# Entire function having simple zero at the Gaussian integers

I am trying to find the entire function with only zeros of order 1 at the Gaussian integers and no other zeros. For that I did the enumeration of Gaussian integers and try to use the same way I used to find the entire function with simple zeros at $n^2$ and no other zeros. That just comes out like a mess. Seriously, I don't know how to get explicit function. Thanks for the help.

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Look up "Weierstrass factorization theorem". I don't know why you say "the entire function ...": it is far from unique. One solution is $$g(z) = z \prod_{w \ne 0} E_{3}(z/w)$$ where the product is over all nonzero Gaussian integers $w$, and $E_3(z)$ is an elementary factor $$E_3(z) = (1-z) \exp\left(z + \frac{z^2}{2} + \frac{z^3}{3} \right)$$

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What is your w in the elementary factor? Do we have to enumerate Gaussian integers to write down in elementary factor? –  Deepak Dec 24 '12 at 5:44
$w$ is a Gaussian integer, and the infinite product is over all Gaussian integers $w$ except $0$. You could write this using an enumeration of the Gaussian integers, if you wish. –  Robert Israel Dec 24 '12 at 6:15
Sounds good. Thank you. –  Deepak Dec 24 '12 at 6:17

You may be interested in the Weierstrass $\sigma$ function, where you take the lattice $\Lambda$ to be the Gaussian integers.

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The Jacobi theta function can be used to produce such an entire function for any lattice in $\mathbb{C}$. For a lattice $\Lambda = \mathbb{Z}+\mathbb{Z}\tau$ with $\operatorname{Im}(\tau) > 0$ the function $$z \mapsto \vartheta(z + \frac{1+\tau}{2};\tau)$$ is entire and has simple zeroes precisely on the lattice $\Lambda$.

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