Linear Fractional Transform
2020-2학기, 대학에서 ‘응용복소함수론’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)
Linear Fractional Transformation
\[w = \frac{az + b}{cz + d}\]Linear Fractional Trans.의 각 케이스를 살펴보자!
(1) $ad - bc = 0$
\[\begin{aligned} ad - bc = 0 \iff \frac{a}{b} = \frac{c}{d} \quad (b, \; d \ne 0) \end{aligned}\]만약 위와 같이 $ad - bc = 0$이라는 조건에서는 $w = C$; const transformation이 된다!
\[w = \frac{\frac{a}{b} z + 1}{\frac{c}{d} z + 1} = 1 \quad (ad - bc = 0)\]이런 const transformation은 우리의 관심이 아니다!
Derivative of Linear Fractional Transform
\[\begin{aligned} \frac{dw}{dz} &= \frac{a \cdot (cz + d) - c \cdot (az + b)}{(cz + d)^2} \\ &= \frac{ad - bc}{(cz + d)^2} \end{aligned}\]우리가 $ad - bc \ne 0$이라고 가정한다면, Linear Fractional Transform은 미분값이 $0$이 되지 않는다!!
some special cases
(i) $c=0$, $d=1$
$w = \dfrac{az + b}{0z + 1} = az + b$; Linear Transformation
(ii) $a=d=0$, $b=c=1$
$w = \dfrac{0z + 1}{1z + 0} = \dfrac{1}{z}$; Inversion Transformation
Inverse of Linear Frational Transformation
\[\begin{aligned} w &= \frac{az + b}{cz + d} \\ w \cdot (cz + d) &= az + b \\ (c \cdot w - a) z &= - d \cdot w + b \\ z &= \frac{-d \cdot w + b}{c \cdot w - a} \end{aligned}\]즉, Inverse Transform 역시 Linear Fractional Trnasformation이다!
Linear Fractional Transform + Extended Complex Plane
$w = T(z) = \frac{az + b}{cz + d}$는 $z = -\frac{d}{c}$에서 정의되지 않는다. Linear Fractional Transform에서도 Extended Complex Plane을 도입하여 $T\left(-\frac{d}{c}\right) = \infty$로 정의한다.
- $T\left(-\frac{d}{c}\right) = \infty$
- $T\left(\infty\right) = \underset{z \rightarrow \infty}{\lim} \frac{az+b}{cz+d} = \frac{a}{c}$
- 만약 $c=0$이라면, $T\left(\infty\right) = \infty$
Properties of L.F. Transform
Statement.
proof.
즉,
\[z \longrightarrow cz + d \longrightarrow \frac{1}{cz+d} \longrightarrow k \cdot \frac{1}{cz+d} + \frac{a}{c}\]따라서 L.F. Trnasform은 Linear Trnasform과 Inversion Transform의 합성이다. $\blacksquare$
Exercise.
Definition.; Fixed points
Let $T(z) = \frac{az+b}{cz+d}$, THEN
\[\begin{aligned} &T(z_0) = \frac{az_0+b}{cz_0+d} = z_0 \\ &\implies c {z_0}^2 + d z_0 = a z_0 + b \\ &\implies c {z_0}^2 + (d-a) z_0 - b = 0 \end{aligned}\]$c {z_0}^2 + (d-a) z_0 - b = 0$는 2차 방정식으로 해가 많아야 2개다.
(i) $c\ne0$ :
at most two fixed points
(ii) $c = 0$:
- $d-a \ne 0$ : $z_0 = -\frac{b}{d-a}$
- $d-a = 0$, $b\ne0$ : no fixed point
- $d-a = 0$, $b=0$ : $T(z) = z$; $\mathbb{C}$
Lemma.
Supp. that $T_1(z_1) = T_2(z_1)$, $T_1(z_2) = T_2(z_2)$, and $T_1(z_3) = T_2(z_3)$
THEN, $$ T_1(z) = T_2(z) $$
즉, 두 Transform이 세 점에서 그 값이 같다면, 두 Transform은 동일하다는 말이다.
proof.
$T^{-1}_1 \circ T_2$ is also a L.F. transform
THEN,
\[\begin{aligned} T^{-1}_1 \circ T_2(z_1) = T^{-1}_1 (z_1) = z_1 \\ T^{-1}_1 \circ T_2(z_2) = z_2 \\ T^{-1}_1 \circ T_2(z_3) = z_3 \\ \end{aligned}\]즉, $T^{-1}_1 \circ T_2$는 fixed point를 3개 가지므로, $T^{-1}_1 \circ T_2(z) = z$이다.
따라서 $T^{-1}_1 (z) = T_2(z)$이다. $\blacksquare$
Theorem.
THEN the mapping $Tz$ is given by $$ \frac{w-w_1}{w-w_3} \cdot \frac{w_2-w_3}{w_2-w_1} = \frac{z-w_1}{z-w_3} \cdot \frac{z_2-w_3}{z_2-w_1} $$ (IF one of these point is the point $\infty$, THEN 그 부분을 약분하여 1로 만들어라.) $$ w_3 = \infty \implies \frac{w-w_1}{\cancel{w-w_3}} \cdot \frac{\cancel{w_2-w_3}}{w_2-w_1} = \frac{w-w_1}{w_2-w_1} $$
proof.
Let’s define two L.F transformation $F(w)$, $G(z)$
THEN,
\[F(w_1) = 0, \quad F(w_2) = 1, \quad F(w_3) = \infty\] \[G(z_1) = 0, \quad G(z_2) = 1, \quad G(z_3) = \infty\]Because $F(w)$, $G(z)$ are L.F. transform, $F^{-1} \circ G$ is also a L.F. transformation.
THEN,
\[\begin{aligned} F^{-1} \circ G(z_1) = F^{-1} (0) = w_1 \\ F^{-1} \circ G(z_2) = F^{-1} (1) = w_2 \\ F^{-1} \circ G(z_3) = F^{-1} (\infty) = w_3 \\ \end{aligned}\]따라서 $F^{-1} \circ G$가 바로 우리가 찾는 L.F. transformation이다!
\[\begin{aligned} F^{-1} \circ G (z) &= w \\ G(z) &= F(w) \\ \frac{z-w_1}{z-w_3} \cdot \frac{z_2-w_3}{z_2-w_1} &= \frac{w-w_1}{w-w_3} \cdot \frac{w_2-w_3}{w_2-w_1} \end{aligned}\]