The Riemann zeta function is defined by the Euler product as :$\zeta(s)=\prod \frac{1}{1-p^{-s}}$ it is converge for $ \Re(s) >1$ , now if we plug instead of ordinary prime $p$ the Gaussian prime of the form $p =i q$ such that $|q|$ is ordinary prime and $|q|=3\bmod 4$ , Now do we still have the product converge for $ \Re(s) >1$,In particular $s = 2n$? For example what about $\zeta(2),\zeta(4),\cdots$?
Clarification: for $s=2$ the product becomes $\frac{1}{(1+3^{-2})(1+7^{-2})(1+11^{-2})\cdots}$ but this is not $\zeta(2)$.
The convergence of the Riemann zeta-function is related to that of the geometric series \[ (1 - x)^{-1} = 1 + x^{2} + x^{3} + \cdots. \]
Remember that the geometric series \[ (1 - x)^{-1} = 1 + x^{2} + x^{3} + \cdots \] converges absolutely for $|x| < 1$.
And, if $s = a + bi$, then \[ |(iq)^{-s}| = |(e^{\pi i /2}q)^{-a - bi}| = q^{-a}e^{\pi b /2} \] and so, for the geometric series of $[1 - (iq)^{-s}]^{-1}$ to converge absolutely, we need to have \[ q^{-a}e^{\pi b /2} < 1 \] for all sufficiently large $q$ primes.
Yes, your zeta-like infinite product converges to whatever it is. But I don't know if it's useful or not.