52.4. EXTENSIONS OF MAXIMUM MODULUS THEOREM 1661

Proof: First of all, for |z|< 1/ |α| ,

φ α ◦φ−α (z)≡( z+α

1+αz

)−α

1−α( z+α

1+αz

) = z

after a few computations. If I show that φ α maps B(0,1) to B(0,1) for all |α|< 1, this willhave shown that φ α is one to one and onto B(0,1).

Consider∣∣φ α

(eiθ)∣∣ . This yields∣∣∣∣ eiθ −α

1−αeiθ

∣∣∣∣= ∣∣∣∣1−αe−iθ

1−αeiθ

∣∣∣∣= 1

where the first equality is obtained by multiplying by∣∣e−iθ

∣∣ = 1. Therefore, φ α maps∂B(0,1) one to one and onto ∂B(0,1) . Now notice that φ α is analytic on B(0,1) becausethe only singularity, a pole is at z = 1/α . By the maximum modulus theorem, it follows

|φ α (z)|< 1

whenever |z|< 1. The same is true of φ−α .It only remains to verify the assertions about the derivatives. Long division gives

φ α (z) = (−α)−1 +(−α+(α)−1

1−αz

)and so

φ′α (z) = (−1)(1−αz)−2

(−α +(α)−1

)(−α)

= α (1−αz)−2(−α +(α)−1

)= (1−αz)−2

(−|α|2 +1

)Hence the two formulas follow. This proves the lemma.

One reason these mappings are so important is the following theorem.

Theorem 52.4.8 Suppose f is an analytic function defined on B(0,1) and f maps B(0,1)one to one and onto B(0,1) . Then there exists θ such that

f (z) = eiθφ α (z)

for some α ∈ B(0,1) .

Proof: Let f (α)= 0. Then h(z)≡ f ◦φ−α (z) maps B(0,1) one to one and onto B(0,1)and has the property that h(0) = 0. Therefore, by the Schwarz lemma,

|h(z)| ≤ |z| .

but it is also the case that h−1 (0) = 0 and h−1 maps B(0,1) to B(0,1). Therefore, the sameinequality holds for h−1. Therefore,

|z|=∣∣h−1 (h(z))

∣∣≤ |h(z)|and so |h(z)| = |z| . By the Schwarz lemma again, h(z) ≡ f

(φ−α (z)

)= eiθ z. Letting z =

φ α , you get f (z) = eiθ φ α (z).