Complex Analysis - Proving a bijection on a closed disk

[semithinking]

Homework Statement

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

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

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

Homework Equations

The Attempt at a Solution

So following from the hint, I can calculate the inverse of \phi_w. 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".)

 

[Dick]

You want to show that if |z|=1 then |phi_w(z)|=1. Use that |z|=1 is equivalent to zz*=1.

 

[semithinking]

So I would substitute zz* into |phi_w(z)| to see if it's 1?

 

[Dick]

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.