$ \prod_{k=0}^\infty \prod_{j=1}^q ( 1- \chi(j) z/(kq+j) ) = (1-z) \prod_{k=2}^\infty ( 1 - \chi(k) z/k) $

Question

Let $\chi$ be a primitive non-principal Dirichlet character with conductor $q>1$. The following equation

$$ \prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- \chi(j) \frac{z}{kq+j} \biggr) = (1-z) \prod_{k=2}^\infty \biggl( 1 - \chi(k) \frac{z}{k} \biggr) $$

appears as (3.6) in https://doi.org/10.1016/j.aam.2018.05.003. How does one go about proving this identity?

Answer 1

If $f$ is $q$-periodic then $$\prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- f(j) \frac{z}{kq+j} \biggr)=\prod_{k=0}^\infty\prod_{j=1}^q \biggl( 1- f(kq+j) \frac{z}{kq+j} \biggr)=\prod_{n=1}^\infty \biggl( 1- f(n) \frac{z}{n} \biggr)$$ It remains to show it converges (locally uniformly, thus to an entire function) iff $f$ has zero mean, by taking the $\log$ and using that for $|z|<N$, $$\sum_{n>2 N} \log(1-f(n) \frac{z}n)=-\sum_{n>2N} (f(n) \frac{z}{n}+O(\frac{z^2}{n^2}))$$ converges uniformly.