Mobius transformation

 

Def 1

Let $\alpha \in \overline{\mathbb{C}}$ such that $\left|\alpha \right| < 1$.

$\phi_{\alpha}(z) = \dfrac{z - \alpha}{1 - \overline{\alpha} z}$

is Mobius transformation.

 

 

 

Thm 1

Let $\phi_{\alpha}(z)$ be Mobius transformation and $A = \left\{z \in \overline{\mathbb{C}} \; | \; \left|z \right| \leq 1 \right\}$. Then

$\phi_{\alpha}(A) = A$

 

  Let $B = \left\{z \in \overline{\mathbb{C}} \; | \; \left|z \right| = 1 \right\}$

  By simple calculation,

  $\phi_{\alpha} (\dfrac{\alpha}{\left|\alpha \right|}) = \dfrac{\alpha}{\left|\alpha \right|}$

  $\phi_{\alpha} (- \dfrac{\alpha}{\left|\alpha \right|}) = - \dfrac{\alpha}{\left|\alpha \right|} $

  $\left|\phi_{\alpha}(1) \right| = \left|\dfrac{1 - \alpha}{1 - \overline{\alpha}} \right| = 1$

 

  Since $\dfrac{\alpha}{\left|\alpha \right|}, - \dfrac{\alpha}{\left|\alpha \right|}, 1, \phi_{\alpha}(1) \in B$,

  $\phi_{\alpha}(B) = B$

  # Linear fractional function - Thm 1

 

  Also

  $\left|\phi_{\alpha}(0) \right| = \left|-\alpha \right| < 1$

 

  Since $0, \phi_{\alpha}(0) \in A \setminus B$,

  $\phi_{\alpha}(A) = A$

 

 

 

Thm 2

Let $\phi_{\alpha}(z)$ be Mobius transformation. Then

$\phi_{\alpha}(z) = z$     $\Rightarrow$     $z = \pm \dfrac{\alpha}{\left|\alpha \right|}$

 

  Since $\dfrac{z - \alpha}{1 - \overline{\alpha} z} = z$,

  $z - \alpha = z - \overline{\alpha} z^{2}$

  $\therefore$  $z^{2} = \dfrac{\alpha}{\overline{\alpha}} = \dfrac{\alpha^{2}}{\alpha \overline{\alpha}} = \dfrac{\alpha^{2}}{\left|\alpha \right|^{2}}$

  $\therefore$  $z = \pm \dfrac{\alpha}{\left|\alpha \right|}$

 

 

 

 

Thm 3

Let $f$ be First-order fractional function and $A = \left\{z \in \overline{\mathbb{C}} \; | \; \left|z \right| \leq 1 \right\}$

$f(A) = A$     $\Rightarrow$     $f = \gamma \phi_{\alpha}$  $(\left|\gamma \right| = 1)$

 

  WLOG, WMA $f(z) = \delta \dfrac{z - \alpha}{z - \beta}$  $(\delta, \alpha, \beta \in \mathbb{C})$

  Let $B = \left\{z \in \overline{\mathbb{C}} \; | \; \left|z \right| = 1 \right\}$

  Since $f(A) = A$ and boundary to boundary,

  $f(B) = B$

  $\therefore$  $\left| f(\dfrac{\alpha}{\left|\alpha \right|}) \right| = 1$

  So there exists $\gamma_{1} \in \mathbb{C}$  s.t.

$\left| \gamma_{1} \right| = 1 \quad \text{and} \quad \gamma_{1} \cdot f(\dfrac{\alpha}{\left|\alpha \right|}) = \dfrac{\alpha}{\left|\alpha \right|}$

 

  Let $g(z) = \gamma_{1} \cdot f(z) = \gamma_{1} \delta \dfrac{z - \alpha}{z - \beta}$

  ( i.e.  $g(\dfrac{\alpha}{\left|\alpha \right|} ) = \dfrac{\alpha}{\left|\alpha \right|} $)

 

 

  Claim 1 : $g(B) = B$

  Since $r_{1}$ is rotation factor and $f(B) = B$,

  $g(B) = B$

 

  end

 

  Let line $L = \overleftrightarrow{\alpha \beta} $

 

 

  Claim 2 : $g(L) = L$

  If $\left|z \right| = 1$,  $\left|g(z) \right|  = \left|\gamma_{1} \right| \left|f(z) \right| = 1$. Therefore

$$\left|\gamma_{1} \delta \right| \left|z - \alpha \right| = \left|z - \beta \right|$$

  This represents a circle with its center lying on the line connecting points $\alpha$ and $\beta$.

  This circle is $B$, so the origin lies on $L$.

 

  

  First-order fractional function $g$ maps $\dfrac{\alpha}{\left|\alpha \right|}, \alpha, \beta \in L$ as follows.

$$\left\{\dfrac{\alpha}{\left|\alpha \right|}, \alpha, \beta \right\} \mapsto \left\{\dfrac{\alpha}{\left|\alpha \right|}, 0, \infty \right\}$$

  Since $\dfrac{\alpha}{\left|\alpha \right|}, 0, \infty \in L$,

$g(L) = L$

 

  end

 

 

  Claim 3 : $g = \phi_{\alpha}$

  Since $-\dfrac{\alpha}{\left|\alpha \right|} \in B \cap L$,

  $g(- \dfrac{\alpha}{\left|\alpha \right|}) \in B \cap L$

  Since $g$ is 1-1 onto,

$g(-\dfrac{\alpha}{\left|\alpha \right|} ) = -\dfrac{\alpha}{\left|\alpha \right|} $

 

  So $g$ maps as follows.

$\left\{\dfrac{\alpha}{\left|\alpha \right|}, -\dfrac{\alpha}{\left|\alpha \right|}, \alpha \right\} \mapsto \left\{\dfrac{\alpha}{\left|\alpha \right|}, -\dfrac{\alpha}{\left|\alpha \right|}, 0 \right\}$

 

  When destinations of three points are determined, the first-order fractional function is

  determined accordingly, and $\phi_{\alpha}$ also maps as described above.

  $\therefore$  $g = \phi_{\alpha}$

 

  end

 

  $\therefore$  $f = \dfrac{1}{\gamma_{1}} \phi_{\alpha} = \gamma \phi_{\alpha}$  $(\left|\gamma \right| = 1)$

 

 

 

Thm 4

If line L satisfies $\phi_{\alpha}(L) = L$, then

$L$ passes through the origin and the point $\alpha$.

 

  Let line $L$ that satisfies $\phi_{\alpha}(L) = L$.

  Since $\infty \in L$,

$$\phi_{\alpha}(\infty) = - \frac{1}{\overline{\alpha}} = - \frac{\alpha}{\left|\alpha \right|^{2}} \in L$$

  Since $\phi_{\alpha}(\frac{1}{\overline{\alpha}}) = \infty \in L$ and $\phi_{\alpha}$ is 1-1 onto,

$$\frac{1}{\overline{\alpha}} = \frac{\alpha}{\left|\alpha \right|^{2}} \in L$$

 

  $\therefore$  $L$ passes through the origin and the point $\alpha$.