When working in the complex plane, often times I would like to scale a disk $|z-z_0|
I'm curious because in reading a proof of Schwarz' lemma, one can map a disk $|z|
When working in the complex plane, often times I would like to scale a disk $|z-z_0|
I'm curious because in reading a proof of Schwarz' lemma, one can map a disk $|z|
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 $$ T \left( \frac{r(ri + z_0)}{ r + i \bar{z}_0} \right) = i. $$