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!
Best Answer
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.