Does $\zeta(s)=\prod \frac{1}{1-p^{-s}}$ converge for $ \Re(s) >1$ for $p= iq $ (Gaussian prime)?what about $\zeta(2),\zeta(4),\cdots$? 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) &gt;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) &gt;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 $|x| &lt; 1$.

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
\[
q^{-a}e^{\pi b /2} &lt; 1
\]
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.
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 $|x| &lt; 1$.

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
\[
q^{-a}e^{\pi b /2} &lt; 1
\]
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.