1. Basic Properties of Elliptic Function
Let ω1, ω2 be two complex numbers such that the ratio ω2/ω1 is not purely real. Without loss of generality, we may assume Im τ > 0, where τ = ω2/ω1. A function f on C is called a doubly periodic function with periods ω1, ω2 if
f (z + ω1) = f (z), f (z + ω2) = f (z).
A doubly periodic function is called an elliptic function if it is meromorphic.
Let P be the parallelogram
P = {xω1+ yω2 : 0 ≤ x < 1, 0 ≤ y < 1}.
We call P a fundamental period-parallelogram for f. The set Λ = {mω1+ nω2 : m, n ∈ Z}
is called the period lattice for f.
Remark. The set Λ is a free abelian group generate by {ω1, ω2}.
Two complex numbers z1, z2 are said to be congruent modulo Λ if z1 − z2 ∈ Λ. The congruence of z1, z2 is expressed by the notation
z1 ≡ z2 mod Λ.
Notice that any two points in P are not congruent. In general, a fundamental parallelogram for Λ is a parallelogram P in C such that distinct points of P are not congruent and
C = [
ω∈Λ
(ω + P ).
A cell P for an elliptic function f is a fundamental parallelogram such that f has no zeros and no poles on ∂P.
Theorem 1.1. An elliptic function without poles is a constant.
Proof. An elliptic function f with poles is a bounded entire function. By Liouville’s theorem,
f must be a constant.
Proposition 1.1. The number of zeros / poles of an elliptic function in any cell is finite.
Proof. Assume {zn} is a sequence of poles of an elliptic function f with zi 6= zj. Since a cell is bounded, Bolzano-Weierstrass theorem implies that {zn} has a limit point. The limit point would be an essential singularity of f which leads to a contradiction to the assumption that f is a meromorphic function. Notice that the zeros of f are poles of 1/f. Since 1/f is again elliptic, 1/f can have only finite many poles in a cell, i.e. f has only finitely many zeros in a cell.
Theorem 1.2. The sum of all residues of an elliptic function in any period parallelogram is zero.
Proof. Let us choose the parallelogram P spanned by {ω1, ω2}. Then C =
[
ω∈L
(P + ω).
1
2
Let C be the closed curve C = ∂P with positive orientation. The reside of an elliptic function f in P is
I
C
f (z)dz = Z
C1
f (z)dz + Z
C2
f (z)dz + Z
C3
f (z)dz + Z
C4
f (z)dz.
Here C1 is the path connecting 0 and ω1, C2 is the path connecting ω1 and ω1+ ω2 and C3 is the point connecting ω1+ ω2 and ω2 and C4 is the path connecting ω2 and 0. We also assume that f has no poles on C. (If f has poles on C, we can use another parallelogram such that the poles of f do not lie on its boundary.) Using the periodicity of f, we obtain
Z
C1
f (z)dz + Z
C3
f (z)dz = Z
C2
f (z)dz + Z
C4
f (z)dz = 0.
This shows that
X
p∈P
resp(f ) = I
C
f (z)dz = 0.
Corollary 1.1. The number of zeros of a nonconstant elliptic function in a period paral- lelogram P is equal to the number of poles in P. The zeros and poles are counted according to their multiplicities.
Proof. Let f be a nonconstant elliptic function. Then f0(z)/f (z) is also an elliptic function.
By argument principle, the number of zeros minus the number of poles of f in P equals to the sum of residues of f in P, in other words,
1 2πi
I
∂P
f0(z)
f (z)dz = #zeros of f (z) in P − #poles of f (z) in P . Theorem 1.2 implies thatH
∂P f0(z)
f (z)dz = 0. This proves our assertion.
Theorem 1.3. The sum of the zeros of a nonconstant elliptic function in a period-parallelogram differs from the sum of its poles by a period.
Proof. Let P, C, and Ci, for i = 1, 2, 3 be as that in Theorem 1.2. Suppose a1, · · · , an and b1, · · · , bn are zeros and poles of an elliptic function f. Then
n
X
i=1
ai−
n
X
j=1
bj = 1 2πi
I
C
f0(z) f (z)dz.
Using the periodicities of f, Z
C1
zf0(z) f (z)dz +
Z
C3
zf0(z) f (z)dz =
Z
C1
(z − (z + ω2))f0(z)
f (z)dz = ω2
Z
C1
f0(z) f (z)dz, Z
C2
zf0(z) f (z)dz +
Z
C4
zf0(z) f (z)dz =
Z
C4
(z − (z + ω1))f0(z)
f (z)dz = ω1 Z
C4
f0(z) f (z)dz.
This implies that
n
X
i=1
ai−
n
X
j=1
bj = 1 2πi
ω2
Z ω2
0
f0(z)
f (z)dz − ω1 Z ω1
0
f0(z) f (z)dz
. One sees that both numbers
1 2πi
Z ω1
0
f0(z)
f (z)dz and 1 2πi
Z ω2
0
f0(z) f (z)dz
3
are integers. In fact, choose a branch of log, we have Z ωj
0
f0(z)
f (z)dz = logf (ωi) f (0). Since f (ωj) = f (0), we find
Z ωj
0
f0(z)
f (z)dz = log 1 = 2njπi, for some integer nj. This implies that
n
X
i=1
ai−
n
X
j=1
bj = mω1+ nω2
for some n, m ∈ Z. This completes the proof of our assertion.
