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Let $P$ be a polynomial and suppose $f : \Bbb{C}\longrightarrow \Bbb{C}$ is a non-constant analytic function such for all $z \in \Bbb{C}, f(z) = f(P(z))$. Clearly when $P$ is linear we can find such $f$. what happens when $P$ is not linear ? Any suggestion would be helpful.

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«Clearly» when $P$ has degree $1$ and a fixed-point which is either attractive or repulsive then every such $f$ is constant. This is the generic situation for polynomials of degree $1$ (especially linear ones). –  Loïc Teyssier Jul 26 '13 at 14:11
    
The linear function $z\mapsto\exp(i\alpha\pi)z$, with $\alpha$ real and irrational, is not covered by Loic's comment but also admits no non-constant $f$ as in the question. –  Andreas Blass Jul 26 '13 at 15:29

2 Answers 2

up vote 3 down vote accepted

Here is a more conventional proof:-) Let $M(r)=\max\{|f(z)|:|z|=r\}$. Maximum principle implies that this function is strictly increasing (unless $f$ is constant). This gives a contradiction because |P(z)|>|z| when $z$ is sufficiently large, and $P$ is of degree greater than $1$.

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Mine was more «dynamical» and yours «analytical». I don't know about «conventional» ;) –  Loïc Teyssier Jul 27 '13 at 16:55

If the degree of $P$ is greater than $1$ then the Julia set $J$ of $P$ is a nonempty perfect compact set of $\mathbb C$, completely invariant by $P$. Obviously $f$ is constant on any orbit $(P^{\circ n}(z))_n$. It is well known that most orbits of points of $J$ are dense in $J$, so that $f$ must be constant on $J$ (and therefore constant everywhere).

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