While I check the proof for entry 10(vii) in Chapter 19 in Ramanujan's notebook, I couldn't understand one equality. It is \begin{equation} \prod_{\zeta}\varphi(q^{1/5}\zeta) = \varphi^6(q)/\varphi(q^5) \end{equation} where $\zeta$ is an arbitrary fifth root of unity and \begin{equation} \varphi(q):=\sum_{n=-\infty}^{\infty}q^{n^2}=(q^2;q^2)_{\infty}(-q;q^2)_{\infty}^2 \qquad, |q|<1\\ (a;q)_{\infty}:= \prod_{n=0}^{\infty}(1-aq^n)\quad. \end{equation} Could anyone help me to understand how it holds?
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First note that $\prod_{\xi} (1 - \xi^n q^n)$ is equal to $(1- q^n)^5$ if $5 | n$ and to $1 - q^{5n}$ otherwise. Moreover $$ \varphi(q) = \prod_{n \geq 1} (1 - q^n)^{e_n} $$ where $e_n = 1,-2,3,$ or $-2$ if $n \equiv 0,1,2$ or $3 \pmod 4$ respectively. In particular $e_n = e_{5n}$ $(*)$ . Thus $$ \prod_{\xi} \varphi(\xi q) = \prod_{n \geq 1} (1 - q^n)^{f_n} $$ where $f_n = 5 e_n \mathbb{1}_{5 \mid n} + e_{\frac{n}{5}} \mathbb{1}_{5 \mid n} - e_{\frac{n}{5}} \mathbb{1}_{5^2 \mid n}$, and thus $f_n = e_n (6 \mathbb{1}_{5 \mid n} - \mathbb{1}_{5^2 \mid n} )$ by $(*)$. On the other hand $$ \varphi(q)^6 / \varphi(q^5) = \prod_{n \geq 1} (1 - q^n)^{g_n} $$ where $g_n = 6 e_n - e_{\frac{n}{5}} \mathbb{1}_{5 \mid n}$, i.e. $g_n = e_n (6 - \mathbb{1}_{5 \mid n} )$ by $(*)$. By noting that $f_n = \mathbb{1}_{5 \mid n} g_{\frac{n}{5}}$, one sees that $$ \prod_{\xi} \varphi(\xi q) = \varphi(q^5)^6 / \varphi(q^{25}). $$ Note that the the only thing about $e_n$ we used is its $4$-periodicity (plus the fact that $5 \equiv 1 \pmod 4$).
