Proposition

Find each Mobius transformation $f$:

$f$ maps $0 \rightarrow 1, 1 \rightarrow \infty, \infty \rightarrow 0$.
$f$ maps $1 \rightarrow 1, -1 \rightarrow i, -i \rightarrow -1$.
$f$ maps the $x$-axis to $y = x$, the $y$-axis to $y = -x$, and the unit circle to itself.

Solution

We will use the cross-ratio formula, which is

\[\begin{align*} \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} \end{align*}\]

sends $(z_1, z_2, z_3)$ to $(0, 1, \infty)$.

1

By setting $z_1 = \infty, z_2 = 0, z_3 = 1$,

\[\begin{align*} \frac{(z - \infty)(0 - 1)}{(z - 1)(0 - \infty)} &= \frac{(z / \infty - 1)(0 - 1)}{(z - 1)(0 - 1)} \\ &= \frac{(0 - 1)(0 - 1)}{(z - 1)(0 - 1)} \\ &= \frac{1}{1 - z}. \end{align*}\]

Dividing $\infty$ by $\infty$ is not exactly rigorous, but it is easy to show that this map satisfies the requirements.

2

We will construct two Mobius transformation:

A Mobius transformation $f_1$ that maps $(1, -1, -i)$ to $(0, 1, \infty)$,
A Mobius transformation $f_2$ that maps $(1, i, -1)$ to $(0, 1, \infty)$,

By using the cross ratio formula,

\[\begin{align*} f_1(z) &= \frac{(i - 1)z + (1 - i)}{-2z - 2i}, \\ f_2(z) &= \frac{-iz + i}{z + 1} \\ \end{align*}\]

$f_2^{-1} \cdot f_1$ is the desired Mobius transformation.

By using the connection between matrices and Mobius transformations,

\[\begin{align*} \begin{bmatrix} -i & i \\ 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix} (i - 1) & (1 - i) \\ -2 & -2i \end{bmatrix} &= \begin{bmatrix} 1 & -i \\ -1 & -i \end{bmatrix} \begin{bmatrix} (i - 1) & (1 - i) \\ -2 & -2i \end{bmatrix} \\ &= \begin{bmatrix} 3i - 1 & -1 - i \\ i + 1 & i - 3 \end{bmatrix}. \end{align*}\]

Thus the desired Mobius transformation is

\[\begin{align*} \frac{(3i - 1)z - (1 + i)}{(i + 1)z + (i - 3)}. \end{align*}\]

3

$f$ simply rotates the complex plane by 45 degrees. Thus $f(z) = e^{\pi i / 2}z = z(1 + i) / \sqrt{2}$.