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 + P ).
ω∈Λ
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=
(P + ω).
ω∈L
1
2
Let C be the closed curve C = ∂P with positive orientation. The reside of an elliptic
function f in P is
I
Z
Z
Z
Z
f (z)dz =
f (z)dz +
f (z)dz +
f (z)dz +
f (z)dz.
C
C1
C2
C3
C4
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
Z
Z
Z
f (z)dz = 0.
f (z)dz +
f (z)dz =
f (z)dz +
This shows that
X
C4
C2
C3
C1
I
f (z)dz = 0.
resp (f ) =
C
p∈P
Corollary 1.1. The number of zeros of a nonconstant elliptic function in a period parallelogram 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 f 0 (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,
I
1
f 0 (z)
dz = #zeros of f (z) in P − #poles of f (z) in P .
2πi ∂P f (z)
H
0 (z)
Theorem 1.2 implies that ∂P ff (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
I
n
n
X
X
f 0 (z)
1
ai −
dz.
bj =
2πi C f (z)
i=1
j=1
Using the periodicities of f,
Z
Z
Z
Z
f 0 (z)
f 0 (z)
f 0 (z)
z
dz +
z
dz =
(z − (z + ω2 ))
dz = ω2
f (z)
f (z)
f (z)
C
C
C
C
Z 1 0
Z 3 0
Z 1
Z 1
f (z)
f (z)
f 0 (z)
z
dz +
z
dz =
(z − (z + ω1 ))
dz = ω1
f (z)
C2 f (z)
C4 f (z)
C4
C4
This implies that
n
X
i=1
ai −
n
X
j=1
1
bj =
2πi
ω2
Z
ω2
One sees that both numbers
Z ω1 0
1
f (z)
dz
2πi 0 f (z)
0
f 0 (z)
dz − ω1
f (z)
and
1
2πi
Z
0
ω2
Z
0
ω1
f 0 (z)
dz,
f (z)
f 0 (z)
dz.
f (z)
f 0 (z)
dz .
f (z)
f 0 (z)
dz
f (z)
3
are integers. In fact, choose a branch of log, we have
Z ωj 0
f (z)
f (ωi )
dz = log
.
f (z)
f (0)
0
Since f (ωj ) = f (0), we find
f 0 (z)
dz = log 1 = 2nj πi,
f (z)
0
for some integer nj . This implies that
Z
ωj
n
X
i=1
ai −
n
X
bj = mω1 + nω2
j=1
for some n, m ∈ Z. This completes the proof of our assertion.