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Let $F\left(z\right)$ be a polynomial of degree $d≥2$ with complex coefficients. Let $D$ be an open disk in the complex plane containing no critical points of $F\left(z\right)$.

Let $f\left(z\right)$ denote the inverse of $F\left(z\right)$; since $F\left(z\right)$ is not one-to-one, $f\left(z\right)$ is a multivalued function with d distinct branches: $f_{1}\left(z\right),f_{2}\left(z\right),...,f_{d}\left(z\right)$. Let $m,n$ be two distinct integers in the set $\left\{ 1,2,...,d\right\}$

Then, does there necessarily exist a function:

$T_{m,n}\left(z\right)=a_{m,n}z+b_{m,n}$

(with the constants $a_{m,n}$ and $b_{m,n}$ being complex numbers to be determined) such that:

$T_{m,n}\left(f_{m}\left(z\right)\right)=f_{n}\left(z\right),\textrm{ }\forall z\in D$

This little question has been something of a thorn in my research. I'd like to think that it is true.

Example: Let $F\left(z\right)=\left(\frac{z}{2}-i\right)^{3}-1$ .

$F$ has three inverses:

$f_{m}\left(z\right)=2i+2\omega^{m-1}\left(z+1\right)^{1/3}$

where $\omega=e^{\frac{2\pi i}{3}}$ and $m=1,2,3$.

These inverses are related to one another by the linear function:

$T_{m,n}\left(z\right)=2i+\omega^{m-n}\left(z-2i\right)$

Any thoughts on whether or not this is true? And how one might go about showing it?

Max
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1 Answers1

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It's not true for cubics in general. Try $$ F(X) = X^3 + c_2 X^2 + c_1 X + c_0$$ The roots of $F(X) - z$ are $f_1, f_2, f_3$ with $$ \eqalign{f_1 + f_2 + f_3 &= -c_2\cr f_1 f_2 + f_1 f_3 + f_2 f_3 &= c_1\cr f_1 f_2 f_3 &= -c_0 - z\cr} $$ Note that the first two equations don't involve $z$. We can eliminate $f_3$ from the first two equations, obtaining $$ f_1^2 + f_1 f_2 + f_2^2 + c_2 (f_1 + f_2) + c_1 = 0$$ But unless this factors into linear factors, you can't have an equation of the form $f_2 = a f_1 + b$. You happened to pick one that factors.

Robert Israel
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