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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!

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    $\begingroup$ If a function is complex differentiable at $z_0$, it need not be complex differentiable at any other point, consider $f(z) = z(1+\overline{z})$. $\endgroup$
    – Daniel Fischer
    May 17, 2015 at 14:12
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    $\begingroup$ The terms "complex differentiable (at some point)", and "analytic at some region are different: the latter implies the former, but the other way around is false as Daniel's example shows. $\endgroup$
    – Timbuc
    May 17, 2015 at 14:16
  • $\begingroup$ @DanielFischer Thanks. Your $f$ is also differentiable in a neighborhood of $0$, but the CR conditions only hold in $0$, so $f$ is only complex differentiable there. You can make that an answer if you want :) $\endgroup$
    – Ant
    May 17, 2015 at 14:47

2 Answers 2

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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.

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  • $\begingroup$ thank you for this comprehensive answer! Much appreciated! :D $\endgroup$
    – Ant
    May 17, 2015 at 15:23
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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.

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  • $\begingroup$ yes but $\{z_0\}$ is closed $\endgroup$
    – Ant
    May 17, 2015 at 14:12
  • $\begingroup$ I thought from your question $f$ was holomorphic – which implies it is differentiable on an open set. $\endgroup$
    – Bernard
    May 17, 2015 at 14:19
  • $\begingroup$ I'm sorry maybe the title should be changed.. However I thought I explained in the text of the question :) $\endgroup$
    – Ant
    May 17, 2015 at 14:20
  • $\begingroup$ That was not quite clear to me. I'll withdraw my answer; $\endgroup$
    – Bernard
    May 17, 2015 at 14:20

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