Weierstrass Factorization Theorem

Theorem
Let $f$ be an entire function.

Let $0$ be a zero of $f$ of multiplicity $m\geq0$.

Let the sequence $\left\langle{a_n}\right\rangle$ consist of the nonzero zeroes of $f$, repeated according to multiplicity.

First Form
Let $\left\langle{p_n}\right\rangle$ be a sequence of non-negative integers for which the series:
 * $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{\dfrac r {a_n} }\right\vert^{1 + p_n}$

converges for every $r \in \R_{> 0}$.

Then there exists an entire function $g$ such that:
 * $\displaystyle f \left({z}\right) = z^m e^{g \left({z}\right)} \prod_{n \mathop = 1}^\infty E_{p_n} \left({\frac z {a_n}}\right)$

where:
 * $ E_{p_n}$ are Weierstrass's elementary factors

and the product converges locally uniformly absolutely on $\C$.

Second Form
Then there exists a sequence $\left\langle{p_n}\right\rangle$ of non-negative integers and an entire function $g$ such that:
 * $\displaystyle f \left({z}\right) = z^m e^{g \left({z}\right)} \prod_{n \mathop = 1}^\infty E_{p_n} \left({\frac z {a_n}}\right)$

where:
 * $ E_{p_n}$ are Weierstrass's elementary factors

and the product converges locally uniformly absolutely on $\C$.

Proof
From Weierstrass Product Theorem, the function:


 * $\displaystyle h \left({z}\right) = z^m \prod_{n \mathop = 1}^\infty E_{p_n} \left({\frac z {a_n}}\right)$

defines an entire function that has the same zeros as $f$ counting multiplicity.

Thus $f / h$ is both an entire function and non-vanishing.

As $f / h$ is both holomorphic and nowhere zero there exists a holomorphic function $g$ such that:
 * $e^g = f / h$

Therefore:
 * $f = e^g h$

as desired.

Also known as
Some sources give this as the Weierstrass factor theorem.

Also see

 * Hadamard Factorization Theorem, a refined version of this
 * Weierstrass Product Theorem