# 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?

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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 $$T \left( \frac{r(ri + z_0)}{ r + i \bar{z}_0} \right) = i.$$

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I'm sorry, how does this justify the formula in my question? –  fao Mar 13 '12 at 20:11
So since $T$ maps circles to circle, we see it must map through $1,-1$ and $i$, and thus must map to the unit circle. Thanks. –  fao Mar 13 '12 at 20:55