What Möbius transformation maps the unit circle $\{z: |z|=1\}$ to the real axis?

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.

hint:

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 $z_1= 1$, $z_2 =i$, $z_3=-1$