Since ℘(z) is an even function, ζ is an odd function, i.e

1. Weierstrass ζ-function

Let ω₁, ω₂ be two complex numbers such that Im τ > 0 where τ = ω₂/ω₁. Let L = Zω1⊕ Zω2 be the lattice generated by {ω1, ω2}. The Weierstrass ζ function is defined by

ζ(z) = 1

z + X

ω∈L\{0}

 1

z − ω+ 1 ω + z²

ω²



It follows from the definition of ζ and ℘(z) that ζ⁰(z) = −℘(z).

Since ℘(z) is an even function, ζ is an odd function, i.e. ζ(−z) = −ζ(z). If we write

℘(z) = 1 z² +

∞

X

n=1

b_nz²ⁿ in a neighborhood of 0, then

ζ(z) = 1 z−

∞

X

n=1

bn

2n + 1z²ⁿ⁺¹

in a neighborhood of 0. By integrating the relation ℘(z + ω1) = ℘(ω), we obtain ζ(z + ω1) = ζ(z)+C, where C is a constant to be determined. Substituting z = −ω₁/2 into this equation and using ζ(−z) = −ζ(z), we find C = 2ζ(ω1/2). Hence we obtain ζ(z+ω1) = ζ(z)+2ζ ^ω₂¹
. In fact, we have

ζ(z + ω_i) = ζ(z) + 2ζω_i 2



, i = 1, 2, 3, where ω₃= ω₁+ ω₂. We denote

η_i = ζω_i 2



, i = 1, 2, 3.

Theorem 1.1. The Weierstrass ζ-function is an odd function of z such that ζ⁰(z) = −℘(z) and ζ(z + ω_i) = ζ(z) + 2η_i for i = 1, 2, 3. It is holomorphic on C \ Λ and has pole at ω ∈ Λ of residue 1.

Theorem 1.2. (Legendre relation)

η1ω2− η₂ω1 = πi.

Let us for simplicity denote Q

ω∈Λ\{0} by Q0

ω and P

ω∈Λ\{0} by P0

ω. Let us define E(z) = (1 − z)e^z+z22 , z ∈ C.

Lemma 1.1. For |z| ≤ 1/2,

|E(z) − 1| ≤ 2|z|³. This lemma implies that the finite product

Y

ω 0E

z ω



converges absolutely and uniformly on |z| ≤ R for every R > 0. This allows us to define an entire function, called the Weierstrass σ-function by

σ(z) = z Y

ω

0Ez ω



= z Y

ω 0

1 − z ω



e^ωz+2ωz2.

1

2

After taking the principal branch of log, it follows from the definition that d

dzlog σ(z) = ζ(z).

The σ function is an odd function of z. Integrating ζ(z + ωi) = ζ(z) + 2ηi, we obtain σ(z + ω_i) = c_iσ(z)e^2ηiz

where ci are constant to be determined. Substituting z = −ωi/2 into the equation and using σ being odd, we find c_i = −e^−2ηiωi. In other words, we obtain

(1.1) σ(z + ωi) = −σ(z)e^2ηi(^z+ωi₂ )

Theorem 1.3. Let f (z) be an elliptic function of period {ω1, ω2}. Let a₁, · · · , an be the zeros and b₁, · · · , b_n be the poles of f (z) repeated according to their multiplicity which lie in a fundamental period parallelogram P so that

n

X

i=1

a_i ≡

n

X

j=1

b_j mod L.

Then there exists a constant c ∈ C such that for all z ∈ C, (1.2) f (z) = c · σ(z − a₁) · · · σ(z − a_n)

σ(z − b1) · · · σ(z − bn).

Proof. Let g(z) be the meromorphic function defined by the right hand side of equation (1.2). Equation (1.1) implies

g(z + ω₁) = (−1)ⁿe^2η1(^nz+n₂^ω1−a1−···−an)σ(z − a₁) · · · σ(z − a_n) (−1)ⁿe^2η1(^nz+n2ω1−b1−···−bn)σ(z − b₁) · · · σ(z − bn)

= g(z).

Similarly, we have g(z + τ ) = g(z). This shows that g(z) is an elliptic function of periods {1, τ }. The quotient f (z)/g(z) is again elliptic without poles. By Lemma ??, f (z)/g(z) is a constant function. This completes the proof of our assertion.