Is following contradictory? Can you give an example?

"Then $f = u + iv$ is complex-differentiable at that point if and only if the partial derivatives of $u$ and $v$ satisfy the Cauchy–Riemann equations (1a) and (1b) at that point. The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point." - 1[1, second paragraph]

They say that $f$ is complex-differentiable iff partial derivatives of $u$ and $v$ satisfy C-R equations, but still it is not enought to ensure complex differentiability at that point.

So do you need extra conditions as wikipedia says or not for $f$ to be complex differentiability? Can you give me example, where function satisfy C-R equations, but is not Complex differentiable at certain point?

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For an explicit example, let $$f(z) = \begin{cases} \exp(-1/z^4) & z \neq 0 \\ 0 & z = 0\end{cases}$$

You can check that $f$ satisfies Cauchy-Riemann's equations everywhere, but $f$ is not real-differentiable (or even continuous) at $z=0$.

You may also be interested in Looman-Menchoff's theorem which shows that it's enough to assume that $f$ is continuous and satisfies Cauchy-Riemanns equations everywhere to conclude that $f$ is holomorphic. (Note that in Looman-Menchoff, it's not enough to assume continuity and CR at a point.)

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You left out the part where it says, "Suppose that $u$ and $v$ are real-differentiable...." It says that in the absence of that assumption, the existence of partials satisfying C-R is not enough to insure complex differentiability. So to find the example you want, you'll first have to find functions with partials but not real-differentiable.

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Maybe I'm missing something, but how can they not be real differentiable if they satisfy C-R? – Thomas Andrews Jun 14 '13 at 12:30
You are missing the possibility that a function $u(x,y)$ can be such that its partials with respect to $x$ and to $y$ both exist at some point, but it isn't differentiable at that point. Review the definition of differentiability for multivariate functions. – Gerry Myerson Jun 14 '13 at 12:37
Yes. So to find that eq. |x| is not real-differentiable at $x=0$, so I think you can't find complex function of whose u and v would not be real-differentiable? – alvoutila Jun 14 '13 at 12:44