Let $f(z)$ be a polynomial with integer coefficients of degree $n$. Also $f(z)$ is irreducible over the integers, and it is not a cyclotomic polynomial.
Let $v$ be a given nonreal complex number.
Consider solving $f(z)=0$ on the complex plane.
For a given $v$ let $z_0$ be the nearest number to it such that
$$f(z_0) = 0$$
Now for what integer polynomials $f(z)$ do we have some $v$ such that
$$f(z_0)=0, |z_0| = 1 $$
??
It might be useful to write
$$z_0^m = \cos(m w) + \sin(m w) i$$
( de moivre formula )
and then setting
$$f( \cos(w) + \sin(w) i ) = 0 $$
This way we solve for a real $w$ if it exists.
By using chebychev polynomials we can get a system of 2 polynomial equations for the real and imag part.
Maybe that helps.
Minimum polynomial probably related too.
How to classify the solutions $f(z)$ ??
I found the answer.
Lets say $f(z)$ is a polynomial with integer coefficients.
Then the zero's of $f(z)$ come in two kinds.
First the real ones.
Secondly pairs of complex conjugates.
Those complex conjugate pairs have ( for every pair ) a minimum polynomial of degree 2.
More precisely if the complex conjugate pairs are on the unit circle , then the minimum polynomial is of the form
$$ z^2 + 2 \cos(\theta) + 1$$
where $\theta$ is the argument of one the zero's.
This implies the minimum polynomial is of the form
$$ z^2 + 2 v + 1$$
where $v$ is a real number between $0$ and $1$.
This implies that $f(z)$ is irreducible ( so $+1$ or $-1$ is not a zero ) then we have the factorization over the reals of $f(z)$ :
$$f(z) = (z^2 + 2 v + 1)^n g(z)$$
for some integer $n$ and $v$ is a real number between $0$ and $1$ and a polynomial $g(z)$ with real coefficients.
We can reduce this by putting $n$ equal to $1$.
So
$$f(z) = (z^2 + 2 v + 1) h(z)$$
So basicly we are looking for a $v$ such that $v$ is a real number between $0$ and $1$ and
$$ f(z) = 0 \mod (z^2 + 2 v + 1) $$
This is just polynomial division and checking for a remainder, but with a parameter.