1844 CHAPTER 58. ELLIPTIC FUNCTIONS

Theorem 58.1.23 Let f : B(a,R)→ C be analytic and let

f (z)−α = (z−a)m g(z) , ∞ > m≥ 1

where g(z) ̸= 0 in B(a,R) . ( f (z)−α has a zero of order m at z = a.) Then there existε,δ > 0 with the property that for each z satisfying 0 < |z−α|< δ , there exist points,

{a1, · · · ,am} ⊆ B(a,ε) ,

such thatf−1 (z)∩B(a,ε) = {a1, · · · ,am}

and each ak is a zero of order 1 for the function f (·)− z.

Corollary 58.1.24 Let Ω be the region above. Consider the set of points, Q = Ω∪Ω′ \{0,1} described by the following picture.

Ω′

−1

Ω

C l2l1

112

Then λ (Q) = C\{0,1} . Also λ′ (z) ̸= 0 for every z in ∪∞

k=−∞(Q+2k)≡ H.

Proof: By Theorem 58.1.22, this will be proved if it can be shown that λ (Ω′) ={z ∈ C : Im(z)< 0} . Consider λ 1 defined on Ω′ by

λ 1 (x+ iy)≡ λ (−x+ iy).

Claim: λ 1 is analytic.Proof of the claim: You just verify the Cauchy Riemann equations. Letting λ (x+ iy)=

u(x,y)+ iv(x,y) ,

λ 1 (x+ iy) = u(−x,y)− iv(−x,y)

≡ u1 (x,y)+ iv(x,y) .

Then u1x (x,y) =−ux (−x,y) and v1y (x,y) =−vy (−x,y) =−ux (−x,y) since λ is analytic.Thus u1x = v1y. Next, u1y (x,y) = uy (−x,y) and v1x (x,y) = vx (−x,y) =−uy (−x,y) and sou1y =−vx.

Now recall that on l1,λ takes real values. Therefore, λ 1 = λ on l1, a set with a limitpoint. It follows λ = λ 1 on Ω′ ∪Ω. By Theorem 58.1.22 λ maps Ω one to one onto theupper half plane. Therefore, from the definition of λ 1 = λ , it follows λ maps Ω′ one to oneonto the lower half plane as claimed. This has shown that λ is one to one on Ω∪Ω′. Thisalso verifies from Theorem 52.6.3 on Page 1665 that λ

′ ̸= 0 on Ω∪Ω′.