A family of polynomials whose roots all have moduli 1

Question

Let $k$ be a natural number. Let $$f_k(x)=(k+1)x(x^{2k}-1)+k(x^{2k+2}-1)=0.$$ Using Mathematica, I find that, when $k\le 13$, all of the roots of $f_k(x)$ have moduli 1.

Does this hold for all $k$? If YES, how to prove it?

Thanks to everyone!

Answer 1

This follows from surprisingly simple properties of the sine function, when we make the substitution $x=e^{it}, t\in[0,2\pi)$.

We see that $$ x^{-(k+1)}f_k(x)=k\left(x^{k+1}-\frac1{x^{k+1}}\right)+(k+1)\left(x^k-\frac1{x^k}\right). $$ So with $x=e^{it}$ we see that $f(e^{it})=0$ if and only if $$ g_k(t)=k\sin((k+1)t)+(k+1)\sin(kt)=0. $$ It is easy to check that $f_k(x)$ has a simple zero at $x=1$ and a triple zero at $x=-1$ corresponding to a simple zero of $g_k(t)$ at $t=0$ and a triple zero at $t=\pi$.

Let's look at the sign changes in the interval $[0,\pi]$ (the other half is symmetric because $g_k(t)$ is an odd function). Fix an integer $j$ in the range $0<j\le k$. At $t_j:=j\pi/(k+1)$ we have $$ \begin{aligned} g_k(t_j)&=k\sin((k+1)j\pi/(k+1))+(k+1)\sin(kj\pi/(k+1))\\ &=(k+1)\sin(kj\pi/(k+1)). \end{aligned} $$ For this range of $j$ we have $\lfloor\frac{kj}{k+1}\rfloor=j-1$. It follows that $t_j\in ((j-1)\pi,j\pi)$, implying that $g(t_j)$ has the same sign as $(-1)^{j+1}$ for all choices $j=1,2,\ldots,k$. Therefore $g_k(t)$ has a zero in all the intervals $(t_j,t_{j+1})$, $j=1,2,\ldots,k-1$, for a total of $k-1$ zeros in the interval $(0,\pi)$. Together with the $(k-1)$ mirror image zeros in the interval $t\in(\pi,2\pi)$ and the four known zeros (simple at $t=0$ and triple at $t=\pi$) we have accounted for $$ N=1+3+2(k-1)=2k+2=\deg f_k(x) $$ zeros of $f_k(x)$ on the unit circle. The claim follows.

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In the plot below $k=7$. The more complicated (blue) wave is $g_7(t)$. The orange simple wave is $8\sin 7t$. Its sign determines the sign of $g_7(t)$ at the points $t_1,\ldots t_7$. These are the points in $(0,\pi)$ where the two waves intersect, and are also tickmarked on the horizontal axis.