1818 CHAPTER 58. ELLIPTIC FUNCTIONS

Now all of these φ (k) in the last sum are contained in {θ (n1 +1) , · · ·} and so the last sumabove is dominated by

≤∞

∑k=n1+1

∣∣aθ(k)∣∣< ε.

Therefore, ∣∣∣∣∣ ∞

∑k=1

aφ(k)−∞

∑k=1

aθ(k)

∣∣∣∣∣ ≤∣∣∣∣∣ ∞

∑k=1

aφ(k)−n2

∑k=1

aφ(k)

∣∣∣∣∣+

∣∣∣∣∣ n2

∑k=1

aφ(k)−n1

∑k=1

aθ(k)

∣∣∣∣∣+

∣∣∣∣∣ n1

∑k=1

aθ(k)−∞

∑k=1

aθ(k)

∣∣∣∣∣ < ε + ε + ε = 3ε

and since ε is arbitrary, it follows ∑∞k=1 aφ(k) = ∑

∞k=1 aθ(k) as claimed. This proves the

theorem.

58.1 Periodic FunctionsDefinition 58.1.1 A function defined on C is said to be periodic if there exists w such thatf (z+w) = f (z) for all z ∈ C. Denote by M the set of all periods. Thus if w1,w2 ∈M anda,b ∈ Z, then aw1 + bw2 ∈ M. For this reason M is called the module of periods.1In allwhich follows it is assumed f is meromorphic.

Theorem 58.1.2 Let f be a meromorphic function and let M be the module of periods.Then if M has a limit point, then f equals a constant. If this does not happen then ei-ther there exists w1 ∈ M such that Zw1 = M or there exist w1,w2 ∈ M such that M ={aw1 +bw2 : a,b ∈ Z} and w1/w2 is not real. Also if τ = w2/w1,

|τ| ≥ 1,−12≤ Reτ ≤ 1

2.

Proof: Suppose f is meromorphic and M has a limit point, w0. By Theorem 57.1.8on Page 1793 there exist analytic functions, p,q such that f (z) = p(z)

q(z) . Now pick z0 suchthat z0 is not a pole of f . Then letting wn → w0 where {wn} ⊆ M, f (z0 +wn) = f (z0) .Therefore, p(z0 +wn) = f (z0)q(z0 +wn) and so the analytic function, p(z)− f (z0)q(z)has a zero set which has a limit point. Therefore, this function is identically equal to zerobecause of Theorem 51.5.3 on Page 1630. Thus f equals a constant as claimed.

This has shown that if f is not constant, then M is discrete. Therefore, there existsw1 ∈ M such that |w1| = min{|w| : w ∈M}. Suppose first that every element of M is areal multiple of w1. Thus, if w ∈ M, it follows there exists a real number, x such thatw = xw1. Then there exist positive integers, k,k+1 such that k ≤ x < k+1. If x > k, thenw− kw1 = (x− k)w1 is a period having smaller absolute value than |w1| which would be acontradiction. Hence, x = k and so M = Zw1.

1A module is like a vector space except instead of a field of scalars, you have a ring of scalars.