The following question was motivated by [this MO-post](https://mathoverflow.net/q/384488/61536).


I hope that the answer should be known to experts (because of very simple formulation)...



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> **Problem.** Let $n\ge 2$. Is the set of complex numbers $\{e^{i\pi k/2^n}:0\le k<2^n\}$ linearly independent over the field of rationals?
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Are half of the $2^n$-th roots of the unit rationally independent?
Denote by $\omega$ the order $2^{n+1}$-th primitve root of unity $\omega=e^{i\pi/2^n}$. The linear dependence of the above set would imply that there is a polynomial of degree at most $2^n-1$ with $\mathbb{Q}$ coefficients which vanishes on $\omega$. But its minimal polynomial is the cyclotomic polynomial $\Phi\_{2^{n+1}}(X)=X^{2^n}+1$ and we are done.


