asked essentially the question I was getting at. While the answers didn't make the existence of the 2nd derivative obvious they did give links to proofs of the existence of the 2nd derivative that don't use any notion of contour or line integral. Instead relying on just looking at difference quotients.

G.T. Whyburn, Developments in topological analysis, 1961 http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf

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