 Solution

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.