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According to Wikipedia, "Moreover, the equations are necessary and sufficient conditions for complex differentiation once we assume that its real and imaginary parts are differentiable real functions of two variables." I've always thought that the C-R equations holding alone isn't sufficient for differentability. You would need the continuity of the partial derivatives, right? The talk page made me more confused.

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2 Answers 2

I agree with Wikipedia. Complex analysis textbooks tend to make much ado about the fact that existence of partials and the CR equations do not imply complex differentiability. But this is merely a reflection of the real analysis phenomenon: partials do not guarantee real differentiability. Once the latter is assumed, the CR equations are equivalent to complex differentiability.

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There are functions whose partial derivatives exist, yet are not continuous, and these are not complex differentiable. For example, see math.stackexchange.com/a/253852/641 –  user641 Jan 20 '13 at 5:57
    
@5PM: Do you have a reference for the statement that real differentiability together with the CR equations implies complex differentiability? –  Jonas Meyer Jan 20 '13 at 6:02
    
@SteveD: The example linked to is of a function that is not real differentiable. –  Jonas Meyer Jan 20 '13 at 6:03
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Ah, sorry, I was focused on the OP's question. A reference for @JonasMeyer is Proposition 0.1 in Ullrich's Complex Made Simple; chapter 0 is an brief exposition of the difference between real and complex differentiability. –  user641 Jan 20 '13 at 6:09
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@Jonas This is just a straightforward fact: we have an $\mathbb R$-linear approximation; what do we need for it to be $\mathbb C$-linear? Exactly the Cauchy-Riemann equations. –  user53153 Jan 20 '13 at 6:17

The question of characterizations of analyticity is rather interesting, though technical. In practice, it is rather rare to be in a setting where the Cauchy-Riemann equations hold and one does not also have assumptions that guarantee the partial derivatives to be continuous.

A good reference for this topic is the paper

J. D. Gray, S. A. Morris. "When is a function that satisfies the Cauchy-Riemann equations analytic?" Amer. Math. Monthly 85 (1978), no. 4, 246-256.

Besides the result mentioned in the answer by user 5PM, here are several results stated in the paper:

The function $f$ given by $f(0)=0$, $f(z)=e^{-z^{-4}}$ for $z\ne0$ satisfies the Cauchy-Riemann equations everywhere, and is not analytic at $0$. This is an example due to Looman.

The Looman-Menchoff theorem gives us a very general sufficient condition: It states that if the partial derivatives of $f$ exist everywhere on a domain and satisfy the Cauchy-Riemann equations, then $f$ is analytic provided that it is continuous.

However, this is not a pointwise result: $f$ given by $f(0)=0$, $f(z)=z^5/|z|^4$ is continuous everywhere, and satisfies the Cauchy-Riemann equations at the origin, but it is not analytic.

One can assume less than continuity of $f$, but the conditions become significantly more involved. You may want to see also

Maynard G. Arsove. "On the definition of an analytic function" Amer. Math. Monthly 62 (1955), 22-25.

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