This post is closely related to Entire function whose square or composition with itself is a polynomial.
In the linked post, the OP asked to prove the following:
If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f^{2}(z)$ is a polynomial, then $f$ is also a polynomial.
If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f(f(z))$ is a polynomial, then $f$ is also a polynomial.
Several proofs were given: see Proving $f(z)$ and $g(z)$ are polynomials for a proof using Casorati-Weierstrass Theorem, and Two confusions on Great Picard Theorem for a proof using Great Picard Theorem.
My question is what if $f$ is not entire in the first place, does $f^{2}(z)$ being a polynomial still imply $f$ being a polynomial? and does $f(f(z))$ being a polynomial imply $f$ be a polynomial?
As the proof depends heavily on the entire property of $f$, I believe the answer for both of my questions is NO.
But I cannot find a counterexample.
For the first one, I tried $f(z):=|z|$, but then $f(z)\neq z^{2}$, because $|z|$ is a complex norm, so that $$f(z)=|z|^{2}=x^{2}+y^{2}\neq z^{2}=x^{2}+2ixy-y^{2}.$$
Without absolute value, I am not sure what else I can try since $f^{2}(z)$ must be a polynomial (so all the fraction without removable singularities seems not work well..)
For the second one, perhaps we can use $f(z):=\dfrac{1}{z}$ which has a pole at $z=0$, and $f(f(z))=z$ is a polynomial.
Is my second example correct? Any idea about the first example? Thank you!
What do you mean with 'not entire' ?
$f(z)=1/z^n$ gives $f(f(z))=z^{n^2}$
with $\phi$ biholomorphic (or just bijective ?) on the right open set then $g(z)= \phi^{-1}(1/\phi(z))$ gives $g(g(z)) = z$,
same for $h(z)= \phi^{-1}(1-\phi(z)),h(h(z))=z$.