Prove that the function
$ \begin{align} \phi (z) = i \dfrac{1 - z}{1 + z} \end{align} $ maps the set $D = \{z \in \mathbb{C}: |z| < 1 \} $ one-to-one onto the set $U = \{ z \in \mathbb{C} : Im(z) > 0 \}$.
(This is exercise 1.9 in "Function Theory..." by Green & Krantz, and is also a claim on the Wikipedia page, though they have a function $W:U \to D $...)
For injective, I suppose that $\begin{align} \phi (z) &= \phi (w) \\ \\ i \dfrac{1 - z}{1 + z}&= i \dfrac{1 - w}{1 + w} \end{align}$ ... and after a few manipulations, end up with $z = w$.
For surjective, I am stuck.
Q: How to I prove "onto"? Do I work with $z, w$ in $(x + i \cdot y)$ form?
Also, how/where do I use that $|z| < 1$ (in $D$ ) or that $Im (\phi (z)) >0$ ?