# Holomorphic function in an open connected set such that for some real number in the domain all the derivatives of the function at that point are real

Let $$D \subseteq \mathbb C$$ be an open connected set such that $$z\in D \implies \bar z\in D$$. Let $$f : D \to \mathbb C$$ be a holomrophic function such that $$\exists z_0 \in D \cap \mathbb R$$ such that $$f^{(n)}(z_0) \in \mathbb R, \forall n \ge 0$$.

How to show that $$\overline {f(\bar z)}=f(z) , \forall z\in D$$ ?

For any neighborhood of $$z_0$$ that is contained in $$D$$, by power series expansion, I can easily see that $$f(z)=\overline {f(\bar z)}$$ in that neighborhood. Bit then, how do I show it globally on $$D$$ ? If I could show $$\overline {f(\bar z)}$$ is holomorphic , then I would be done, but I can't even show that.

Both $$f$$ and $$z\mapsto\overline{f\left(\overline z\right)}$$ are analytic functions. You have already proved that are equal in an open set. But then they are equal everywhere, by the identity theorem and because $$D$$ is connected.
The function $$z\mapsto\overline{f\left(\overline z\right)}$$ is analytic because, if $$f(x+yi)=u(x,y)+v(x,y)i$$, then$$\overline{f\left(\overline{x+yi}\right)}=u(x,-y)-v(x,-y)i$$and it is now easy to check the the Cauchy Riemann equations hold for every point of the domain.
• please see my question body ... I am trying to show $\overline {f(\bar z)}$ is indeed holomorphic ... so it would be appreciated if you could give a sketch for that ... Commented May 8, 2019 at 23:13