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.

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What is the general form of a rational function which is bijective on the unit disk?I&#039;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&#039;t know how to show that higher combinations of them are bijective.

Jake Mcpherson · Accepted Answer

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      |    &amp;lt;  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.

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

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