10. Theorem 2.12.3 (Weierstrauss)

Let $(a_{n})$ be a sequence of complex numbers without accumulation points. Then there exists an entire function with zeros 1;;precisely at $a_{n}$ , given by the formula:

z^{m} \prod_{n \in N, a_{n \neq 0}} [(1 - \frac{z}{a_{n}}) \exp (\sum_{k = 1}^{N_{n}} \frac{1}{k} {(\frac{z}{a_{n}})}^{k})]

Furthermore, this function is unique up to multiplication by an entire function $g (z)$ such that $g$ does not vanish. Same thing as multiplication by $e^{ϕ (z)}$ for an entire function $ϕ$ .

Remark

The $z^{m}$ accounts for the zeros of the sequence $a_{n}$ . note that the sequence $a_{n}$ may have repeat terms, which accounts for the multiplicities of the zeros.

PROOF:

the function $\ln (1 - z) = - \sum_{k = 1}^{\infty} \frac{z^{k}}{k}$ converges absolutely locally uniformly in $B_{\frac{1}{2}} (0)$ .
For every $n$ , choose $N_{n}$ such that for any $z \in B_{\frac{1}{2}} (0)$ :

| \ln (1 - z) + \sum_{k = 1}^{N_{n}} \frac{z^{k}}{k} | < \frac{1}{2^{n}}

Since $a_{n}$ has no finite accumulation points, for $r > 0$ , there exists $n_{0} \in N$ such that for all $n \geq n_{0}$ , we have that $2 r < | a_{n} |$ . Then for any $z \in B_{r} (0)$ and $n \geq n_{0}$ , we have that $| \frac{z}{a_{n}} | < \frac{1}{2}$ .
Since $\frac{z}{a_{n}} \in B_{\frac{1}{2}} (0)$ , we can plug it into the above equality. Notice that 1;;by an analog of weierstrauss M-test, this means that:

\sum_{n = 1}^{\infty} [\ln (1 - \frac{z}{a_{n}}) + \sum_{k = 1}^{N_{n}} \frac{1}{k} {(\frac{z}{a_{n}})}^{k}]

Converges absolutely and locally uniformly.
5. Deduce the claim using the previous theorem