Roots of a polynomial of $n^{\rm th}$ degree having phase $e^{\frac{i2\pi}{m}}$

Question

Having a polynomial $p_n(x)$ of degree $n$ with real coefficients and only complex solutions (i.e., no real solutions) $x_1,\dots,x_n$ and $0 < \arg(x_k) < 2\pi$ for all $k\in\{1,...,n\}$, is it generally possible that some solution $x_k$ fulfills $\arg\left(x_k^m\right)=0$ (in other words it be positively real) if $m<n$ ?

I presume it is not possible, but can it be proven?

Answer 1

It seems trivial to me, or I missed something.

Take any phase $\theta$ you want, pick a real number $r$, then

$p(x) = [\text{some other factors}](x^2+1)^2\cdot(x-re^{i\theta})(x-re^{-i\theta})$ would be a solution.