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I want to construct an entire function which vanishes at points $n+in$ for all $n$ integers. I'm looking for the most simple entire function which satisfies this condition, in the sense that the integers $p_n$ in weierstrass elementary factor should be as small as possible.

If i enumerate non-zero points of the form $n+in$ as $\{z_n\}$ i get $$f(z)=z\prod_{n=1}^{+\infty}E(\frac{z}{z_n},p_n)$$

which is entire with zeros where i want, for suitable integers $p_n$.

But i don't know hot to determine $p_n$'s. Moreover i don't know which is the right way to enumerate $n+in$ in this case. I need integers $p_n$ such that $$\sum_{n=1}^{+\infty}\frac{1}{|z_n|^{p_n+1}}<+\infty$$ How can i find them?

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$$f(z) = \sin \left(\dfrac{\pi}{1+i}z \right)$$ – user17762 Jan 9 '13 at 22:12

If you indeed want $\sum_{n}|z_n|^{-p_n-1}<\infty$, then enumeration makes no difference. Just set $z_n=n+in$. Now you see that $p_n\equiv 0$ does not work, while $p_n\equiv 1$ does. What is the "minimal" sequence of $p_n$? Well, there isn't one. Whatever sequence you come up with can be made "smaller" by replacing one of the nonzero numbers $p_n$ with a $0$.

However, it is not necessary to have $\sum_{n}|z_n|^{-p_n-1}<\infty$ in order for the product to converge. This is what the example of Marvis and Jonas demonstrates: in it $p_n\equiv 0$, yet the product converges conditionally with the enumeration $1,-1,2,-2,3,-3\dots $ (times $1+i$).

Moral: the Weierstrass theorem does not exhaust all possibilities for constructing entire functions with prescribed zeros.

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ok,thank you very much, but suppose i want a sequence p_n such that we have convergence of weierstrass product, and p_n equal to a unique integer p. Is there a minimal p such that i can find such an entire function using weierstrass factorization? – bateman Jan 9 '13 at 22:31
@bateman Then $p_n\equiv 1$, of course. Because $1/n^2$ converges while $\sum 1/n$ does not. – user53153 Jan 9 '13 at 22:53
tell me if you think this is right: using W factorization, our function is $f(z)=z\prod_{z_n} E(\frac{z}{z_n},1)=z(\prod_{n=1}^{\infty}(1-\frac{z}{n+in})e^{\frac{z}{n+in}}\pr‌​od_{n=1}^{\infty}(1+\frac{z}{n+in})e^{-\frac{z}{n+in}})=z\prod (1-\frac{z^2}{(1+i)^2 n^2})$ Now, from product factorization of sine (see for example Lang p. 379) i get that this is a constant times $sin(\frac{\pi z}{1+i})$ which is Jarvis example – bateman Jan 10 '13 at 8:46

i tried to post a comment but there were problems, do you think the following is correct?

UUsing Weierstrass factorization, our function is $$f(z)=z\prod_{z_n} E(\frac{z}{z_n},1)=z\prod_{n=1}^{\infty}(1-\frac{z}{n+in})e^{\frac{z}{n+in}}\prod_{n=1}^{\infty}(1+\frac{z}{n+in})e^{-\frac{z}{n+in}}=z\prod (1-\frac{z^2}{(1+i)^2 n^2})$$Now, from product factorization of sine (see for example Lang p. 379) i get that this is a constant times $sin(\frac{\pi z}{1+i})$ which is Jarvis example

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Yes, that works. – user53153 Jan 10 '13 at 13:38

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