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. $