Entry 98

Level 6. Given $q = e^{2\pi i \tau}$, Ramanujan's theta function $f(a,b)$, its one-parameter version $f(-q) = f(-q,-q^2)$, and the q-Pochhammer symbol. We have the sum-product identities,

$$\begin{align}A(q) &= \frac{f(-q,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}(q,q^2)_n} {(-q,q)_{2n+1} (q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-3})(1-q^{6n-3})} = q^{-1/4}\frac{\eta(\tau)\eta^2(6\tau)}{\eta^2(2\tau)\eta(3\tau)}\\ B(q) &=  \frac{f(-q^3,-q^3)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}(q,q^2)_n} {(-q,q)_{2n} (q^2;q^2)_n} \;=\; \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-1})(1-q^{6n-5})} = q^{1/12}\frac{\eta^2(3\tau)}{\eta(2\tau)\eta(6\tau)}\end{align}$$ where $\alpha_6 = \dfrac{(1-q^{n})(1-q^{3n})}{(1-q^{2n})^2}$ 

Let $a=q^{1/4}A(q)$ and $b= q^{-1/12}\,B(q)$. Then $(a, b)$ for appropriate $\tau$ are eta quotients like in Level 4 and are actually radicals.