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.

Please help.


1 Answer 1


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.

  • $\begingroup$ 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 ... $\endgroup$
    – user102248
    Commented May 8, 2019 at 23:13
  • $\begingroup$ I've edited my answer. What do you think now? $\endgroup$ Commented May 8, 2019 at 23:18

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