Is there a Möbius transformation that scales disks to the unit disk? When working in the complex plane, often times I would like to scale a disk $|z-z_0|<R$ to the unit disk. I would first translate $z_0$ to the origin, but after that, what can we multiply by to scale the radius down from $R$ to $1$?
I'm curious because in reading a proof of Schwarz' lemma, one can map a disk $|z|<r$ to the unit disk with some point $z_0$ mapping to $0$. The transformation is given by
$$
\frac{r(z-z_0)}{r^2-\bar{z}_0z}
$$
but I don't understand how that formula comes up. How does one come up with it?
 A: To take a disk with center $z_0$ and radius $R$ to the unit disk, 
$$   \frac{z - z_0}{R}       $$ 
Meanwhile, about the Schwarz item, the basic fact is that a Möbius transformation takes lines or circles to lines or circles. Also, the transformation is defined once the (distinct) images of three points are specified). If $z_0$ is not real, the picture is not entirely transparent. 
Now, $$ | r + z_0 | = | r + \bar{z}_0 |    $$ as $r$ is real. Therefore
$$ \left| \frac{r + z_0}{ r + \bar{z}_0}   \right| = 1,  $$ and 
 $$ \left| \frac{r(r + z_0)}{ r + \bar{z}_0}   \right| = r.  $$
Writing the transformation as
$$  T(z) =  \frac{r(z - z_0)}{ r^2 - \bar{z}_0 z},     $$
we can check
$$  T \left(  \frac{r(r + z_0)}{ r + \bar{z}_0}   \right) = 1.  $$
Very similar, we have
$$  T \left(  \frac{-r(r - z_0)}{ r - \bar{z}_0}   \right) = -1.  $$
A third point suffices. Note
$$  \overline{r i + z_0} = - r i + \bar{z}_0,  $$ while
$$ i \; \; \overline{r i + z_0} =  r  + i \bar{z}_0,  $$ so
 $$ \left| \frac{r(r i + z_0)}{ r + i \bar{z}_0}   \right| = r,  $$ while
$$  T \left(  \frac{r(ri + z_0)}{ r + i \bar{z}_0}   \right) = i.  $$
