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Complex Analysis - Proving a bijection on a closed disk

  • #1

Homework Statement



For each [tex] w \in \mathbb{C} [/tex] define the function [tex] \phi_w [/tex] on the open set [tex] \mathbb{C}\backslash \{\bar{w}^{-1}\} [/tex] by [tex] \phi_w (z) = \frac{w - z}{1 - \bar{w}z}[/tex], for [tex] z \in \mathbb{C}\backslash \{\bar{w}^{-1}\} \back[/tex].

Prove that [tex] \phi_w : \bar{D} \mapsto \bar{D} [/tex] is a bijection on the closed disk [tex] \bar{D} [/tex] for [tex] w \notin \bar{D} [/tex].

Hint: Compute the inverse of [tex] \phi_w[/tex] and prove first that both [tex] \phi_w[/tex] and [tex](\phi_w)^{-1} [/tex] map the circle [tex]\mathbb{T} [/tex] into itself.

Homework Equations





The Attempt at a Solution



So following from the hint, I can calculate the inverse of [tex] \phi_w [/tex]. But I before I can start, I'm getting lost even at how I would show mapping the function and it's inverse to a circle...(I'm even more confused by "into itself".)
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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You want to show that if |z|=1 then |phi_w(z)|=1. Use that |z|=1 is equivalent to zz*=1.
 
  • #3
So I would substitute zz* into |phi_w(z)| to see if it's 1?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
So I would substitute zz* into |phi_w(z)| to see if it's 1?
No. You want to show (phi_w(z))(phi_w(z))*=1 for any z such that zz*=1.
 

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