Weierstrass factorization theorem

There are several different statements of this theorem, but in essence this theorem will allow us to prescribe zeros and their orders of a holomorphic function. It also allows us to factor any holomorphic function intoa product of zeros and a non-zero holomorphic function. We will need to knowhere how an infinite product converges. It can then be shownthat if ${\prod }_{k=1}^{\mathrm{\infty }}{f}_k(z)$ converges uniformly and absolutely (http://planetmath.org/AbsoluteConvergenceOfInfiniteProduct)on compact subsets, then it converges to a holomorphic function given that all the $f_k(z)$ are holomorphic. This is what we will by the infinite product in what follows.

Note that once we can prescribe zeros of a function then we can also prescribe the poles as well and get a meromorphic function just by dividingtwo holomorphic functions $f/h$ where $f$ will contribute zeros, and$h$ will make poles at the points where $h(z)=0$. So let’s start with the existence statement.

Theorem (Weierstrass Product).

Let $G\mathrm{\subset }\mathrm{C}$ be a domain, let $\mathrm{\{}{a}_k\mathrm{\}}$ be a sequence ofpoints in $G$ with no accumulation points in $G$, and let $\mathrm{\{}{n}_k\mathrm{\}}$ be anysequence of non-zero integers (positive or negative).Then there exists a function$f$ meromorphic in $G$ whose poles and zeros are exactly at the points $a_k$and the order of the pole or zero at $a_k$ is $n_k$(a positive order stands for zero, negative stands for pole).

Next let’s look at a more specific statement with more . Forone let’s start looking at the whole complex plane and further let’s forget about poles for now to make the following formulas simpler.

Definition.

We call

$E_0(z)$$:=1-z,$

$E_p(z)$$:=(1-z)e^{z+\frac{1}{2}{z}^2+\mathrm{\cdots }+\frac{1}{p}{z}^p}\mathit{\hspace{1em}\hspace{1em}}\text{for }p\ge 1,$

an elementary factor.

Now note that for some $a\in ℂ\backslash \backslash \{0\}$, $E_p(z/a)$ has a zero (zero of order 1) at $a$.

Theorem (Weierstrass Factorization).

Suppose $f$ be an entire function and let $\mathrm{\{}{a}_k\mathrm{\}}$ be the zeros of $f$such that $a_k\mathrm{\ne }\mathrm{0}$ (the non-zero zeros of $f$). Let $m$ be the order ofthe zero of $f$ at $z\mathrm{=}\mathrm{0}$ ($m\mathrm{=}\mathrm{0}$ if $f$ does not have a zero at $z\mathrm{=}\mathrm{0}$). Thenthere exists an entire function $g$ and a sequence of non-negativeintegers $\mathrm{\{}{p}_k\mathrm{\}}$ such that

$$f(z)=z^m{e}^{g(z)}\prod _{k=1}^{\mathrm{\infty }}{E}_{p_k}\left(\frac{z}{a_k}\right).$$

Note that we can always choose $p_k=k-1$ and the product above will converge as needed, but we may be able to choose better $p_k$ for specific functions.

Example.

As an example we can try to factorize the function $\sin \mathit{}\mathrm{(}\pi \mathit{}z\mathrm{)}$, which has zeros at all the integers. Applying the Weierstrass factorization theorem directly we get that

$$\sin (\pi z)=z{e}^{g(z)}\prod _{k=-\mathrm{\infty },k\ne 0}^{\mathrm{\infty }}\left(1-\frac{z}{k}\right)e^{z/k},$$

where $g\mathit{}\mathrm{(}z\mathrm{)}$ is some holomorphic function. It turns out that $e^{g\mathit{}\mathrm{(}z\mathrm{)}}\mathrm{=}\pi$,and rearranging the product we get

$$\sin (\pi z)=z\pi \prod _{k=1}^{\mathrm{\infty }}\left(1-\frac{z^2}{k^2}\right).$$

This is an example where we could choose the $p_k\mathrm{=}\mathrm{1}$ for all $k$ and thuswe could then get rid of the ugly parts of the infinite product. For calculations in this example see Conway [1].