# Polynomials with roots having the same module and linear dependent arguments

Is it possible for a polynomial with integer coefficients to have some of its roots: $$m_1e^{i\theta_1 \pi}, m_2e^{i\theta_2 \pi}, \ldots, m_ke^{i\theta_k \pi}$$ such that there exist nonzero integers $a_1, a_2, \ldots, a_k$ and an integer $c$ with:

• $m_j = m_{j'}$ for all $j, j'$,
• The $\theta_j$'s are distinct,
• $a_1\theta_1 + a_2\theta_2 + \cdots + a_k\theta_k = c$.

Is it possible in the case that all the $\theta_j$'s are transcendental?

I have some opposing feelings regarding this question, and went in different directions with no luck, so any help is welcome.

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Have you had a chance to look at the answer I posted? Any comments, further questions? – Gerry Myerson Aug 12 '12 at 0:08

There are many examples. I'll start with the most trivial, and work my way up to the more interesting.

The polynomial $(x-1)^2$ has roots $e^{0\pi i}$ and $e^{2\pi i}$, and all three bullet points are satisfied. The $\theta_j$ are distinct, but the roots aren't; from here on, assume $0\le\theta_j\lt2$ to avoid this kind of example.

The polynomial $x^2-1$ has roots $e^{0\pi i}$ and $e^{1\pi i}$, and all three bullet points are satisfied.

If a polynomial with integer coefficients has any nonreal roots at all, then they come in complex conjugate pairs, $re^{i\pi\theta}$ and $re^{i\pi(2-\theta)}$, and $(1)\theta+(1)(2-\theta)=2$, so all three bullet points are satisfied. Moreover, $\theta$ will generally be transcendental.

If two of the roots are complex conjugates, then their moduli will be equal, their thetas will add to the integer zero, and the thetas can be transcendental.

If we rule out complex conjugate pairs, any polynomial that has a root of unity as a root will do, since all roots of unity are of the form $e^{2\pi ip/q}$, so the corresponding $\theta$s are rational. Similarly if it has roots $c\zeta$, $c$ a fixed real, $\zeta$ a root of unity.

EDIT: Here's another way to generate examples. Let $\alpha=re^{\pi i\theta}$ be any algebraic number, let $p/q$ be any rational, then $\beta=re^{\pi i(\theta+(p/q))}$ is also algebraic, and there is a polynomial that has $\alpha$ and $\beta$ among its roots, and those two roots meet all the bullet points. This generates so many examples, I am struggling to see how to make an interesting question out of it.

Now let's look at the question more generally. Let $r_j=me^{\pi i\theta_j}$. $$a_1\theta_1+a_2\theta_2+\cdots+a_k\theta_k=c$$ implies $$r_1^{a_1}r_2^{a_2}\cdots r_k^{a_k}=\pm m^a$$ where $a=a_1+a_2+\dots+a_k$. I don't know if there's any approach to get at all the examples of this.

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