58.1. PERIODIC FUNCTIONS 1829

℘′ (z)2 =

4z6 −

24G2

z2 −80G3 + · · ·

4℘(z)3 = 4(

1z2 +3G2z2 +5G3z4 · · ·

)3

=4z6 +

36z2 G2 +60G3 + · · ·

and finally

60G2℘(z) =60G2

z2 +0+ · · ·

where in the above, the positive powers of z are not listed explicitly. Therefore,

℘′ (z)2−4℘(z)3 +60G2℘(z)+140G3 =

∞

∑n=1

anzn

In deriving the equation it was assumed |z| < |w| for all w = aw1 + bw2 where a,b areintegers not both zero. The left side of the above equation is periodic with respect to w1and w2 where w2/w1 is not a real number. The only possible poles of the left side are at0, w1, w2, and w1 +w2, the vertices of the parallelogram determined by w1 and w2. Thisfollows from the original formula for ℘(z) . However, the above equation shows the leftside has no pole at 0. Since the left side is periodic with periods w1 and w2, it follows it hasno pole at the other vertices of this parallelogram either. Therefore, the left side is periodicand has no poles. Consequently, it equals a constant by Theorem 58.1.4. But the right sideof the above equation shows this constant is 0 because this side equals zero when z = 0.Therefore, ℘ satisfies the differential equation,

℘′ (z)2−4℘(z)3 +60G2℘(z)+140G3 = 0.

It is traditional to define 60G2 ≡ g2 and 140G3 ≡ g3. Then in terms of these new quantitiesthe differential equation is

℘′ (z)2 = 4℘(z)3−g2℘(z)−g3.

Suppose e1,e2 and e3 are zeros of the polynomial 4w3− g2w− g3 = 0. Then the aboveequation can be written in the form

℘′ (z)2 = 4(℘(z)− e1)(℘(z)− e2)(℘(z)− e3) . (58.1.8)

58.1.4 A Modular FunctionThe next task is to find the ei in 58.1.8. First recall that ℘ is an even function. That is℘(−z) =℘(z). This follows from 58.1.6 which is listed here for convenience.

℘(z)≡ 1z2 + ∑

w ̸=0

(1

(z−w)2 −1

w2

)(58.1.9)