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Let's consider the rational functions whose numerator and denominator of the function term are coprime.

Which kinds of rational functions of one variable have an inverse relation that contains a branch that is a rational function?

Which kinds of polynomial functions of one variable have an inverse relation that contains a branch that is a rational function?

I assume the degree of the numerator and the degree of the denominator of the function term has to be less than or equal to $1$. Some calculations with algebraic equations with undetermined parameters as coefficients seem to show that. But I'm not sure.

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If $R$ is a rational function and $S$ a branch of $R^{-1}$ in an open set $U \subset \Bbb C$ then $S(R(z)) = z$ in $U$. If $S$ is also a rational function then it follows that $S(R(z)) = z$ globally (as meromorphic functions).

It follows that $S$ is injective and therefore has degree one. Then $R$ has degree one as well.

So the only rational functions with a (local) rational branch of the inverse are rational functions of degree one (which are the Möbius transformations).

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  • $\begingroup$ "$S$ is injective and therefore has degree one." Why $S$ must be degree one? For example, $S(z)=1/z$, $S^{-1}=1/z$ is rational. $\endgroup$
    – hbghlyj
    Nov 1, 2022 at 14:52
  • $\begingroup$ @hbghlyj: $S(z) = 1/z$ is injective and has degree one. $\endgroup$
    – Martin R
    Nov 1, 2022 at 14:53
  • $\begingroup$ You are right. The definition of degree of $\frac{p(x)}{q(x)}$ where $\gcd(p(x),q(x))=1$ is $\max(\deg p,\deg q)$. (Related: math.stackexchange.com/questions/3389265) Thanks! $\endgroup$
    – hbghlyj
    Nov 1, 2022 at 15:01

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