Kenneth Kuttler

1794 CHAPTER 57. INFINITE PRODUCTS

Proof: From Theorem 57.1.8 there are analytic functions, f ,g such that Q = fg . There-

fore, the zero set of the function, f (z)−cg(z) has a limit point in Ω and so f (z)−cg(z) = 0for all z ∈Ω. This proves the corollary.

57.2 Factoring A Given Analytic FunctionThe next theorem is the Weierstrass factorization theorem which can be used to factor agiven analytic function f . If f has a zero of order m when z = 0, then you could factor outa zm and from there consider the factorization of what remains when you have factored outthe zm. Therefore, the following is the main thing of interest.

Theorem 57.2.1 Let f be analytic on C, f (0) ̸= 0, and let the zeros of f , be {zk} ,listedaccording to order. (Thus if z is a zero of order m, it will be listed m times in the list, {zk} .)Choosing nonnegative integers, pn such that for all r > 0,

∞

∑n=1

(r|zn|

)pn+1

< ∞,

There exists an entire function, g such that

f (z) = eg(z)∞

∏n=1

Epn

(zzn

). (57.2.12)

Note that eg(z) ̸= 0 for any z and this is the interesting thing about this function.Proof: {zn} cannot have a limit point because if there were a limit point of this se-

quence, it would follow from Theorem 51.5.3 that f (z) = 0 for all z, contradicting thehypothesis that f (0) ̸= 0. Hence limn→∞ |zn|= ∞ and so

∞

∑n=1

(r|zn|

)1+n−1

=∞

∑n=1

(r|zn|

< ∞

by the root test. Therefore, by Theorem 57.1.4

P(z) =∞

∏n=1

Epn

(zzn

)a function analytic on C by picking pn = n−1 or perhaps some other choice. ( pn = n−1works but there might be another choice that would work.) Then f/P has only removablesingularities in C and no zeros thanks to Theorem 57.1.4. Thus, letting h(z) = f (z)/P(z) ,Corollary 51.7.23 implies that h′/h has a primitive, g̃. Then(

he−g̃)′

= 0

and soh(z) = ea+ibeg̃(z)

for some constants, a,b. Therefore, letting g(z) = g̃(z) + a + ib, h(z) = eg(z) and thus57.2.12 holds. This proves the theorem.

1794 CHAPTER 57. INFINITE PRODUCTSProof: From Theorem 57.1.8 there are analytic functions, f,g such that Q = f There-fore, the zero set of the function, f (z) — cg (z) has a limit point in Q and so f (z) —cg(z) =0for all z € Q. This proves the corollary.57.2 Factoring A Given Analytic FunctionThe next theorem is the Weierstrass factorization theorem which can be used to factor agiven analytic function f. If f has a zero of order m when z = 0, then you could factor outaz” and from there consider the factorization of what remains when you have factored outthe z”. Therefore, the following is the main thing of interest.Theorem 57.2.1 Let f be analytic on C, f (0) 4 0, and let the zeros of f, be {z,},listedaccording to order. (Thus if z is a zero of order m, it will be listed m times in the list, {zx} .)Choosing nonnegative integers, Py such that for allr > 0,co ( r yrLia) <=n=1 |Zn|There exists an entire function, g such thatf (2) =e TJ Ep, (=) ; (57.2.12)n=1 SnNote that e8) 4 0 for any z and this is the interesting thing about this function.Proof: {z,} cannot have a limit point because if there were a limit point of this se-quence, it would follow from Theorem 51.5.3 that f(z) = 0 for all z, contradicting thehypothesis that f (0) 4 0. Hence lim, ,.0 |Z,| = 0 and sooo r I+n-1 oo r nEle) Ea)n=1 [Zn n=1 [<nby the root test. Therefore, by Theorem 57.1.4- zP(e) =[] E>, (=)n=1 <na function analytic on C by picking p, =n— 1 or perhaps some other choice. ( pp =n—1works but there might be another choice that would work.) Then f/P has only removablesingularities in C and no zeros thanks to Theorem 57.1.4. Thus, letting h(z) = f(z) /P(z),Corollary 51.7.23 implies that h’/h has a primitive, g. Then(ne®) ~0h (z) _ el tib o8(<)for some constants, a,b. Therefore, letting g(z) = &(z) +a+ib, h(z) = e8@) and thus57.2.12 holds. This proves the theorem.and so