Zero of a complex polynomial satisfying one of three assertions.

Let $n$ be a positive integer greater than $1$. Prove that if $x$ is a zero of $ X^n+1+(-1)^n(X+1)^n$ then $|x|=1$ or $|x+1|=1$ or $|x+1|=|x|$.

My initial thought was to study the cases $n=2,3,4$ (that confirm indeed the result) in order to establish a general approach, but I didn't succeed to do this.

The set of roots of this polynomial has several symmetries:

• It is symmetric in the line $\operatorname{Im}(z) = 0$.
• It is symmetric in the line $\operatorname{Re}(z) = -\tfrac{1}{2}$.
• It is symmetric under $z \mapsto z^{-1}$ ($z \neq 0$).

Note that the last symmetry maps the line $\operatorname{Re}(z) = -\tfrac{1}{2}$ to $\{z \mid |z+1|=1\}$ and vice versa.

If $n$ is odd then $\{-1,0\}$ are roots. Let $\zeta$ be a primitive third root of unity. Then $\zeta + 1$ is a primitive sixth root of unity. If $3 \nmid n$ then $\zeta^{\pm 1}$ are roots and if $n \equiv 1, 4$ ($\bmod$ $6$) then $\zeta^{\pm 1}$ are double roots. (One can check this by considering the derivative at $\zeta^{\pm 1}$.)

If we can show that the polynomial has sufficiently many zeroes on the unit circle then its symmetries may suffice to locate all zeroes, which will then automatically satisfy the required conditions.

Substitute $X \leftarrow e^{2it}$ to get

$$\begin{eqnarray} e^{2i n t} + 1 + (-1)^n(e^{2it} + 1)^n &=& e^{2int} + 1 + (-1)^ne^{int}(2\cos(t))^n \\ &=& e^{int}\left(2\cos(nt) + 2^n(-\cos(t))^n \right) \end{eqnarray} $$

and therefore $e^{2it}$ is a zero precisely if $$ \cos(nt) + 2^{n-1}(-\cos(t))^n = 0.$$

If $t \in [\tfrac{\pi}{3}, \tfrac{\pi}{2}]$ then $|2^{n-1}\cos(t)^n| \leq \tfrac{1}{2}$ while $\cos(n t)$ oscillates between $-1$ and $1$. Therefore there must be a root between any two points where $\cos(nt)$ goes from $-1$ to $1$ or vice versa. Now $\cos(n t)= \pm 1$ precisely if $nt=m \pi$ for some integer $m$. By symmetry, each root in the open interval accounts for six roots of the polynomial. Let's check what happens by considering six different cases $n$ $\bmod$ $6$.

$n = 6k$: $\cos(n t) = \pm 1$ when $2k \leq m \leq 3k$ so $k+1$ times and therefore at least $k$ roots on the open interval. All $6k$ roots are accounted for.

$n=6k+1$. At least double roots at $\zeta^{\pm 1}$ and simple roots at $\{-1,0\}$. Now $\cos(n t) = \pm 1$ when $2k + 1 \leq m \leq 3k$ so $k$ times accounting for at least $k-1$ roots. So all $6k$ roots are accounted for.

$n=6k+2$. At least simple roots at $\zeta^{\pm 1}$. Now $\cos(n t) = \pm 1$ when $2k + 1 \leq m \leq 3k + 1$ so $k + 1$ times accounting for at least $k$ roots. So all $6k + 2$ roots are accounted for.

$n=6k+3$. At least simple roots at $\{-1,0\}$. Now $\cos(n t) = \pm 1$ when $2k + 1 \leq m \leq 3k + 1$ so $k + 1$ times accounting for at least $k$ roots. So all $6k + 2$ roots are accounted for.

$n=6k+4$. At least double roots at $\zeta^{\pm 1}$. Now $\cos(n t) = \pm 1$ when $2k + 2 \leq m \leq 3k + 2$ so $k + 1$ times accounting for at least $k$ roots. So all $6k + 4$ roots are accounted for.

$n=6k+5$. At least simple roots at $\zeta^{\pm 1}$ and $\{-1,0\}$. Now $\cos(n t) = \pm 1$ when $2k + 2 \leq m \leq 3k + 2$ so $k + 1$ times accounting for at least $k$ roots. So all $6k + 4$ roots are accounted for.

So in all cases we conclude that the roots satisfy the required conditions.