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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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It's impossible for a function $f:\mathbb{C}\to\mathbb{C}$ to be analytic in just one point. By definition, if $f$ is analytic at $z_0$, then there's an open neighbourhood $U$ of $z_0$ such that $f$ is given by a power series $\sum_{n=0}^\infty a_n \left( z-z_0 \right)^n$ on $U$, and so is analytic on $U$. –  Chris Eagle Oct 10 '11 at 21:40
    
@Ray Navarrete: Google "Looman–Menchoff theorem", which I think is relevant to your question (I'm not very knowledgeable about complex analysis), such as this web page: en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem –  Dave L. Renfro Oct 10 '11 at 21:48
    
@Ray Navarrete: Then again, after seeing the comments and Julián Aguirre's answer, maybe the Looman-Menchoff theorem is overkill for what you want! –  Dave L. Renfro Oct 10 '11 at 22:00
    
First of all, thank you all. Chris, I know that there are some complex functions that are analytic at one point but whose power series at that point is not equal to the function itself (other than at the point). This happens when the radius of convergence is equal to zero. Has anyone heard about this? Related functions that comes to mind are what are called test functions, which are zero everywhere except at some open interval but are infinitely differentiable (this I've only seen in real functional analysis, I don't know if this applies here). –  Ray Navarrete Oct 11 '11 at 6:29
    
Thanks Dave, that result is definitely useful. I guess I can now restate my question to 'if you know $f$ is continuous in some neighborhood of $z$ and that the Cauchy-Riemann equations hold at the point $z$, could this equations also hold in some neighborhood of $z$? –  Ray Navarrete Oct 11 '11 at 6:33

1 Answer 1

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). –  Ray Navarrete Oct 11 '11 at 6:46

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