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$$\displaystyle\prod_{n=1}^{+\infty}(1-e^{-2\pi n}e^{2\pi iz})$$

I've proven that this product converges uniformly on compact subsets of complex plane since the serie $\sum_{n=0}^{+\infty}|\frac{e^{2\pi iz}}{e^{2\pi n}}|$ does.

Now i'm interested to zeros of $F$, the entire function to which the product converges. How can i find them? Can i say that all the zeros of $F$ are those complex numbers $z$ such that $e^{-2\pi n}e^{2\pi i z}=1$? If yes, zeros are of the form

$$z_{n,k}=-in+k$$

$n,k$ integers. Do you think it's correct?