 Conjugate symmetry of trigonometric polynomial coefficients

A trigonometric polynomial is real on the unit circle precisely when its finite coefficients have the displayed symmetry. To prove necessity, conjugate its values on the circle, use $\overline{z}=z^{-1}$, and compare coefficients. Multiplying a coefficient difference by $z^d$ gives an ordinary polynomial vanishing at infinitely many points, so it is zero. Symmetry also gives $p(z^{-1})=\overline{p(\overline{z})}$ for every nonzero complex $z$.