Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 280

Therefore, the integral over C from ρ to i equals the integral from − 1 /ρ =1+ ρ

to 1 /i = i , which is the negative of the integral from i to 1 + ρ . Therefore,

the two parts cancel.
All that remains are the parts of C near ρ ,1+ ρ , i ,and i∞ .Near i ,we
have f ( z )=( z − i ) k g ( z )forsome k ,with g ( i ) =0 ,∞ . Therefore,

−

f ( z )
f ( z ) i + g ( z )

k

g ( z ) .

= (9.17)

z −

The integral over the small semicircle near i is

f ( i + e iθ ) 1

2 πi

f ( i + e iθ ) ie iθ dθ,

(9.18)

θ

where θ ranges from slightly more than π to slightly less than 0. (Note that
C is traveled clockwise. Because of the curvature of the unit circle, the limits
are 0 and π only in the limit as → 0.) Substitute (9.17) into (9.18) and
let 0. Since g /g is continuous at i , the integral of g /g goes to 0. The

integral of k/ ( z

→

− i ) yields

0

1
2 πi

1
2 k = −

1
2 ord i ( f ) .

ki dθ = −

θ = π

Similarly, the contributions from the parts of C near ρ and 1 + ρ addupto
− (1 / 3)ord ρ ( f ) (we are using the fact that f ( ρ )= f ( ρ + 1), by (9.16)).
Finally, the integral along the top part of C is

−

1
2

f ( t + iN )
f ( t + iN )

1
2 πi

dt.

t = 2

Since f ( τ )= q n ( a n + a n +1 q + ··· ), we have

f ( τ )
f ( τ )

=2 πin + 2 πia n +1 q +

···

.

a n + ···

The second term goes to 0 as q → 0, hence as N →∞ . The limit of the
integral as N →∞ is therefore

−

1
2

1
2 πi

2 πin dt = −n = − ord i∞ ( f ) .

t = 2

Combining all of the above calculations yields the theorem.

COROLLARY 9.18
If z ∈ C ,then there isexactlyone τ ∈F su ch that j ( τ )= z .