Can a hermitian, rational polynomial have non-zero odd and real coefficients in the numerator/denominator? Assume that we have a rational polynomial of the form:
$$\chi\left(\omega\right)=\frac{\sum_{n=0}\left(c_n+ic_n^{\dagger}\right)\omega^{n}}{\sum_{n=0}\left(d_n+id_n^{\dagger}\right)\omega^{n}}$$
where $\omega$, $c_n$, $c_n^{\dagger}$, $d_n$, $d_n^{\dagger}$ are all real and $i$ is imagiary i.
Let us further assume that $\chi\left(\omega\right)$ is hermitian:
$$\chi\left(-\omega\right)=\chi^{*}\left(\omega\right)$$
where $\chi^{*}\left(\omega\right)$ is the complex conjugate of $\chi\left(\omega\right)$.
Can I conclude that the following must be true:
$c_n=d_n=0$ for $n$ is odd,
$c^{\dagger}_n=d^{\dagger}_n=0$ for $n$ is even.
i.e. that $\chi\left(\omega\right)$ can always be brought into the following form
$$\chi\left(\omega\right)=\frac{\sum_{n=0}c_{2n}\omega^{2n}+i\sum_{n=0}c_{2n+1}^{\dagger}\omega^{2n+1}}{\sum_{n=0}d_{2n}\omega^{2n}+i\sum_{n=0}d^{\dagger}_{2n+1}\omega^{2n+1}}$$
Note: this is obviously true for a non-rational polynomial, but I have trouble proving that the above form is the only way to express a Hermitian rational polynomial.
 A: For an easier typing, let me use $X$ instead of $\omega$. Saying that $X$ is real essentially amounts to working with polynomials (as opposed to polynomial functions) - if this line makes no sense to you, just ignore it.
We can always write
$$\chi \left( X \right) = \frac {\sum _n \left( c_n + {\rm i} c_n ^\dagger\right) X^n} {\sum _n \left( d_n + {\rm i} d_n ^\dagger \right) X^n}$$
as
$$\chi \left( X \right) = \frac {\left( \sum _n c_{2n} X^{2n} + {\rm i} \sum _n c_{2n+1} ^\dagger X^{2n+1} \right) + \left( {\rm i} \sum _n c_{2n} ^\dagger X^{2n} + \sum _n c_{2n+1} X^{2n+1} \right)} {\left( \sum _n d_{2n} X^{2n} + {\rm i} \sum _n d ^\dagger _{2n+1} X^{2n+1} \right) + \left( {\rm i} \sum _n d_{2n} ^\dagger X^{2n} + \sum _n d _{2n+1} X^{2n+1} \right)}.$$
Let us introduce the notations
$$P = \sum _n c_{2n} X^{2n} + {\rm i} \sum _n c_{2n+1} ^\dagger X^{2n+1} \\
R = {\rm i} \sum _n c_{2n} ^\dagger X^{2n} + \sum _n c_{2n+1} X^{2n+1} \\
Q = \sum _n d_{2n} X^{2n} + {\rm i} \sum _n d ^\dagger _{2n+1} X^{2n+1} \\
S = {\rm i} \sum _n d_{2n} ^\dagger X^{2n} + \sum _n d _{2n+1} X^{2n+1}, $$
so that
$$\chi = \frac {P + R} {Q + S} .$$
Note that, given how we have grouped terms, we have
$$P(-X) = P^* (X), \quad Q(-X) = Q^* (X), \quad R(-X) = -R^* (X), \quad S(-X) = -S^* (X) .$$
Using the above relations,
$$\frac {P(-X) + R(-X)} {Q(-X) + S(-X)} = \chi (-X) = \chi^* (X) = \frac {P^* (X) + R^* (X)} {Q^* (X) + S^* (X)} = \frac {P(-X) - R(-X)} {Q(-X) - S(-X)} ,$$
whence after cross-multiplying, reducing the identical terms grouping the remaining ones and changing $-X$ into $X$, we get
$$SP = QR .$$


*

*If $P = 0$, then $QR = 0$.
1.1. If $Q = 0$, then
$$\chi = \frac R S = \frac {{\rm i} R} {{\rm i} S}$$
and the numerator and denominator have exactly the required form.
1.2. If $Q \ne 0$, then $R = 0$, so $\chi = 0$, which again has the required form (trivially).

*If $P \ne 0$, then
$$ S = \frac {QR} P .$$
2.1. If $P + R = 0$, then $\chi = 0$ has the required form.
2.2. If $P + R \ne 0$, then
$$\chi = \frac {P + R} {Q + \frac {QR} P} = \frac {P (P + R)} {Q (P + R)} = \frac P Q$$
which again has the required form.
