This is part of home exercise, and it annoys me whole two days(definitely more time invested than supposed to).

Suppose we have two palindomic polynomials of same degree, $p(x)$ and $q(x)$. (Palindromic polynomial is polynomial which equals to own reciprocal polynomial)

Let $z\in\mathbb{C}$. If $|z|=1$, then $\frac{p(z)}{q(z)}\in\mathbb{R}$

For even degree things are sort of clear(for example purpose, lets say degree is 4, any even degree works same way):$p(x)=Ax^4+Bx^3+Cx^2+Bx+A$ and $q(x)=A'x^4+B'x^3+C'x^2+B'x+A'$

Lets transform $p(x)$, $p(x)=(\frac{A}{C}x^2+\frac{B}{C}x^1+1+\frac{B}{C}x^{-1}+\frac{A}{C}x^{-2})\frac{1}{Cx^2}$

Similary $q(x)=(\frac{A'}{C'}x^2+\frac{B'}{C'}x^1+1+\frac{B'}{C'}x^{-1}+\frac{A'}{C'}x^{-2})\frac{1}{C'x^2}$

Now complex arguments of $\frac{A}{C}x^2+\frac{B}{C}x^1+1+\frac{B}{C}x^{-1}+\frac{A}{C}x^{-2}$ and $\frac{A'}{C'}x^2+\frac{B'}{C'}x^1+1+\frac{B'}{C'}x^{-1}+\frac{A'}{C'}x^{-2}$ are $0$ (for $|x|=1$), so those are real, which leaves argument of $\frac{C'}{C}$ which is also real, and hence $p(x)/q(x)$ also real.

(Actually thinking of $C=0$ brings that $x^{-2}$ works even better than $Cx^{-2}$)

Why does it work with odd degree?

(Polynomials are in $\mathbb{R}[x]$ ring)

  • $\begingroup$ Are you sure there aren't additional constraints? $\mathrm{i}$ and $\mathrm{1}$ are palindromic polynomials, but $\dfrac{\mathrm{i}}{1} \not\in \Bbb{R}$. $\endgroup$ Mar 20, 2016 at 19:34
  • 1
    $\begingroup$ maybe coefficients should be real... $\endgroup$
    – Jef
    Mar 20, 2016 at 19:36

1 Answer 1


Assuming the polynomial coefficients are real, ...

Observe that $\dfrac{p(z)}{q(z)} = \dfrac{z^{\deg p}p(1/z)}{z^{\deg{q}}q(1/z)}$, so $\dfrac{p(z)q(1/z)}{q(z)p(1/z)} = z^{\deg p - \deg q} = z^0 = 1$. But (if you keep the powers of $z$ with their polynomials) this is a ratio of palindromic polynomials of even degree, which you know works (because products of palindromes are palindromes). Suppose we didn't know that ...

We want to show something about arguments when $|z| = 1$, so we replace $z \mapsto \mathrm{e}^{\mathrm{i} \theta}$ and notice $$ p(1/z) = p(\mathrm{e}^{-\mathrm{i}\theta}) = p(\overline{\mathrm{e}^{\mathrm{i}\theta}}) = \overline{p(\mathrm{e}^{\mathrm{i}\theta})} = \overline{p(z)} $$ and similarly for $q(1/z)$. So \begin{align*} 1 &= \frac{p(z)q(1/z)}{q(z)p(1/z)} \\ &= \frac{p(z)\overline{q(z)}}{q(z)\overline{p(z)}} \\ &= \frac{\mathrm{e}^{2 \mathrm{i} \arg(p(x))}}{\mathrm{e}^{2 \mathrm{i} \arg(q(x))}} \\ &= \mathrm{e}^{2 \mathrm{i} (\arg(p(x)) - \arg(q(x)))} \text{.} \end{align*} This means $\arg(p(x)) - \arg(q(x)) = 0 + 2\pi k$ for some integer $k$. But then their ratio is real.

(This is perhaps less surprising if you symmetrize the degrees first and notice that contributions to the arg from one term are cancelled by its palindromic friend. For instance $Az^2 + A = z(A\mathrm{e}^{\mathrm{i} \theta}+A\mathrm{e}^{-\mathrm{i} \theta})$ and $Az^3 + A = z^{3/2}(A\mathrm{e}^{\mathrm{i} (3/2)\theta}+A\mathrm{e}^{-\mathrm{i} (3/2)\theta})$. So the only contribution to the arg of $p$ and $q$ comes from the power of $z$ prefactor, which cancel in the ratio when their degrees are the same.)


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