On $\Bbb{R}$, we can construct the convergent infinite product $$F_{\Bbb{Z}}(x) := x \prod_{n \geq 1} \prod_{|k| = n} \left(1 - \frac{x}{k} \right) = x \prod_{n \geq 1} \left(1 - \frac{x^2}{n^2} \right),$$ the second of which converges absolutely for all $x$ (by comparing logs with the series $\sum_{n \geq 1} \frac{1}{n^2} = \zeta(2)$), and get a nonzero analytic function which has zeroes exactly at the integers, no poles or singularities, and $\Bbb{Z}$-translation symmetry: $F_{\Bbb{Z}}(x+k) = F_{\Bbb{Z}}(x)$ for all $k \in \Bbb{Z}$ and all $x \in \Bbb{R}$. (In fact, $F_{\Bbb{Z}}(x)$ is an infinite product representation for $\sin( \pi x)$.)
Now Liouville's theorem (any bounded entire function is constant) forbids us from constructing an analogous function holomorphic on all of $\Bbb{C}$ with zeroes exactly at the Gaussian integers, no poles or singularities, and $\Bbb{Z}[i]$ translational symmetry on all of $\Bbb{C}$.
However, I can easily write down an analogous convergent infinite product which has zeroes at all of $\Bbb{Z}[i]$:
\begin{align*}
F_{\Bbb{Z}[i]}(z) &:= z \prod_{n \geq 1} \prod_{|a+bi| = n} \left(1 - \frac{z}{a + bi} \right) \\
&= z \prod_{\substack{a \geq 1 \\
-a < b \leq a}} \left(1 - \frac{z^4}{(a+bi)^4} \right) \
\end{align*}
the second of which converges absolutely for all $z \in \Bbb{C}$ (by comparing logs with the series $\sum_{a \geq 1} \frac{2a}{a^4} = 2 \zeta(3)$) to a nonzero function holomorphic on all of $\Bbb{C}$.
My question is, what is the behavior of $F_{\Bbb{Z}[i]}(z)$ under translation by elements of $\Bbb{Z}[i]$? I know we can't have $F_{\Bbb{Z}[i]}(z + a + bi) = F_{\Bbb{Z}[i]}(z)$ for all $z$ and for any $a + bi \in \Bbb{Z}[i]$, by Liouville, but I'm interested in seeing the algebra of why the "obvious" argument for translation symmetry fails, and how translation by Gaussian integers actually affects the function (since it can't leave it invariant). I played with it a little bit myself but was unable to calculate what was happening directly from the definition, and nothing immediately jumped out at me from related results on elliptic functions.
If the half period values $\frac{F'}{F}(1/2)=p, \frac{F'}{F}(i/2)=q$ then (using that the derivative of $F'/F$ is the elliptic function associated with the lattice so is periodic with periods $1,i$, integrating and using that $F'/F$ is odd) we have:
$\frac{F'}{F}(z+1)=\frac{F'}{F}(z)+2p, \frac{F'}{F}(z+i)=\frac{F'}{F}(z)+2q$
(eg since $(F'/F)'(z+1)=(F'/F)'(z)$ we get that $\frac{F'}{F}(z+1)=\frac{F'}{F}(z)+A$; putting $z=-1/2$ gives $A=2\frac{F'}{F}(1/2)$ etc)
Hence by integration and using that $F$ is odd we get:
$F(z+1)=-e^{2p(z+1/2)}F(z)$
$F(z+i)=-e^{2q(z+i/2)}F(z)$
(as above we get $F(z+1)=Be^{2pz}F(z)$, putting $z=-1/2$ gives $Be^{-p}=-1$ etc)