# Where is $f(x+iy)=x^3+y^3$ complex differentiable?

$Let f(z) = (Re(z))^3 + (Im(z))^3$

(a) At which points (if any) is f differentiable? Find the expression of f' at those points.

(b) Draw a picture of the subset of C consisting of those points at which f is differentiable. Hence decide at which points (if any) f is analytic.

$f(z) = x^3 + y^3$

(a) Using Cauchy Riemann equations I found that f is only differentiable at (0,0). And f' is 0 at that point.

(b) f is nowhere analytic as it is not differentiable in an $\epsilon$ neighbourhood of z.

Is that correct?

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Whether (a) is correct could depend on how you reasoned. In general, a function can satisfy the Cauchy-Riemann equations at a point without being differentiable at that point, but not conversely. That is, differentiability at a point does imply that the Cauchy-Riemann equations hold at that point, so using CR alone is enough for showing that $f$ is not differentiable away from $(0,0)$. But it is not enough for showing that $f$ is differentiable at $(0,0)$. For that, a good method is to explicitly compute the derivative using the definition, $f'(0)=\lim\limits_{(x,y)\to (0,0)}\frac{f(x+iy)-f(0)}{x+iy}$, which is $0$ as you indicated.
@JohnMcDonald: Actually, with more care (using additional hypotheses that apply in this case), CR can be sufficient, it just isn't in general enough so further reasoning would be needed. There are pointers to partial converses to differentiability$\implies$CR on Wikipedia. –  Jonas Meyer Feb 14 '12 at 21:37