Kenneth Kuttler

416 CHAPTER 16. MAPPING THEOREMS

3. Suppose you have f ∈M (C) , not M(Ĉ). Show that if f has finitely many zeros

{α1, · · · ,αn} and poles {β 1, · · · ,β m} , then there is an entire function g(z) such that(z−α1)

r1 ···(z−αn)rn

(z−β 1)p1 ···(z−β m)

pm eg(z) = f (z) where pi is the order of the pole at β i and ri is theorder of the zero at α i. Hint: First show f (z) = ∏

nk=1 (z−αk)

rk h(z) where h(z) ismeromorphic but has no zeros. Then h(z) has the same poles with the same ordersas f (z). Then h(z)−∑

mi=1 Si (z) = l (z) where l (z) is entire and the Si (z) are the

principal parts h(z) corresponding to β i. Argue now that

f (z) =∏

nk=1 (z−αk)

∏mk=1 (z−β k)

pk(q(z))

where q(z) is analytic on C and can’t have any zeros. Next use Problem 15 on Page399.

4. Let w1,w2,w3 be independent periods for a meromorphic function f (z). This meansthat if ∑

3i=1 aiwi = 0 for each ai an integer, then each ai = 0. Hint: At some point

you may want to use Lemma 16.1.4.

(a) Show that if ai is an integer, then ∑3i=1 aiwi is also a period of f (z).

(b) Let PN be periods of the form ∑3i=1 aiwi for ai an integer with |ai| ≤ N. Show

there are (2N +1)3 such periods.

(∑

3i=1 |wi|

),N(∑

3i=1 |wi|

)]2 ≡ QN .

(d) Between the cubes of any two successive positive integers, there is the squareof a positive integer. Thus (2N)3 < M2 < (2N +1)3. Show this is so. It is easyto verify if you show that (x+1)3/2− x3/2 > 2 for all x≥ 2 showing that thereis an integer m between (n+1)3/2 and n3/2. Then squaring things, you get theresult.

(e) Partition QN into M2 small squares. If Q is one of these, show its sides are nolonger than((

2N(∑

3i=1 |wi|

))2

)1/2

≤

((N(∑

3i=1 |wi|

))2

(2N)3

)1/2

≤ CN1/2

(f) You have (2N +1)3 points which are contained in M2 squares where M2 issmaller than (2N +1)3 . Explain why one of these squares must contain twodifferent periods of PN .

(g) Suppose the two periods are ∑3i=1 aiwi and ∑

3i=1 âiwi, both in Q which has sides

of length no more than C/N1/2. Thus the distance between these two periodshas length no more than

√2C/√

N. Explain why this shows that there is asequence of periods of f which converges to 0. Explain why this requires f tobe a constant.

This result, that there are at most two independent periods is due to Jacobi fromaround 1835. In fact, there are nonconstant functions which have two independentperiods but they can’t be bounded.

416CHAPTER 16. MAPPING THEOREMS3. Suppose you have f € .@(C), not. #@ (¢) . Show that if f has finitely many zeros{Q@1,-+- ,Q@} and poles {8 ,,---,B,,}, then there is an entire function g(z) such thatem ©) = f(z) where p; is the order of the pole at B; and r; is theorder of the zero at a@;. Hint: First show f (z) = []{_) (z— @)/* A(z) where h(z) ismeromorphic but has no zeros. Then /1(z) has the same poles with the same ordersas f(z). Then h(z) — Yi", S;(z) =1(z) where /(z) is entire and the S;(z) are theprincipal parts h(z) corresponding to B;. Argue now that— Mar & = oe)"Thi1 (¢— By)”where q(z) is analytic on C and can’t have any zeros. Next use Problem 15 on Page399.f(z) (q(z))Let w1,w2,w3 be independent periods for a meromorphic function f (z). This meansthat if ye , aiw; = O for each a; an integer, then each a; = 0. Hint: At some pointyou may want to use Lemma 16.1.4.(a) Show that if a; is an integer, then yh ajw; is also a period of f (z).(b) Let Py be periods of the form yh , aw; for a; an integer with |a;| <<.N. Showthere are (2N + 1)? such periods.(c) Show Py © [=N (E44 wil) VLE Iwil)]? = Ow.(d) Between the cubes of any two successive positive integers, there is the squareof a positive integer. Thus (2N)* < M? < (2N+1)*. Show this is so. It is easyto verify if you show that (x+ 1)°/? —33/2 > 2 for all x > 2 showing that thereis an integer m between (n+ 1)3/? 3/2result.and n°/~. Then squaring things, you get the(e) Partition Qy into M7 small squares. If Q is one of these, show its sides are nolonger than( Oh mi) a (“ hi el) easM (2N)3(f) You have (2N+ 1)3 points which are contained in M? squares where M? issmaller than (2N + 1). Explain why one of these squares must contain twodifferent periods of Py.(g) Suppose the two periods are yh ajw; and a Gjw;, both in Q which has sidesof length no more than C/N '/2. Thus the distance between these two periodshas length no more than V2C/\/N. Explain why this shows that there is asequence of periods of f which converges to 0. Explain why this requires f tobe a constant.This result, that there are at most two independent periods is due to Jacobi fromaround 1835. In fact, there are nonconstant functions which have two independentperiods but they can’t be bounded.