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This must be something really easy/standard, but for some reason I can't see what's wrong with this argument.

Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be any complex function, and write $f = u + iv$, with $u, v : \mathbb{C} \rightarrow \mathbb{R}$.

I know that if $f$ is $C^1$ in the real sense and $f$ satisfies the Cauchy-Riemann equations, then $f$ is holomorphic. My definition of holomorphic here is that the limit $\lim_{\zeta \rightarrow 0} \frac{f(z + \zeta) - f(z)}{\zeta}$ exists and is a complex number. This is not hard to prove, as I outline below, but I'm having trouble seeing where in this proof I'm actually using the continuity of the derivative of $f$. Here's the argument:

Proof: Let $z_0 = x_0 + iy_0 \in \mathbb{C}$, and let $A = u_x(x_0, y_0)$, $B = u_y(x_0, y_0)$, $C = v_x(x_0, y_0)$ and $D = v_y(x_0, y_0)$.

Since $u$ is differentiable at $(x_0, y_0)$, we can write

$\epsilon_1(h, k) = u(x_0 + h, y_0 + k) - u(x_0, y_0) - Ah - Bk,$ where $\epsilon_1(h, k)/\|(h, k)\|$ goes to zero as $(h, k) \rightarrow 0$. Now, for a moment there I thought this was the part where you need continuity of the partials $u_x$ and $u_y$, but note that they're being computed at a fixed point $(x_0, y_0)$, and the fact that $\epsilon_1$ goes to zero above is just the definition of differentiability, isn't it?

If we accept this and do the same expansion for $v$, then it's easy to see that the aforementioned limit will exist and will be a complex number, because of the Cauchy-Riemann equations.

So, again, my question is:

Where in the above proof are we using that $f$ is $C^1$, and not just differentiable?

Thanks!

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    In other words, there is nothing wrong with your $p$roof, but maybe you are confused about the word *differentiable*. For functions of one real variable, it means the same as having a derivative. Not so for functions of *several* variables.2012-10-09

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EDIT:I misunderstood your definition.

I'ld like to point out When is a function satisfying the Cauchy-Riemann equations holomorphic? instead.


Example Consider the function$f(z)=e^{-\frac{1}{z^4}} \ \ \forall z \neq 0$ $f(0)=0$

This function satisfies cauchy Riemann equation but is not holomorphic in any open neighbourhood of $0$.

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    Thanks for the link, which contains relevant information. The only problem is that my question is really about my proof, not about $\mathbb{C}$-differentiability or conditions implying analiticity/holomorphy (I changed the title). By the way, this seems to be the same proof outlined there, but unfortunately nobody there said anything about it.2012-10-08