Consider the following function: $G(z) = \prod\_{n \in \mathbb{Z}} {1 \over{\tanh^2\left(\pi\left(z-n\right)\right)}}$.


By constuction, it has poles at $z=m+in$ with $m,n \in \mathbb{Z}^2$.


Additionally, it is doubly periodic such that for any other value $G(z)=G(z+1)=G(z+i)$.


Now, strong computational evidence shows that it admits the following special values:


$G({i \over 2}) = 0, G({1 \over 2}+i{1 \over 4}) = 1$ and $G({1 \over 2}+i{1 \over 2}) = \frac{\sqrt{2}}{2}$


However, I have no idea how to find a proof for this.


Anyone can help? 



Special values of a doubly periodic meromorphic function 
As a meromorphic doubly-periodic function, it is an elliptic function in the classical sense. You didn't mention the fact that it is an even function, but that also helps. The poles are double, so that up to addition of a constant this must be a constant multiple of the Weierstrass P-function for the lattice of Gaussian integers. Since the P-function in this case is 0 at a known point, your third value gives the additive constant as (presumably) 0.5. 


Everything about this function is standard in the theory of complex multiplication. However you came up with it, the formulae are probably written down somewhere.


'''Edit''': What I wrote before is probably fine, but starting from your formula it probably makes sense to use the Weierstrass sigma function to express the infinite product first. We're in the case where one of the roots of the cubic in the equation for the Weierstrass P-function is 0, and that means the P-function itself is easy to express in terms of the sigma function. 


