I came across this question whilst doing some research into complex analysis, and I just can't see what to do!

Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\overline{z})}$ is holomorphic, whilst $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.

I know that holomorphic means that the function is differentiable everywhere, and I need to apply the Cauchy-Riemann equations somehow, but I'm not sure how to approach this.


2 Answers 2


Since $f$ is entire, it has a Taylor expansion (centered at $z_0\in \mathbb{C}$) $$f(z)=\sum_{n\ge 0}a_n(z-z_0)^n$$ Hence, $$\overline{ f(\bar z)}=\overline{\sum_{n\ge 0}a_n(\overline{z}-z_0)^n}=\sum_{n\ge 0}\overline{a_n}(z-\overline{z_0})^n$$ and we know this series converges since you can show it has the same radius of convergence as the original series. Therefore, $\overline{ f(\bar z)}$ is holomorphic.


So you want to show that if $f(z)$ is holomorphic, then $\overline{f(\bar z)}$ is holomorphic too.

I think it will be easiest not to split into real and imaginary parts -- so no Cauchy-Riemann -- but instead work directly from the definition of differentiability.

A natural guess would be that $\frac{d}{dz} \overline{f(\bar z)}$ would be $\overline{f'(\bar z)}$. Can you show that this is in fact the case?

For the second part, perhaps show that if $g(z)$ and $\overline{g(z)}$ are both holomorphic, then $g$ is constant. (Here, using Cauchy-Riemann feels more promising).

  • $\begingroup$ thanks, I've done the first part, but I don't see how to do the second part, or why it shows what I want to get $\endgroup$
    – Tom
    Nov 4, 2013 at 12:03
  • $\begingroup$ @Tom: If $\overline{f(\overline{z})}$ and $f(\overline{z})$ are analytic, then $f(\overline{z})$ is constant. Cauchy-Riemann equations can help you see that. $\endgroup$
    – Francis
    Nov 4, 2013 at 12:10

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