For $\alpha\in\mathbb C$ with $\mathrm{Im}(\alpha)>0$, consider the infinite product $$f(z) = \prod_{n\in\mathbb Z} \frac{z-n-\bar\alpha}{z-n-\alpha} \times \frac{-z-n-\alpha}{-z-n-\bar\alpha}.$$ This infinite product is well-defined since each term is $1+O(1/n^2)$ for $n\to\infty$ for fixed $z\in\mathbb C$.
Is there any closed form for $f$?
Hint
Let $\beta=\bar\alpha$ $$F(z)=\prod_{n=0}^p \frac{n-z+\beta }{n-z+\alpha}\,\,\frac{n+z+\alpha}{n+z+\beta }$$ $$F(z)=\frac{\prod_{n=0}^p (n-z+\beta) } {\prod_{n=0}^p (n-z+\alpha) }\frac{\prod_{n=0}^p (n+z+\alpha) } {\prod_{n=0}^p (n+z+\beta) }$$ Using Pochhammer symbols $$F(z)=\frac{(z+\alpha ) (z-\beta ) }{(z-\alpha ) (z+\beta ) }\,\,\color{red}{\frac{(\alpha +1+z)_p (\beta +1-z)_p}{ (\alpha +1-z)_p (\beta +1+z)_p}}$$ Take logarithms of the pieces of the red part and use $$\log \left[(A)_p\right]=p \,(\log (p)-1)+\left(A-\frac{1}{2}\right) \log (p)+\log \left(\frac{\sqrt{2 \pi }}{\Gamma (A)}\right)+$$ $$\frac{6 (A-1) A+1}{12 p}-\frac{(A-1) A (2 A-1)}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ Remembering that $X=e^{\log(X)}$ you will easily obtain the asymptotics of $F(z)$.