Form of most general transformation of the upper half plane to the unit disk.

Question

In David Blair's book on Inversion Theory, he write that the transformation $$ T(z)=e^{i\theta}\frac{z-z_0}{z-\bar{z}_0} $$ is the most general transformation mapping the upper half plane to the unit circle, provided $z_0$ is in the upper half plane. If $z_0$ is in the lower half plane, then the upper half plane is mapped to the exterior of the circle. Why does the location of $z_0$ affect the map this way?

Answer 1

The points in the unit circle satisfy $|z|<1$. Thus when $z_0$ is in the upper half plane, points in the upper half plane will be closer to $z_0$ than to $\bar{z_0}$, that is $|z-z_0|<|z-\bar{z_0}|$. This means that the fraction in $T(z)$ is less than $1$ precisely when $z$ is in the upper half plane. (The $e^{i\theta}$ factor has magnitude one and thus is only a rotation.)