Assume that $P(z)$, $Q(z)$ are complex polynomials such that $P(S)=Q(S)$, where $S=\{z\colon |z|=1\}$ (equality is understood in the sense of sets, but I do not know the answer even for multisets). Does it follow that there exist polynomial $f(z)$, positive integers $m,n$ and complex number $w\in S$ such that $P(z)=f(z^n)$, $Q(z)=f(wz^m)$?

It is motivated by this question (and if above claim is true, it actually implies much more than asked therein.) I started a new question with algebraic geometry tag, since it looks reasonable and may pay attention of right people rather than comments to an old post.

  • $\begingroup$ Not related, but FWIW the following gives the general form of 2 polynomials having pointwise equal absolute values on the unit circle: Rational function with absolute value 1 on unit circle. $\endgroup$
    – dxiv
    Jan 24, 2016 at 1:26
  • $\begingroup$ Is it possible to prove it in the special case when the images $P(S)$ and $Q(S)$ are a non-intersecting curves? $\endgroup$ Jan 24, 2016 at 1:43
  • $\begingroup$ P.S. From the link posted in my previous comment it also follows that if $P(z) = \lambda z^n$ then $Q(z) = \mu z^m$ with $|\mu| = |\lambda|$. Which of course is still quite far from the conjecture proposed here. $\endgroup$
    – dxiv
    Jan 24, 2016 at 1:45

3 Answers 3


This is a special case of the main theorem in the paper by I. N. Baker, J. A. Deddens, and J. L. Ullman, A theorem on entire functions with applications to Toeplitz operators, Duke Math. J. Volume 41, Number 4 (1974), 739-745.

They proved a similar statement for arbitrary entire functions.


One can also argue algebraically.

First of all, the unit circle $S^1$ in $\mathbf C$ is a real algebraic curve in $\mathbf R^2$. Its complexification $S^1_{\mathbf C}$ in $\mathbf C^2$ is the complex projective line minus two complex conjugate points. A convenient model for $S^1_{\mathbf C}$ is $\mathbf{P}_{\mathbf C}^1\setminus\{0,\infty\}$, with complex conjugation acting as $z\mapsto1/\bar z$. The algebraic endomorphisms of the latter are the endomorphisms of the form $z\mapsto wz^m$ with $w$ a nonzero complex number and $m$ an integer. Such an endomorphism commutes with complex conjugation if and only if $|w|=1$. This will be useful below.

Another useful fact is that the set of algebraic morphisms from $S_{\mathbf C}^1$ into $\mathbf C^2$ that commute with complex conjugation can be identified with the $\mathbf C$-algebra $\mathbf C[z,1/z]$ of Laurent polynomials. Indeed, both coincide with the set of real polynomial maps from $S^1$ into $\mathbf R^2=\mathbf C$.

Now, the complex polynomial $P\in\mathbf C[z]$ is a real polynomial endomorphism of $\mathbf R^2$, and complexifies to a complex polynomial endomorphism $P_{\mathbf C}$ of $\mathbf C^2$. Assuming that $P$ is nonconstant, the image of $S_{\mathbf C}^1$ by $P_{\mathbf C}$ is an affine complex algebraic curve in $\mathbf C^2$ since $P$ is a polynomial. Moreover, it is real in the sense that it is stable for complex conjugation on $\mathbf C^2$. To put it otherwise, it is the complexification $C_{\mathbf C}$ of a real algebraic curve $C$ in $\mathbf R^2$. The curve $C$ contains the real curve $P(S^1)$, but is not necessarily equal to it. The complement of $P(S^1)$ in $C$ is a finite set of points. The normalization $\tilde C_{\mathbf C}$ of $C_{\mathbf C}$ is an open affine subset of ${\mathbf P}_{\mathbf C}^1$. Since $P$ is a polynomial, the complement of $\tilde C_{\mathbf C}$ in ${\mathbf P}_{\mathbf C}^1$ is a doubleton. We may assume that $\tilde C_{\mathbf C}$ is ${\mathbf P}_{\mathbf C}^1\setminus\{0,1\}$. The complex conjugation on $C_{\mathbf C}$ induces a complex conjugation on ${\mathbf P}_{\mathbf C}^1\setminus\{0,1\}$. Since $P(S^1)$ is compact, this complex conjugation is the same as the one above, i.e., there is an isomorphism from $\tilde C_{\mathbf C} $ to $S_{\mathbf C}^1$ that commutes with complex conjugation. Using the facts recalled above, it follows that there are a complex Laurent polynomial $f\in\mathbf C[z,1/z]$ and a nonzero integer $n$ such that $P(z)=f(z^n)$ on $S^1$, and $f_\mathbf{C}$ is a birational morphism from $S_{\mathbf C}^1$ to $C_{\mathbf C}$. Of course, one may assume that $n$ is a natural number. The Laurent polynomial $f$ is then a true polynomial. Since $f_{\mathbf C}$ is birational, one has a rational map $f_{\mathbf C}^{-1}\circ Q_{\mathbf C}$ of $S_{\mathbf C}^1$ into itself. This rational map is a true endomorphism since $S_{\mathbf C}^1$ is nonsingular, $f_{\mathbf C}$ is the normalization morphism from $S_{\mathbf C}^1$ to $C_{\mathbf C}$ and $Q$ is a polynomial. It follows that there are a nonzero integer $m$ and a complex number $w$ with $|w|=1$ such that $f^{-1}\circ Q(z)=wz^m$ on $S^1$, i.e. $Q(z)=f(wz^m)$ on $S^1$. Of course, the integer $m$ is natural and one has $Q(z)=f(wz^m)$ on $\mathbf C$.

  • $\begingroup$ Oops, sure. Deleted my comment. $\endgroup$
    – Qfwfq
    Jan 25, 2016 at 22:11

Am I missing something, or does not the following work?

The difference $R(z):=P(z)-Q(z)$ then 0 on the unit circle. By the maximum modulus principle, $R(z)$ is $0$ in the entire unit disk, and then $R(z)$ must be the constant $0$, implying that $P(z)=Q(z)$.

  • 6
    $\begingroup$ We only know $P(S) = Q(S)$ (as sets), not $P(z) = Q(z)$ for all $z \in S$. $\endgroup$ Jan 24, 2016 at 1:39
  • $\begingroup$ Ah, they agree as SETS, my mistake. $\endgroup$ Jan 24, 2016 at 1:41
  • 1
    $\begingroup$ Actually the argument is even a bit simpler here, if a polynomial has infinitely many zeros, it is the zero polynomial. No need for any real theorems. I also read the question this way at first. $\endgroup$ Jan 25, 2016 at 2:11
  • $\begingroup$ @RichardRast: That's why I should not do math while having fever... $\endgroup$ Jan 25, 2016 at 4:06
  • 2
    $\begingroup$ This is why I should write not only formally correctly, but reader-friendly. $\endgroup$ Jan 25, 2016 at 15:17

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