Let p be a prime number and let $\epsilon\in\mathbb{C}$ denote a primitive p-th root of unity. Let $(x_{0},\dots,x_{p-2})\in\mathbb{Z}^{p-2}\setminus\{(0,\dots,0)\}.$ Consider a sum $S=\sum_{j=0}^{p-2}x_{j}\epsilon^{j}.$ It is clear that $S\neq0,$ since $p$ is prime. Is there a reasonable lower bound for |S| depending on $p$ but not on $x_{0},\dots,x_{p-2}?$ In other words how small can be values of a function $$(x_{0},\dots,x_{p-2})\mapsto|\sum_{j=0}^{p-2}x_{j}\epsilon^{j}|$$?
An easy computation shows that for $p=3$ this is $1$ and in general for any $x_{0},\dots,x_{p-2}$ we have $$\frac{1}{p-1}\sum_{k=1}^{p-1}|\sum_{j=0}^{p-2}x_{j}\epsilon^{kj}|^{2}\geq1$$ but i don't know if this is helpful. Also one can show that for any $p$ we have $|1-\epsilon|\geq\frac{2}{p}$ and therefore for $x_{0}=\dots=x_{k}=1,x_{k+1}=\dots=x_{p-2}=0$ $|S|\geq\frac{1}{p},$ since $1+\dots+\epsilon^{k}=\frac{1-\epsilon^{k+1}}{1-\epsilon}.$
I was thinking about this for a while but I don't know how to aproach this. I will be thankful for any hint.
Let $a\in\Bbb{C}$ be a non-real complex number, then $\Bbb{Z}+a\Bbb{Z}$ is a lattice in $\Bbb{C}$, its fundamental domain is the parallelogram $P= [0,1)+a[0,1)$ which is (pre)compact. If $a^2\not\in \Bbb{Q}+a\Bbb{Q}$ then for each $n$ there is a unique $u_n+v_na\in \Bbb{Z}+a\Bbb{Z}$ such that $na^2-(u_n+v_na)\in P$ and $z_n= na^2-(u_n+v_na)$ is a sequence of distinct points in $P$, it has a convergent subsequence $z_{n_j}$ which satisfies $\lim_{j\to\infty} |z_{n_j}-z_{n_{j+1}}|=0$.
And hence, if $\inf \ \{ |b|,b\in \Bbb{Z}[a]-0\}> 0$ then $a^2\in \Bbb{Q}+a\Bbb{Q}$ which means that $a$ is an imaginary quadratic algebraic number, when $a=\zeta_n$ it means $n\in 3,4,6$.