 Solution

For $z\ne 0$, the coefficient symmetry yieldsThus $P$ equals its reversed-conjugate polynomial. If $\zeta \ne 0$ is a root of a polynomial, with $P(\zeta )=0$, this identity gives $P(1/\overline{\zeta })=0$, provingThe reciprocal-conjugate root pairing also preserves the multiplicity of a root: reversal and complex conjugation take each factor associated with $\zeta$ to the corresponding factor associated with $1/\overline{\zeta }$, with the same exponent.

Since $P(0)=p_{-d}\ne 0$, no zero root of a polynomial occurs, so the reciprocal operation is always defined on the roots of a polynomial. Roots off the unit circle are paired on opposite sides of it; roots of a polynomial on the unit circle are fixed by this operation. Pairing alone does not force even multiplicity at those fixed roots of a polynomial; that extra fact uses nonnegativity.