Can you determine a Möbius transformation that maps unit circle $\{z: |z|=1\} \rightarrow$ real axis. I.e., how would you find one? Would this transformation be uniquely determined?

The Möbius transformations are the maps of the form: $$ f(z)= \frac{az+b}{cz+d}.$$

  • 2
    $\begingroup$ As a small additional note, it's easy to see that the transformation can't be uniquely determined; if $f(z)$ is a transformation mapping the unit circle to the line, then any transformation of the form $f(wz), |w|=1$ (that is, the composition of 'the' transformation with a rotation of the unit circle) also obviously maps the unit circle to the line, because it maps the unit circle to itself and then maps the circle to the line; similarly, any transformation of the form $xf(z),\mathcal{I}(x)=0$ also maps the unit circle to the line, because it maps the circle to the line and the line to itself. $\endgroup$ – Steven Stadnicki Mar 19 '13 at 15:58


Try to find a Möbius transformation that sends

$1$ to $0$ , $i$ to $ 1$ and $-1$ to $\infty$

Möbius transformation for three points $z_1$, $z_2$, $z_3$ is given by the formula below

$$\frac{z-z_1}{z-z_3}\cdot \frac{z_2-z_3}{z_2-z_1}$$

in this case $z_1= 1$, $z_2 =i$, $z_3=-1$

  • $\begingroup$ Why do those 3 distinct points determine the transformation? thanks $\endgroup$ – sarah Mar 19 '13 at 14:42
  • $\begingroup$ since three points determine a unique circle you can find the prove here math.stackexchange.com/questions/16634/… $\endgroup$ – jim Mar 19 '13 at 14:45
  • $\begingroup$ Also where would -i be mapped to? $\endgroup$ – sarah Mar 19 '13 at 14:45
  • $\begingroup$ How would one know those are the images of the 3 points? why are the images 0,1, infinity? thanks alot $\endgroup$ – sarah Mar 19 '13 at 14:50
  • $\begingroup$ i have edited the post $\endgroup$ – jim Mar 19 '13 at 14:55

Try sending $1$ to $0$ and $-1$ to $\infty$: $$ \frac{z-1}{z+1}\tag{1} $$ Compute the real part of $(1)$: $$ \begin{align} \frac12\left(\frac{z-1}{z+1}+\frac{\bar{z}-1}{\bar{z}+1}\right) &=\frac12\frac{(z\bar{z}+(z-\bar{z})-1)+(z\bar{z}-(z-\bar{z})-1)}{|z+1|^2}\\ &=\frac{|z|^2-1}{|z+1|^2}\tag{2} \end{align} $$ Thus, if $|z|=1$, $(2)$ says that $\mathrm{Re}\left(\frac{z-1}{z+1}\right)=0$; that is, $(1)$ sends the unit circle to the imaginary axis. Therefore, $$ i\frac{z-1}{z+1}\tag{3} $$ sends the unit circle to the real axis.


Two important basic facts about Möbius functions:

  • Pick three distinct points ($\infty$ is allowed). Any Möbius function is completely determined by its action on those three points. Conversely, if you pick any action on those three points (that would send them to distinct images), it can be extended to a Möbius function.
  • Möbius functions map lines and circles to lines and circles.

A useful geometric fact is:

  • any three distinct points determine a unique line or circle.

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