Rogers polynomials

Remarks

Expanding each factor by the $q$-binomial theorem gives the explicit form $$C_n(x; \beta \mid q) = \sum_{k=0}^{n} \frac{(\beta; q)_k\, (\beta; q)_{n-k}}{(q; q)_k\, (q; q)_{n-k}}\, e^{i(n - 2k)\theta}.$$ The terms for $k$ and $n - k$ pair to $2\cos((n - 2k)\theta)$ times the same coefficient. So $C_n$ is a polynomial in $x$ of degree $n$, with leading coefficient $2^n \frac{(\beta; q)_n}{(q; q)_n} \ne 0$.

Example: $\beta = q$. Then $\frac{(q; q)_k (q; q)_{n-k}}{(q; q)_k (q; q)_{n-k}} = 1$, so $C_n(x; q \mid q) = \sum_{k=0}^{n} e^{i(n-2k)\theta} = \frac{\sin((n+1)\theta)}{\sin\theta}$, the Chebyshev polynomial $U_n(x)$ of the second kind.

Non-example: $\beta = 1$ fails the condition $\abs{\beta} < 1$. There the right side of the generating function is $1$, so all coefficients of $t^n$ with $n \ge 1$ vanish, and there is no polynomial of degree $n$.

For $\beta = 0$, $C_n(x; 0 \mid q) = \frac{H_n(x \mid q)}{(q; q)_n}$, with the continuous q-Hermite polynomials. For $0 < q < 1$ and $-1 < \beta < 1$, the $C_n$ are orthogonal on $[-1, 1]$ for the weight $\frac{(e^{2i\theta}, e^{-2i\theta}; q)_\infty}{(\beta e^{2i\theta}, \beta e^{-2i\theta}; q)_\infty}$ with respect to $d\theta$, $x = \cos\theta$. As $q \to 1$, $C_n(x; q^\lambda \mid q) \to C_n^{(\lambda)}(x)$, the Gegenbauer polynomial.