Why does $\frac{z-i}{z+i}$ map the unit disk onto an open half-plane? 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?
 A: You can use the fact that the inverse function is $\varphi^{-1}(z)=-i\frac{z+1}{z-1}$, and that for $z$ in the left half plane, $|z-1|^2$ is always greater than $|z|^2+1$ (expand $(z-1)\overline{(z-1)}$, and see how $\text{Re}(z)$ appears). Meanwhile, $|z+1|^2 < |z|^2 + 1$ for similar reasons. So, the norm-square of $\varphi^{-1}(z)$ on the left half plane always has a larger denominator than numerator; thus its norm-square is always less than 1, and therefore so is its norm. So, the image of $\varphi^{-1}$ on the left half plane is contained in the unit disc, and we’re done.
Here’s a more geometric way of looking at it: $\varphi(z) = \frac{i-z}{-i-z}$, i.e. the ratio of the two “vectors” from $z$ to $\pm i$. This ratio is the complex number $s$ that makes a similar triangle with $1$ in the complex plane, upon identifying $i$ with $s$, $-i$ with $1$, and $z$ with the origin. We know that for any $s$ in the left open half plane, the angle $s$ makes with $1$ through the origin is between $\pi /2$ and $3\pi /2$. As such, since the triangles are similar, the angle through $0$ is the same as the angle through $z$ and so the point $z$ must lie within the unit circle. So, we’ve constructed a point in the unit disc that yields an arbitrary point in the left half plane.
If any step is unclear here let me know! :)
A: 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.
A: It is onto because you can write its inverse by inverting the matrix
$$\begin{pmatrix} 1 & -i \\ 1 & i\end{pmatrix}.$$
