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

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?

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Yes I believe you are correct, except just be careful to note that $n \geq 1$ – Eric Haengel Feb 18 '13 at 18:59

Yes, by definition of convergence for products, we still have that the product is $=0$ iff one of the factors is $=0$

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it is very strange since i found this exercise on a web site, and the author of the exercise say that, if $z_k$ are the zeros of $F$, then $\sum_{k}\frac{1}{|z_k|^2}$ diverges... – Federica Maggioni Feb 18 '13 at 19:11
That's true actually. If we let $A = \lbrace (n, k) \in \mathbb{Z}^2 : n \geq 1 \rbrace$, then that sum you mentioned is equal to $\sum_{(n, k) \in A} 1 / (n^2 + k^2)$. It may seem at first that this sum will converge, but it does not. – Eric Haengel Feb 18 '13 at 19:27
Indeed, as a very rough approximation there are $\sim 2\pi r$ such pairs at distance $r\le\sqrt{n^2+k^2}\le r+1$ from the origin, thus contributing $\sim \frac 1r$, i.e. the sum diverges essentially like the harmonic series. – Hagen von Eitzen Feb 18 '13 at 20:59

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