414 CHAPTER 16. MAPPING THEOREMS

Lemma 16.6.5 F in the above lemma is nonempty and η is a positive real number.

Proof: Since Ω ̸= C it follows there exists ξ /∈ Ω.Thus 1z−ξ

,z− ξ are both analytic

and since Ω has the square root property, there exists analytic φ with φ (z) =√

z−ξ andφ is not constant, so φ (Ω) is an open connected set, not a single point. Also φ is oneto one. Pick a ∈ φ (Ω) ,a ̸= 0 and za such that a =

√za−ξ . Then consider 0 < r ≡

inf{∣∣∣√z−ξ +a

∣∣∣} . Is r > 0? If so, we can let ψ (z) = r√z−ξ+a

and obtain ψ maps Ω to

B(0,1). If this is not so, there exists zn,√

zn−ξ +a→ 0 and so

zn−ξ +2a√

zn−ξ +a2→ 0 so zn−ξ −2a2 +a2→ 0

and so there exists z = limn→∞ zn and z−ξ = a2 = za−ξ so z = za. But this is impossiblebecause it requires that

√za−ξ +a = 0 so

√za−ξ = −a ̸= a. Note that ψ just defined,

is one to one. Thus F is nonempty.For ψ ∈F , let γ be a small circular contour of radius r about 0 and B(0,1)⊆Ω.

ψ′ (0) =

12πi

∫γ

ψ (w)w2 dw,

∣∣ψ ′ (0)∣∣≤ (1/2π)2πr(1/r2)= 1/r

thus η < ∞. Consider the special ψ of Claim 1. ψ (z) = r√z−ξ+a

. Then

ψ (z)(√

z−ξ +a)= r

and so

ψ′ (z)

(√z−ξ +a

)+ψ (z)

(1

2√

z−ξ

)= 0

and so ψ ′ (0)(√−ξ +a

)=−ψ (0)

(1

2√−ξ

). Now from the construction, ψ (0) ̸= 0 and

also∣∣∣√−ξ +a

∣∣∣≥ r > 0 so ψ ′ (0) ̸= 0 which shows η > 0. ■

Lemma 16.6.6 There is an analytic function h∈F such that |h′ (0)|= h′ (0) = η . Alsoh(0) = 0. Thus if 0 ∈Ω⊊C, for Ω having the square root property, h(Ω) = B(0,1) .

Proof: By Theorem 16.6.1, there exists a sequence, {ψn}, of functions in F and an an-alytic function h, such that |ψ ′n (0)| → η and ψn→ h,ψ ′n→ h′, uniformly on each compactsubset of Ω. It follows ∣∣h′ (0)∣∣= lim

n→∞

∣∣ψ ′n (0)∣∣= η > 0 (16.7)

Now let |ω| = 1 and let ωh′ (0) = |h′ (0)| . Thus {ωψn} could be used in place of {ψn}and we can assume h′ (0) = |h′ (0)|= η and for all z ∈Ω,

|h(z)|= limn→∞|ψn (z)| ≤ 1. (16.8)

By 16.7, h is not a constant. Therefore, in fact, |h(z)| < 1 for all z ∈ Ω in 16.8 by theopen mapping theorem because h(Ω) is a region (open and connected).

It follows from Lemma 16.6.2 that h is one to one. In particular h−1 is analytic on h(Ω)by the open mapping theorem. Why is h(0) = 0?