Any algebraic number with modulus $1$ is root of a polynomial with positive coefficients

Question

Given a complex number on the unit circle $e^{i\theta}\neq 1$ that is the root of some polynomial in $\mathbb Z[x]$, can we always construct a polynomial $p(x)$ with positive integer coefficients such that $e^{i\theta}$ is a root of $p(x)$?

Answer 1

$\frac{2+i 2^{1/4}}{2-i2^{1/4}}$ is on the unit circle and it is a root of the same polynomials in $\Bbb{Q}[x]$ as $\frac{2+2^{1/4}}{2-2^{1/4}}>0$ which can't be the root of a polynomial with non-negative coefficients.

Answer 2

Nope, a counter-example are the complex numbers $\mu_{\pm} =-\frac{1 \pm i\sqrt{\varphi}}{\varphi}$.

It is easy to check $|\mu_{\pm}| = 1$ and $\mu_{\pm}$ are the roots of the irreducible polynomial $$q(x) = x^4 -2x^3 -2x^2 -2x + 1$$ This means $\mu_{\pm}$ are algebraic integers.

On the other hand, $q(x)$ have positive real roots $\nu_{\pm} = \varphi \pm \sqrt{\varphi}$. If any $\mu_{\pm}$ is root of $p(x) \in \mathbb{Z}[x]$, then $q(x) | p(x) \implies p(\nu_{\pm}) = 0$. This implies coefficients of $p(x)$ cannot be all positive.