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## 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".)