= Solution
Since $p$ is real on the <unit circle>,
$$
\sum_{k=-d}^dp_kz^k
=\overline{p(z)}
=\sum_{k=-d}^d\overline{p_k}z^{-k}
=\sum_{k=-d}^d\overline{p_{-k}}z^k.
$$
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>:
$$
\boxed{p_{-k}=\overline{p_k}\quad(0\leq k\leq d)}.
$$
For arbitrary nonzero complex $z$, <coefficient> substitution now gives precisely
$$
\boxed{p(z^{-1})=\overline{p(\overline z)}}.
$$
Equivalently $p(1/\overline z)=\overline{p(z)}$. 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 $p_d\ne0$. Then $p_{-d}=\overline{p_d}\ne0$, so the ensuing <polynomial> $P=z^dp$ 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.
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