1830 CHAPTER 58. ELLIPTIC FUNCTIONS

Thus

℘(−z) =1z2 + ∑

w̸=0

(1

(−z−w)2 −1

w2

)

=1z2 + ∑

w̸=0

(1

(−z+w)2 −1

w2

)=℘(z) .

Therefore, ℘(w1− z) = ℘(z−w1) = ℘(z) and so −℘′ (w1− z) = ℘′ (z) . Letting z =w1/2, it follows℘′ (w1/2) = 0. Similarly,℘′ (w2/2) = 0 and℘′ ((w1 +w2)/2) = 0. There-fore, from 58.1.8

0 = 4(℘(w1/2)− e1)(℘(w1/2)− e2)(℘(w1/2)− e3) .

It follows one of the ei must equal ℘(w1/2) . Similarly, one of the ei must equal ℘(w2/2)and one must equal ℘((w1 +w2)/2).

Lemma 58.1.14 The numbers, ℘(w1/2) ,℘(w2/2) , and ℘((w1 +w2)/2) are distinct.

Proof: Choose Pa, a period parallelogram which contains the pole 0, and the pointsw1/2, w2/2, and (w1 +w2)/2 but no other pole of ℘(z) . Also ∂P∗a does not contain anyzeros of the elliptic function, z→℘(z)−℘(w1/2). This can be done by shifting P0 slightlybecause the poles are only at the points aw1 +bw2 for a,b integers and the zeros of ℘(z)−℘(w1/2) are discrete.

0w1

w2

w1 +w2

a

If℘(w2/2)=℘(w1/2) , then℘(z)−℘(w1/2) has two zeros, w2/2 and w1/2 and sincethe pole at 0 is of order 2, this is the order of ℘(z)−℘(w1/2) on Pa hence by Theorem58.1.7 on Page 1820 these are the only zeros of this function on Pa. It follows by Corollary58.1.9 on Page 1822 which says the sum of the zeros minus the sum of the poles is in M,w12 + w2

2 ∈M. Thus there exist integers, a,b such that

w1 +w2

2= aw1 +bw2

which implies (2a−1)w1 +(2b−1)w2 = 0 contradicting w2/w1 not being real. Similarreasoning applies to the other pairs of points in {w1/2,w2/2,(w1 +w2)/2} . For example,