I'm trying to find the contour integral over the unit circle $C$:
$$\oint_C \frac{p^2(z)}{|p(z)|^2} \frac{dz}{iz}$$ where
$$p(z) = a - z \sum_{n=-N}^{N} c_n z^n \\a, c_n, z \in \mathbb{C}$$
Specifically $c_n$ are the coefficients from a Fourier transform of some real sample points. As such $c_n = \overline{c}_{-n}$.
I'm guessing the integral can be expressed in terms of the roots of $p(z)$, but I'm not sure of the best way to proceed.
Edit
Working on it more, I've made what I think is partial progres:
Letting $r(z) = \sum_{n=-N}^N c_n z^n$, noting that $\overline{r(z)} = r(z)$, and with some polynomial long division:
$$\oint_C \frac{p^2(z)}{|p(z)|^2} \frac{dz}{iz} = \oint_C \frac{p(z)}{\overline{p(z)}} \frac{dz}{iz} \\= \oint_C \frac{a - z \cdot r(z)}{\overline{a} - \frac{1}{z} r(z)} \frac{dz}{iz} \\= \oint_C (z^2 + \frac{\overline{a}z^3 - az}{r(z) - \overline{a} z}) \frac{dz}{iz} \\= -i \oint_C (z + \frac{\overline{a}z^2 - a}{r(z) - \overline{a} z}) dz \\= -i \oint_C \frac{\overline{a}z^2 - a}{r(z) - \overline{a} z} dz $$
But what to do now? I was thinking I could find the roots of $r(z) - \overline{a}z$ and find the residues at these roots, but I'm not sure if that has a nice format? There'd be lots of repeated factors in the denominator of each residue, for instance.
Since $z$ is on the unit circle, $\bar z =1/z$. Thus \begin{align*} & \frac{{p^2 (z)}}{{\left| {p(z)} \right|^2 }}\frac{1}{{iz}} = \frac{{p(z)}}{{\overline {p(z)} }}\frac{1}{{iz}} = \frac{{a - z\sum\limits_{n = - N}^N {c_n z^n } }}{{\bar a - \bar z\sum\limits_{n = - N}^N {c_{ - n} \bar z^n } }}\frac{1}{{iz}} = \frac{{a - z\sum\limits_{n = - N}^N {c_n z^n } }}{{\bar a - \dfrac{1}{z}\sum\limits_{n = - N}^N {c_{ - n} z^{ - n} } }}\frac{1}{{iz}} \\ & = \frac{{a - z\sum\limits_{n = - N}^N {c_n z^n } }}{{\bar a - \dfrac{1}{z}\sum\limits_{n = - N}^N {c_n z^n } }}\frac{1}{{iz}} = \frac{{a - z\sum\limits_{n = - N}^N {c_n z^n } }}{{\bar a - \dfrac{1}{z}\sum\limits_{n = - N}^N {c_n z^n } }}\frac{1}{{iz}} = \frac{1}{i}\frac{{zp(z)}}{{\bar az^2 - a + p(z)}}. \end{align*} Therefore the integral is $$ 2\pi \sum {{\mathop{\rm Res}\nolimits} \frac{{zp(z)}}{{\bar az^2 - a + p(z)}}} . $$ The sum is over the residues inside the unit circle.