Can Rational Composed with Non-Rational Function Be Rational?

Question

Suppose $f(z)$ is continuous and surjective onto $\mathbb{C}$, but not rational. Let $R$ and $Q$ be non-constant rational functions. Can we have $R(f(z))=Q(z)$?

Clearly this is possible if $R$ and $Q$ can be constant. This came up in something I was working on and I want to say no but can’t prove it. Specifically I want to know if there can be a rational $R$ such that $$R(z)=\frac{1}{f(z)}+R(f(z))$$ In case that has a different answer from the general case.

Answer 1

Yes.

Let $f(z) = z^{m/n}$ and $R(z) = Q(z^n)$ for any rational function $Q$. If $n\nmid m$, then $f(z)$ is not rational but both $R(z)$ and $R(f(z)) = Q(z^m)$ are rational.

If $m > n$, then setting $\operatorname{dom}(f) = \Bbb C \setminus (-\infty, 0)$ and choosing the principal branch of $\sqrt[n]z$ provides a continuous surjective function.