Conjugate symmetry of trigonometric polynomial coefficients ($p_{-k}=\overline{p_k}$)

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 $z$.

For a nonzero real-on-the-circle trigonometric polynomial of actual order $d$, the polynomial $P(z)=z^{d}p(z)$ satisfies $P(z)=z^{2d}\overline{P(1/\overline{z})}$. Its constant term is the nonzero conjugate of its leading coefficient. Hence its nonzero roots of a polynomial occur in pairs $\zeta ,1/\overline{\zeta }$ with the same multiplicity of a root. The coefficient reversal and complex conjugation preserve the exponents of the paired factors. Fixed roots of a polynomial on the unit circle require nonnegativity, rather than this symmetry alone, to have even multiplicity.