On the Big Rudin there is the conformal map
$$\varphi(z) = \frac {1+z}{1-z}$$
which sends $\{-1, 0, 1\}$ to $\{0, 1, \infty\}$.
The book says:
The segment $(-1, 1)$ maps onto the positive real axis. The unit
circle $T$ passes through $-1$ and $1$, hence $\varphi(T)$ is a
straight line passing through $\varphi(-1) = 0$. Since $T$ makes a
right angle in $-1$ with the real axis, $\varphi(T)$ is the imaginary
axis.
Up until now it's clear.
But then he says:
Since $\varphi(0) = 1$, it follows that $\varphi$ is a conformal
one-to-one mapping of the open unit disc onto the open right half
plane.
I lost him. Why does all that implies that $\varphi$ maps onto the open right half plane?
Best Answer
Since $\varphi$ is a Möbius transformation it sends circles onto circles. The Orientation Principle (see Chapter III, 3.21 p.53 on John Conway: Functions of one complex variable I. Second edition) states that if $\varphi$ maps the circle $C_1$ onto the circle $C_2$ then everything on one side of $C_2$ must be send exclusively to only one side of $C_2$ (the sides depend on the orientation given).
In this case, we have that the unit circle $T$ is sent onto the imaginary axis (which is a circle on the Riemann Sphere), thus the imaginary axis is the boundary of $\varphi(T)$, that is everything on the inside of $T$ must be send only to the left side or only to the right side of the imaginary axis.
To check what side (that is to check what orientation is taken), we only need to check where $\varphi$ sends any point inside $T$. Then since a point inside the unit circle $T$, in this case $0$ is sent to a point on the right half plane $\varphi(0)=1$, we get that $\varphi(T)$ must be the right half plane.
As a corollary we get that $\varphi(\mathbb{C} \setminus T)$ is the left half plane, and you an check it for example by computing $\varphi(2)=-3$