What is the general form of a rational function which is bijective on the unit disk? I'm stuck on t

Layla Velazquez

Layla Velazquez

Answered question

2022-07-01

What is the general form of a rational function which is bijective on the unit disk?
I'm stuck on this problem. If I let R ( z ) = a 0 ( z a n ) ( z a 1 ) ( z b m ) ( z b 1 ) , then exactly one of the a i must fall in the unit disk and none of the b i can be in the unit disk.
Any hints on how to proceed? I see simple functions like z work, but I don't know how to show that higher combinations of them are bijective.

Answer & Explanation

Jake Mcpherson

Jake Mcpherson

Beginner2022-07-02Added 23 answers

The only bijective holomorphic functions from the unit disk onto itself are of the form e i θ ψ α , where θ is real and
ψ α ( z ) = z α 1 α ¯ z
with | α | < 1. For a proof, consider any bijective holomorphic function f from the unit disk onto itself. Choose an appropriate value of α in the disk, and then apply the Schwarz lemma to f ψ α and its inverse.
We use the fact that a holomorphic bijection has a holomorphic inverse. First, a computation shows that ψ α maps the unit disk D into itself, and since the inverse can be computed explicitly as ψ α , it is bijective. Now assume f : D D which is bijective, and choose the unique point α D satisfying f ( α ) = 0. Consider g = f ψ α , and note that ψ α ( 0 ) = α. By the preceding remarks, g is a holomorphic bijection of D which fixes the origin. By the Schwarz lemma, one has | g ( 0 ) | 1 with equality if and only if g is a rotation. The inverse map h = g 1 also maps D into itself, and satisfies h ( 0 ) = 0, so we have | h ( 0 ) | 1 as well. But h ( 0 ) = 1 / g ( 0 ), so we find that | g ( 0 ) | = 1 after all. By the case of equality in the Schwarz lemma, g is a rotation, and therefore f = e i θ ψ α for some real θ.

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