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Let $$f(z)=\begin{cases}\frac{x^{3}(1+i)-y^{3}(1+i)}{x^{2}+y^{2}}, & z\neq0\\0, & z=0 \end{cases}$$

I want to show that the Cauchy-Riemann equations are satisfied at the origin but $f$ is not analytic. Please, how can I show this?
I already know that the satisfaction of Cauchy-Riemann equations is the neccessary and sufficient condition for a function $f(z)$ to be analytic, then how is it possible to show that the fuction satisfies C-R equations at the origin but not analytic?

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    $\begingroup$ Doesn't $u(x, y) = v(x, y) = -u(y, x)$? This would mean that $\frac{\partial u}{\partial x} = -\frac{\partial u}{\partial y}$. So this would satisfy the second CR equation, but not the first. $\endgroup$
    – Jared
    Apr 16, 2014 at 23:18
  • $\begingroup$ What does this means If it satisfy one condition of C-R equation? $\endgroup$
    – user143681
    Apr 17, 2014 at 4:55

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the satisfaction of Cauchy-Riemann equations is the necessary and sufficient condition for a function $f$ to be analytic

More precisely, the satisfaction of Cauchy-Riemann equations on an open set is the necessary and sufficient condition for $f$ to be analytic on that set.

The exercise is meant to demonstrate that C-R equations can hold at one point without the function being analytic at that point. You can check that all partials are zero at the origin, so C-R equations are satisfied there. But they do not hold anywhere else.

If $f$ were analytic at $0$, it would (by definition) be represented by a power series in a neighborhood of $0$. But then C-R equations would hold in that neighborhood, which is not the case.

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