1746 CHAPTER 55. COMPLEX MAPPINGS

If h(z0) ̸= 0,consider φ h(z0)◦h where φ α is the fractional linear transformation defined

in Lemma 55.3.3. By this lemma it follows φ h(z0)◦h ∈F . Now using the chain rule,∣∣∣∣(φ h(z0)

◦h)′(z0)

∣∣∣∣ =∣∣∣φ ′h(z0)

(h(z0))∣∣∣ ∣∣h′ (z0)

∣∣=

∣∣∣∣∣ 1

1−|h(z0)|2

∣∣∣∣∣ ∣∣h′ (z0)∣∣

=

∣∣∣∣∣ 1

1−|h(z0)|2

∣∣∣∣∣η > η

Contradicting the definition of η . This proves Claim 2.Claim 3: The function, h just obtained maps Ω onto B(0,1).Proof of Claim 3: To show h is onto, use the fractional linear transformation of Lemma

55.3.3. Suppose h is not onto. Then there exists α ∈ B(0,1) \ h(Ω) . Then 0 ̸= φ α ◦ h(z)for all z ∈Ω because

φ α ◦h(z) =h(z)−α

1−αh(z)

and it is assumed α /∈ h(Ω) . Therefore, since Ω has the square root property, you canconsider an analytic function z→

√φ α ◦h(z). This function is one to one because both

φ α and h are. Also, the values of this function are in B(0,1) by Lemma 55.3.3 so it is inF .

Now letψ ≡ φ√

φα◦h(z0)◦√

φ α ◦h. (55.3.13)

Thusψ (z0) = φ√

φα◦h(z0)◦√

φ α ◦h(z0) = 0

and ψ is a one to one mapping of Ω into B(0,1) so ψ is also in F . Therefore,

∣∣ψ ′ (z0)∣∣≤ η ,

∣∣∣∣(√φ α ◦h)′(z0)

∣∣∣∣≤ η . (55.3.14)

Define s(w)≡ w2. Then using Lemma 55.3.3, in particular, the description of φ−1α = φ−α ,

you can solve 55.3.13 for h to obtain

h(z) = φ−α ◦ s◦φ−√

φα◦h(z0)◦ψ

=

 ≡F︷ ︸︸ ︷φ−α ◦ s◦φ−

√φα◦h(z0)

◦ψ

(z)

= (F ◦ψ)(z) (55.3.15)

NowF (0) = φ−α ◦ s◦φ−

√φα◦h(z0)

(0) = φ−1α (φ α ◦h(z0)) = h(z0) = 0