Are complex differentiable function in a point analytic? I know that if a function $f$ is complex differentiable in a neighborhood of $z_0$, then we say it's holomorphic in $z_0$ and it's also analytic in a neighborhood of $z_0$.
But suppose that I know that $f'(z_0) = 0$ in the complex sense. I don't know what happens in a neighborhood of $z_0$; so can I say that the function $f$ is still analytic in $z_0$? 
I don't think we can conclude that there is a neighborhood of $z_0$ such that the function is analytics, but then again I'm not sure.
Are there maybe additional condition to assume to ensure that the function is analytic?
Thank you!
 A: Complex differentiability at a point $z_0$ doesn't imply anything about complex differentiability at any other point. A function like $z\mapsto z\cdot \overline{z}$ is complex differentiable only at $z_0 = 0$, but it is infinitely often real differentiable, even real-analytic, on the whole plane. An example like $g(z) = \overline{z}(1-\lvert z\rvert^2)$ shows that a (real-analytic) function can be complex differentiable at each of a set of non-isolated points without being (complex) analytic anywhere.
There are some criteria that ensure a function is analytic at a point $z_0$ without a priori demanding that it be complex differentiable on a neighbourhood of $z_0$, for example it is an easy consequence of Morera's theorem that a continuous function on the unit disk $D$ is analytic at $0$ if it is complex differentiable on $D\setminus \mathbb{R}$, but such criteria require far far stronger hypotheses than just complex differentiability at a "small" set of points.
A: By Cauchy's theory, the notions of holomorphic function in an open set $U$ of $\mathbf C$ and of analytic function in $U$ are the same. Remember derivability is a much stronger condition in $\mathbf C$ than in $\mathbf R$ – for instance a differentiable function of a real variable need not to be $\mathcal C^\infty$, and even if it be, it need not be the sum of its Taylor series.
