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to obtain
ω 2 =2 I ( e 3
e 1 , e 3
e 2 )
π
M ( e 3 − e 1 , e 3 − e 2 ) .
=
The proof of the formula for ω 1 uses similar reasoning to obtain
e 2 K ( 1 − k 2 )
2 i
e 3
e 1 + e 3
ω 1 =
2 i
e 3 − e 1 + e 3 − e 2 I (1 ,k )
=2 iI ( e 3 − e 1 + e 3 − e 2 , e 3 − e 1 e 3 − e 2 ) .
=
If we let
a = e 3 − e 1 + e 3 − e 2 ,
b = e 3 − e 1 e 3 − e 2 ,
then (9.20) yields
a 1 = e 3 − e 1 ,
1 = e 2 − e 1 .
Proposition 9.24 therefore implies that
ω 1 =2 iI ( e 3 − e 1 , e 2 − e 1 )
πi
M ( e 3
e 1 , e 2
=
e 1 ) .
Example 9.2
Consider the elliptic curve E given by
y 2 =4 x 3
4 x.
Then e 1 = 1 ,e 2 =0 ,e 3 =1,so
πi
M ( 2 , 1) = i 2 . 62205755429211981046483959 ...
ω 1 =
π
M ( 2 , 1) =2 . 62205755429211981046483959 ....
ω 2 =
Therefore, the fundamental parallelogram for the lattice is a square. This also
follows from the fact that E has complex multiplication by Z [ i ]. See Chapter
10. The number 2 . 622 ... can be shown (see Exercise 9.8) to equal
1
dx
1 − x 4
= Γ(1 / 4)Γ(1 / 2)
2Γ(3 / 4)
,
1
where Γ is the gamma function (for its definition, see Section 14.2). This is
a special case of the Chowla-Selberg formula, which expresses the periods of
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