メビウス変換とその逆変換
$$\begin{eqnarray} \phi\left(z\right)&=&\frac{z a + b}{ z c + d} \\\phi^{-1}\left(z\right)&=&\frac{z \alpha + \beta}{ z \gamma + \delta} \\\\\phi\left(\phi^{-1}\left(z\right)\right) &=&\frac{\frac{z \alpha + \beta}{ z \gamma + \delta} a + b}{ \frac{z \alpha + \beta}{ z \gamma + \delta} c + d} \\&=&\frac{z \alpha a + \beta a + b\left( z \gamma + \delta\right)}{z \alpha c+ \beta c + d\left( z \gamma + \delta\right)} \\&=&\frac{z \alpha a + \beta a + z \gamma b+ \delta b}{z \alpha c+ \beta c + z \gamma d+ \delta d} \\&=&\frac{z \left( \alpha a +\gamma b\right)+ \beta a+ \delta b}{z \left(\alpha c+ \gamma d\right)+\beta c + \delta d} \\&=&z \end{eqnarray}$$ $$\begin{cases} \begin{eqnarray} \alpha a +\gamma b&=&1 \\\beta a + \delta b&=&0 \\\alpha c + \gamma d &=&0 \\\beta c + \delta d&=&1 \end{eqnarray} \end{cases}$$ $$\begin{cases} \begin{eqnarray} \alpha&=&\frac{1-\gamma b}{a} \\\beta&=&\frac{-\delta b}{a} \\\gamma&=&\frac{-\alpha c}{d} \\\delta&=&\frac{1-\delta d}{c} \end{eqnarray} \end{cases}$$
\(\alphaについて\)
$$ \begin{eqnarray}
\alpha&=&\frac{1-\gamma b}{a}
\\&=&\frac{1-\frac{-\alpha c}{d} b}{a}
\\&=&\frac{1}{a}-\frac{\frac{-\alpha c}{d} b}{a}
\\&=&\frac{1}{a}+\frac{bc}{ad}\alpha
\\\alpha -\frac{bc}{ad}\alpha&=&\frac{1}{a}
\\\left(1 -\frac{bc}{ad}\right)\alpha&=&\frac{1}{a}
\\ad\cdot\left(1 -\frac{bc}{ad}\right)\alpha&=&ad\cdot\frac{1}{a}
\\\left(ad -bc\right)\alpha&=&\frac{\cancel{a}d}{\cancel{a}}=d
\\\alpha&=&\frac{d}{ad-bc}
\end{eqnarray}$$
同様に\(\delta\)
$$ \begin{eqnarray}
\delta&=&\frac{1-\delta d}{c}&=&\frac{a}{ad-bc}
\end{eqnarray}$$
\(\gamma\)について
$$ \begin{eqnarray}
\gamma&=&\frac{-\alpha c}{d}=\frac{- c}{d}\alpha
\\&=&\frac{- c}{\cancel{d}}\frac{\cancel{d}}{ad-bc}
\\&=&\frac{- c}{ad-bc}
\end{eqnarray}$$
同様に\(\beta\)
$$ \begin{eqnarray}
\beta&=&\frac{-\delta b}{a}=\frac{- b}{a}\delta
\\&=&\frac{- b}{\cancel{a}}\frac{\cancel{a}}{ad-bc}
\\&=&\frac{- b}{ad-bc}
\end{eqnarray}$$
$$ \begin{eqnarray}
\phi^{-1}\left(z\right)&=&\frac{z \alpha + \beta}{ z \gamma + \delta}
\\&=&\frac{z \frac{d}{\cancel{ad-bc}} + \frac{- b}{\cancel{ad-bc}}}{ z \frac{-c}{\cancel{ad-bc}} +
\frac{a}{\cancel{ad-bc}}}
\\&=&\frac{z\,d+(-b)}{z\,(-c)+a}
\end{eqnarray}$$
