Weierstrass Factorization Theorem

Theorem
Let $f$ be an entire function.

Let $f \left({z}\right) = 0$ at $z = 0$ to order $m$ where $m \ge 0$.

Let the sequence $\left\langle{a_n}\right\rangle$ be the other values of $z$ for which $f \left({z}\right)$ vanishes repeated according to multiplicity.

Then there exists an entire function $g$ and a sequence of integers $\left\langle{p_n}\right\rangle$ 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 defined as:


 * $E_n \left({z}\right) = \begin{cases} \left({1 - z}\right) & : n = 0 \\

\left({1 - z}\right) \exp \left({\dfrac {z^1} 1 + \dfrac {z^2} 2 + \cdots + \dfrac {z^n} n}\right) & : n \ne 0 \end{cases}$

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.