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

I know that this product defines an entire function $F$. I must show that the order of growth of $F$ is finite, at most 2. My definition of order of an entire function is $$\rho=\inf\{\lambda>0: \displaystyle\sup_{|z|=r} |f(z)|=O(e^{r^{\lambda}}),r\rightarrow\infty\}$$ Can someone give me an help?

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Maybe Jensen's formula could be of help: – Eric Haengel Feb 18 '13 at 21:27

The zeros of $F$ are $n+ki$ for $n=1,2,\dots$, $k\in\mathbb{Z}$. If $p>2$ then $$ \sum_{n=1}^\infty\sum_{k=-\infty}^{\infty}\frac{1}{|n+ki|^p}<\infty. $$ This shows that the order of $F$ is at most $2$.

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sorry, i can't understand. What i knew was that the convergence of the double series you wrote tells me that the convergence exponent of zeros of $F$ is 2, but the order of $F$ is at least the convergence exponent of zeros, not at most....what is wrong in what i know? – Federica Maggioni Feb 19 '13 at 7:51
Nothing. I should not answer questions after midnight. – Julián Aguirre Feb 19 '13 at 18:13

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