1744 CHAPTER 55. COMPLEX MAPPINGS

55.3.2 Regions With Square Root PropertyTheorem 55.3.6 Let Ω ̸= C for Ω a region and suppose Ω has the square root property.Then for z0 ∈ Ω there exists h : Ω→ B(0,1) such that h is one to one, onto, analytic, andh(z0) = 0.

Proof: Define F to be the set of functions, f such that f : Ω→ B(0,1) is one to oneand analytic. The first task is to show F is nonempty. Then, using Montel’s theorem it willbe shown there is a function in F , h, such that |h′ (z0)| ≥ |ψ ′ (z0)| for all ψ ∈F . Whenthis has been done it will be shown that h is actually onto. This will prove the theorem.

Claim 1: F is nonempty.Proof of Claim 1: Since Ω ̸=C it follows there exists ξ /∈Ω. Then it follows z−ξ and

1z−ξ

are both analytic on Ω. Since Ω has the square root property, there exists an analytic

function, φ : Ω→ C such that φ2 (z) = z− ξ for all z ∈ Ω, φ (z) =

√z−ξ . Since z− ξ

is not constant, neither is φ and it follows from the open mapping theorem that φ (Ω) is aregion. Note also that φ is one to one because if φ (z1) = φ (z2) , then you can square bothsides and conclude z1−ξ = z2−ξ implying z1 = z2.

Now pick a ∈ φ (Ω) . Thus√

za−ξ = a. I claim there exists a positive lower bound to∣∣∣√z−ξ +a∣∣∣ for z ∈Ω. If not, there exists a sequence, {zn} ⊆Ω such that√

zn−ξ +a =√

zn−ξ +√

za−ξ ≡ εn→ 0.

Then √zn−ξ =

(εn−

√za−ξ

)(55.3.6)

and squaring both sides,

zn−ξ = ε2n + za−ξ −2εn

√za−ξ .

Consequently, (zn− za) = ε2n − 2εn

√za−ξ which converges to 0. Taking the limit in

55.3.6, it follows 2√

za−ξ = 0 and so ξ = za, a contradiction to ξ /∈ Ω. Choose r > 0

such that for all z ∈Ω,∣∣∣√z−ξ +a

∣∣∣> r > 0. Then consider

ψ (z)≡ r√z−ξ +a

. (55.3.7)

This is one to one, analytic, and maps Ω into B(0,1) (∣∣∣√z−ξ +a

∣∣∣ > r). Thus F is notempty and this proves the claim.

Claim 2: Let z0 ∈Ω. There exists a finite positive real number, η , defined by

η ≡ sup{∣∣ψ ′ (z0)

∣∣ : ψ ∈F}

(55.3.8)

and an analytic function, h ∈F such that |h′ (z0)|= η . Furthermore, h(z0) = 0.Proof of Claim 2: First you show η < ∞. Let γ (t) = z0 + reit for t ∈ [0,2π] and r

is small enough that B(z0,r) ⊆ Ω. Then for ψ ∈F , the Cauchy integral formula for thederivative implies

ψ′ (z0) =

12πi

∫γ

ψ (w)

(w− z0)2 dw