Let $(z\_i)$ be a square-summable sequence which is even summable but not absolute summable, i.e. $\sum\_{i=1}^{\infty} \vert z\_i \vert = \infty$,$\sum\_{i=1}^{\infty} \vert z\_i \vert^2 < \infty$ and $\sum\_{i=1}^{\infty} z\_i$ exists. I would like to ask if the following function $$f(\mu):=\prod\_{i=1}^{\infty}(1+\mu^2 \vert z\_i \vert^2 - 2 \mu \Re(z\_i))$$
is necessarily entire?


I must say that I do not even know if this product exists away from the real axis. On the real axis it is clear that it exists by using that $(1+x) \le e^x$ such that
$\vert f(\mu) \vert \le e^{\mu^2 \sum\_i \vert z\_i \vert^2 -2\mu \Re \sum\_i z\_i }.$


Assuming it was entire, does there exist a similar growth bound on $\vert f(\mu) \vert$ as the one I obtained on the real axis?



Is this infinite product entire?
This function is (on the real line, at least) the product of
$$ \exp( \mu^2 \sum\_{i=1}^\infty |z\_i|^2 - 2 \mu \Re(\sum\_{i=1}^\infty z\_i)) \quad (1)$$
and the [Hadamard type product](https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem)
$$ \prod\_{i=1}^\infty E\_1( 2 \mu\Re z\_i - \mu^2 |z\_i|^2) \quad(2)$$
where $E\_1$ is the first elementary factor
$$ E\_1(z) := (1-z) \exp(z).$$
The expression (1) is clearly entire in $\mu$; the product (2) is locally uniformly convergent (from the standard bound $E\_1(z) = 1+O(|z|^2)$ when $|z| \leq 1$) and so is also entire. A refinement of this analysis (using for instance the upper bound $|E\_1(z)| \leq \exp(O(|z|^2))$ for all $z$) also gives growth bounds comparable to the ones you already located in the real case.


An alternate factorisation is


$$ \exp( - 2 \mu \Re \sum\_{i=1}^\infty z\_i ) \prod\_{i=1}^\infty E\_1(\mu z\_i) E\_1(\mu \overline{z\_i}).$$
Thus this function is an order two entire function with zeroes precisely at $1/z\_i, 1/\overline{z\_i}$ (counting multiplicity), which specifies the function uniquely up to quadratic exponential factors (such as (1)) by the [Hadamard factorisation theorem](https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).


