58.1. PERIODIC FUNCTIONS 1819

Now suppose there exists w2 ∈ M which is not a real multiple of w1. You can letw2 be the element of M having this property which has smallest absolute value. Now letw ∈ M. Since w1 and w2 point in different directions, it follows w = xw1 + yw2 for somereal numbers, x,y. Let |m− x| ≤ 1

2 and |n− y| ≤ 12 where m,n are integers. Therefore,

w = mw1 +nw2 +(x−m)w1 +(y−n)w2

and sow−mw1−nw2 = (x−m)w1 +(y−n)w2 (58.1.1)

Now since w2/w1 /∈ R,

|(x−m)w1 +(y−n)w2| < |(x−m)w1|+ |(y−n)w2|

=12|w1|+

12|w2| .

Therefore, from 58.1.1,

|w−mw1−nw2| = |(x−m)w1 +(y−n)w2|

<12|w1|+

12|w2| ≤ |w2|

and so the period, w−mw1− nw2 cannot be a non real multiple of w1 because w2 is theone which has smallest absolute value and this period has smaller absolute value than w2.Therefore, the ratio w−mw1−nw2/w1 must be a real number, x. Thus

w−mw1−nw2 = xw1

Since w1 has minimal absolute value of all periods, it follows |x| ≥ 1. Let k≤ x < k+1 forsome integer, k. If x > k, then

w−mw1−nw2− kw1 = (x− k)w1

which would contradict the choice of w1 as being the period having minimal absolute valuebecause the expression on the left in the above is a period and it equals something whichhas absolute value less than |w1|. Therefore, x = k and w is an integer linear combinationof w1 and w2. It only remains to verify the claim about τ .

From the construction, |w1| ≤ |w2| and |w2| ≤ |w1−w2| , |w2| ≤ |w1 +w2| . Therefore,

|τ| ≥ 1, |τ| ≤ |1− τ| , |τ| ≤ |1+ τ| .

The last two of these inequalities imply −1/2≤ Reτ ≤ 1/2.This proves the theorem.

Definition 58.1.3 For f a meromorphic function which has the last of the above alterna-tives holding in which M = {aw1 +bw2 : a,b ∈ Z} , the function, f is called elliptic. Thisis also called doubly periodic.

Theorem 58.1.4 Suppose f is an elliptic function which has no poles. Then f is constant.