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Suppose that a complex valued function $f:\mathbb{C}\to\mathbb{C}$ is complex differentiable at $z$, that is, \lim\limits_{\Delta z \to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z} = f'(z) exists and is finite. Suppose also that $f$ is continuous in some neighborhood of $z$ (or some open ball around $z$). Would this imply that $f$ is analytic at $z$ (or maybe in a neighborhood around it)? Can you give a counterexample? Can you think of a function that is analytic at a single isolated point?

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    @Ray Navarrete: Regarding the various ways to be $C^{\infty}$ at a point and not being analytic at that point, see my very extensive comments and lists of references at http://mathforum.org/kb/message.jspa?messageID=387148 and http://mathforum.org/kb/message.jspa?messageID=3871492011-10-11

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Let $f(z)=|z|^2=x^2+y^2$. $f$ is continuous on $\mathbb{C}$ (even $C^\infty$ as a function of the real variables $x$ and $y$), f'(0)=0 (in the complex sense), but $f$ is not complex differentiable at any $z\ne0$.

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    Thank you! I can't believe I hadn't thought about this function (I assumed it would be hard to find one given that finding a function that is differentiable at 0 but not even continues outside 0 was pretty hard).2011-10-11