Is there a "nice" function with zeros at lattice points in $\Bbb C$?

Question

All functions will be considered on $\Bbb C$:

The function $\sin(\pi z)$ has zeroes precisely at the lattice $\Lambda = \Bbb Z$ in the complex plane. Further, it is nice in the sense that: $$\frac{\sin(\pi z)}{\pi z} = \prod_{n\in \Lambda\backslash\{0\}}\bigg(1 - \frac{z^2}{n^2}\bigg).$$ Now, let us take $\Lambda = \Bbb Z[i] = \{m+ni : m,n\in \Bbb Z\}$. Is there a similarly nice function $f(z)$ satisfying: $$f(z) = z^k\prod_{n\in \Lambda\backslash\{0\}}\bigg(1 - \frac{z^4}{n^4}\bigg)$$ for some $k$? Note that I am raising $z$ to the power $4$ instead of $2$ now. Similarly, is there any nice function $g(z)$ for the lattice: $$\Lambda = \Bbb Z[\omega] = \{m + n\omega : m,n \in \Bbb Z\}$$ where $\omega$ is a primitive cube root of $1$. I expect this $g$ to satisfy: $$g(z) = z^k\prod_{n\in \Lambda\backslash\{0\}}\bigg(1 - \frac{z^6}{n^6}\bigg)$$ for some $k$?

I would also like to know about the more general case of $\Lambda$ being the ring of integers in an arbitrary imaginary quadratic number field $K$.

(Motivation: I would like to use this function to compute the zeta function of the number field $K$ at even integers mimicking Euler's Proof of $\zeta(2k)$.)

Answer 1

The reason ${\sin(\pi z)\over\pi z}$ is so nice is because of some tricks with the $\Gamma$-function. You can get the same result for the functions you note, it'll just involve a product of $\Gamma$-functions at assorted roots of unity just like $\Gamma(z)\Gamma(1-z)$.

The computations aren't difficult, the key facts are just

$${1\over\Gamma(z)} = ze^{\gamma z}\prod_{z=1}^\infty \left(1 +{z\over n}\right)\cdot e^{-z/n}$$

and

$$1+\zeta_n +\zeta_n^2+\ldots +\zeta_n^{n-1} =0$$

where $\zeta_n$ is a primitive $n^{th}$ root of 1.