TITLE: Why does $\frac{z-i}{z+i}$ map the unit disk onto an open half-plane?
QUESTION [5 upvotes]: I've been self studying complex analysis, and I read that there is a canonical conformal bijection
$$\varphi(z) = \frac{z-i}{z+i}$$
from the open unit disc to the open half-plane.
I see why $\varphi$ is analytic, and I was able to show that it is one to one by taking differences and counting zeros. I cannot see why $\varphi$ is onto, however, why is this true?

REPLY [2 votes]: Here is how you can solve the problem using basic properties of Möbius transformations:

Möbius transformations map circles to circles or extended lines. Here $\varphi(-i) = \infty$ and $\varphi(i) = 0$. It follows that $\varphi$ maps the unit circle $S = \{ z : |z| = 1 \}$ onto a line through the origin.

Möbius transformations preserve angles. Here $\varphi(i\Bbb R) \subset i \Bbb R$ and the unit circle $S$ intersects the imaginary axis at a right angle. It follows that $\varphi(S)$ intersects the imaginary axis at a right angle as well.

Combining the previous two results, we see that $\varphi(S) = \Bbb R \cup \{ \infty \}$.

Möbius transformations are bijective conformal mappings from the extended complex plane (aka Riemann sphere) onto itself. The boundary of the unit disk $\Bbb D$ is mapped to the extended real line. It follows that the unit disk itself is conformally mapped onto either the left or the right half-plane.

$\varphi(0) = -1$,  therefore $\varphi$ maps the unit disk conformally onto the left half-plane.