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1. Basic Properties of Elliptic Function Let Ω1,Ω2 Be Two Complex Numbers

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1. Basic Properties of Elliptic Function Let Ω1,Ω2 Be Two Complex Numbers 1. Basic Properties of Elliptic Function Let !1;!2 be two complex numbers such that the ratio !2=!1 is not purely real. Without loss of generality, we may assume Im τ > 0; where τ = !2=!1: A function f on C is called a doubly periodic function with periods !1;!2 if f(z + !1) = f(z); f(z + !2) = f(z): A doubly periodic function is called an elliptic function if it is meromorphic. Let P be the parallelogram P = fx!1 + y!2 : 0 ≤ x < 1; 0 ≤ y < 1g: We call P a fundamental period-parallelogram for f: The set Λ = fm!1 + n!2 : m; n 2 Zg is called the period lattice for f: Remark. The set Λ is a free abelian group generate by f!1;!2g: Two complex numbers z1; z2 are said to be congruent modulo Λ if z1 − z2 2 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 = (! + P ): !2Λ 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 fzng is a sequence of poles of an elliptic function f with zi 6= zj: Since a cell is bounded, Bolzano-Weierstrass theorem implies that fzng 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 f!1;!2g: Then [ C = (P + !): !2L 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 !1;C2 is the path connecting !1 and !1 + !2 and C3 is the point connecting !1 + !2 and !2 and C4 is the path connecting !2 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 + f(z)dz = f(z)dz + f(z)dz = 0: C1 C3 C2 C4 This shows that X I resp(f) = f(z)dz = 0: p2P C 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 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, 1 I f 0(z) dz = #zeros of f(z) in P − #poles of f(z) in P: 2πi @P f(z) H f 0(z) Theorem 1.2 implies that @P 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 n X X 1 I f 0(z) a − b = dz: i j 2πi f(z) i=1 j=1 C Using the periodicities of f; Z f 0(z) Z f 0(z) Z f 0(z) Z f 0(z) z dz + z dz = (z − (z + !2)) dz = !2 dz; C1 f(z) C3 f(z) C1 f(z) C1 f(z) Z f 0(z) Z f 0(z) Z f 0(z) Z f 0(z) z dz + z dz = (z − (z + !1)) dz = !1 dz: C2 f(z) C4 f(z) C4 f(z) C4 f(z) This implies that n n X X 1 Z !2 f 0(z) Z !1 f 0(z) a − b = ! dz − ! dz : i j 2πi 2 f(z) 1 f(z) i=1 j=1 0 0 One sees that both numbers 1 Z !1 f 0(z) 1 Z !2 f 0(z) dz and dz 2πi 0 f(z) 2πi 0 f(z) 3 are integers. In fact, choose a branch of log; we have Z !j f 0(z) f(! ) dz = log i : 0 f(z) f(0) Since f(!j) = f(0); we find Z !j f 0(z) dz = log 1 = 2njπi; 0 f(z) for some integer nj: This implies that n n X X ai − bj = m!1 + n!2 i=1 j=1 for some n; m 2 Z: This completes the proof of our assertion. .
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