Weierstrass Factorization Theorem

Theorem
Let $f$ be an entire function.

Suppose $f(z)=0$ at $z=0$ to order $m$ where $m ≥ 0$.

Let the sequence $\left\langle{a_n}\right\rangle$ be the other values of $z$ for which $ƒ(z)$ 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


 * $\displaystyle f(z)=z^me^{g(z)} \prod_{n \mathop = 1}^\infty E_{p_{n}} \left({\frac z a_{n}}\right)$

where $ E_{p_{n}}$ are Weierstrass's elementary factors defined by:


 * $E_{n}(z)=\begin{cases}(1-z)&{\text{if }}n=0,\\(1-z)\exp \left({\frac {z^{1}}{1}}+{\frac  {z^{2}}{2}}+\cdots +{\frac  {z^{n}}{n}}\right)&{\text{otherwise}}.\end{cases}$

Proof
The function


 * $\displaystyle h(z)=z^m \prod_{n \mathop = 1}^\infty E_{p_{n}} \left({\frac z a_{n}}\right)$ : see Weierstrass Product Theroem.

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.