Prove an entire function is constant on complex plane

Question

Let $f(z)$, $F(z)$ be two analytic functions on $\Bbb C$ satisfies $f(z)=F(\overline{f(z)})$. Here $\overline{f(z)}$ is the complex conjugate of $f(z)$. Prove that $f(z)$ is constant on $\Bbb C$.

I tried to use Cauchy-Riemann equations but I didn't work, and I don't know what else I can do. Any help would be appreciated.

Answer 1

If $F$ is constant, then we are done. Otherwise suppose $f$ is not constant. Then $V := f(\mathbb{C})$ is open by the open mapping theorem. Since $F$ is not constant, $F'$ is not identically $0$ on $V$. Hence there exists an open ball $B(z_0, r) \subset V$ such that $F' \ne 0$ on this ball. By the inverse function theorem, $F$ is invertible with a holomorphic inverse in some (connected) neighborhood of $z_0$ in $B(z_0, r)$. Denote this inverse by $G$. Then $$ G(f(z)) = \overline{f(z)} $$ in the aforementioned neighborhood. However, by the Cauchy-Riemann equations, $ \overline{f(z)}$ is not holomorphic unless $f$ is constant.Thus $f$ is constant in this neighborhood. By analytic continuation, $f$ is constant everywhere.

Answer 2

Geometrically the issue is that analytic functions are orientation preserving and conjugation is orientation reversing. So if $f$ is non-constant, e.g. where $f(0)=w$ and $\gamma=r\cdot \exp\big(2\pi i\cdot t\big)$ for $t\in [0,1]$, $r\gt 0$ small enough, the Argument Principle says
$0 \lt n\big(f\circ \gamma,w\big)=\sum_{z, F(z)=w}v_F(z)\cdot n\big(\overline f\circ \gamma, z\big)\leq 0$
as the valency $v_F(z)\in \mathbb N$ and $n\big(\overline f\circ \gamma, z\big)\leq 0$ for any $z\in \mathbb C-[\bar f\circ \gamma]$
(i.e. all $z$ in the complement of the image of $\overline f\circ \gamma$)
and this is impossible.

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proof of orientation reversal for conjugation
let $f$ be any analytic function in a simply connected Domain, $\gamma$ a closed curve and the winding number $n\big(f\circ \gamma, w\big)$ be well defined. Define $g(z):=f(z)-w$. Then
$0=n\Big(\big(g(z)\cdot \overline g(z)\big)\circ \gamma,0\Big)=n\Big(g\circ \gamma,0\Big)+n\Big(\overline g\circ \gamma,0\Big)=n\Big(f\circ \gamma,w\Big)+n\Big(\overline f\circ \gamma,\overline w\Big)$
the Left Hand Side follows because $g(z)\cdot \overline g(z)\in \mathbb R_{\gt 0}$

$\implies-n\Big(\overline f\circ \gamma,\overline w\Big)= n\Big(f\circ \gamma,w\Big)\geq 0 \text{ (Argument Principle)}$
and the LHS holds for arbitary $\overline w$ in the complement of the image of $\overline f\circ \gamma$

Answer 3

A proof outline. suppose $f(z)=F(\bar{f(z)})$ and $f,F$ analytic.

You can use the Cauchy-Riemann equations to prove that if a function is analytic, its complement is not, unless it's constant.

The inverse of a holomorphic function is holomorphic.

So $F^{-1}(f(x))=\bar{f(x)}$, but we have a holomorphic function on the left and one that is not on the right, unless its constant.