I have the expression
$$ \frac{x(-c z^{n+1} (-1)^k + c x^{n+1})}{(x-ze^{i \theta})(x - ze^{-i\theta})},$$
where $b$ and $c$ are non-zero complex numbers, $z = \frac{b}{\sqrt{bc}}$ and $\theta = \frac{k\pi}{n+1}$. I am confident that for any $n \in \mathbb{N}_{\geq 1}$ and $k \in \{1,2,\ldots,n\}$ this fraction can be written as a polynomial in $x$ (for non-zero denominator), since it should be equal to some generating function, but I haven't proven this conclusively yet. Therefore, I would like to prove that we can rewrite this fraction in the (slightly more specific since $y_0 = 0$) form
$$ Y(x) = y_1x + y_2x^2 + \cdots + y_nx^n, \ y_i \in \mathbb{C}.$$
For $n \in \{1,2,3\}$ Mathematica has provided $Y$, but I was hoping for something more general. Is there some way to either find $Y$ explicitly or prove such a $Y$ exists for all such $n,k$ (possibly with more help from Mathematica)?
Let $f(x):=-c z^{n+1} (-1)^k + c x^{n+1}$.
Such a $Y$ exists since we have $$\begin{align}f(ze^{\pm i \theta})&=-c z^{n+1} (-1)^k + c(ze^{\pm i \theta})^{n+1} \\\\&=-c z^{n+1} (-1)^k + cz^{n+1}e^{\pm i \theta\ (n+1)} \\\\&=-c z^{n+1} (-1)^k + cz^{n+1}e^{\pm i k\pi} \\\\&=cz^{n+1}\bigg(-(-1)^k + e^{\pm i k\pi}\bigg) \\\\&=0\end{align}$$