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.