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A trigonometric polynomial nonnegative on the unit circle has a polynomial modulus-square factor of polynomial degree at most its trigonometric order. For a nonzero trigonometric polynomial of actual order $d$, conjugate symmetry of trigonometric polynomial coefficients yields reciprocal-conjugate root pairing for $P=z^{d}p$. Even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial permits pairing all $2d$ roots of a polynomial. Select representatives ${\zeta }_1,\dots ,{\zeta }_d$ and useIt follows that $p(z)=c{\prod }_i|z-{\zeta }_i{|}^2$ with $c=(-1{)}^d{p}_d/{\prod }_i\overline{{\zeta }_i}$. Evaluating away from the roots of a polynomial gives $c>0$. Thus $q(z)=\sqrt{c}{\prod }_i(z-{\zeta }_i)$ works. Constant and zero trigonometric polynomials have constant or zero factors. Different selections of roots of a polynomial and constant phases can give different factors.

Since $p$ is real on the unit circle,Multiply the difference by $z^d$. It is an ordinary polynomial of polynomial degree at most $2d$ vanishing at every point of the unit circle. A nonzero polynomial has only finitely many roots of a polynomial, so all its coefficients vanish. This proves the conjugate symmetry of trigonometric polynomial coefficients:For arbitrary nonzero complex $z$, coefficient substitution now gives preciselyEquivalently $p(1/\overline{z})=\overline{p(z)}$. The complex conjugation on the right applies to the whole value, including the coefficients; it cannot simply be discarded away from the circle.

The PDF additionally assumes $p_d\ne 0$. Then $p_{-d}=\overline{p_d}\ne 0$, so the ensuing polynomial $P=z^{d}p$ has polynomial degree exactly $2d$ and nonzero constant term. These clauses and this subpart are absent from the damaged TeX, but present in the PDF.