Question

state and prove weirerstrass factorization theorem.

Answer 1

Let f be an entire function.

Let 0 be a zero of f of multiplicity m≥0.

Let the sequence ⟨an⟩ consist of the nonzero zeroes of f, repeated according to multiplicity.

First Form

Let ⟨pn⟩ be a sequence of non-negative integers for which the series:
∑n=1∞∣∣∣ran∣∣∣1+pn
converges for every r∈R>0.

Then there exists an entire function g such that:
f(z)=zmeg(z)∏n=1∞Epn(zan)
where:
Epn are Weierstrass's elementary factors
and the product converges locally uniformly absolutely on C.


Second Form

Then there exists a sequence ⟨pn⟩ of non-negative integers and an entire function gsuch that:
f(z)=zmeg(z)∏n=1∞Epn(zan)
where:
Epn are Weierstrass's elementary factors
and the product converges locally uniformly absolutely on C.

proof
From Weierstrass Product Theorem, the function:
h(z)=zm∏n=1∞Epn(zan)
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 gsuch that:
eg=f/h
Therefore:
f=egh
as desired.