式展開

cos(z)=2

$$\begin{eqnarray} \cos\left(z\right)&=&2\;\cdots\;z\in\mathbb{C} \end{eqnarray}$$

$$\begin{eqnarray} \cos\left(z\right)&=&\frac{ e^{iz}+e^{-iz} }{2} \\&=&\frac{ X+X^{-1} }{2}\;\cdots\;X=e^{iz} \\\frac{ X+X^{-1} }{2}&=&2 \\X+X^{-1}&=&2\cdot2 \\X+X^{-1}&=&4 \\X-4+X^{-1}&=&0 \\X\cdot\left(X-4-X^{-1}\right)&=&X\cdot0 \\X^2-4X+1&=&0 \\X&=&\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot 1}}{2\cdot1} \\&=&\frac{4\pm\sqrt{12}}{2} \\&=&\frac{4\pm2\sqrt{3}}{2} \\&=&2 \pm \sqrt{3} \\X=e^{iz}&=&2 \pm \sqrt{3} \\iz&=&\log{\left(2\pm \sqrt{3}\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(2\pm \sqrt{3}\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(2\pm \sqrt{3}\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(2\pm \sqrt{3}\right)}+2\pi n \\&=&-i\log{\left(2\pm \sqrt{3}\right)}+2\pi n \end{eqnarray}$$

$$\begin{eqnarray} \frac{ e^{iz}+e^{-iz} }{2} &=& \frac{ e^{iz}+e^{i(-z)} }{2} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) +\left(\cos{\left(-z\right)}+i\sin{\left(-z\right)}\right) }{2} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) +\left(\cos{\left(z\right)}-i\sin{\left(z\right)}\right) }{2} \\&=& \frac{ \cos{\left(z\right)}+\cancel{i\sin{\left(z\right)}} +\cos{\left(z\right)}\cancel{-i\sin{\left(z\right)}} }{2} \\&=& \frac{ \cancel{2}\cos{\left(z\right)} }{\cancel{2}} \\&=&\cos{\left(z\right)} \end{eqnarray}$$

sin(z)=2

$$\begin{eqnarray} \sin\left(z\right)&=&2\;\cdots\;z\in\mathbb{C} \end{eqnarray}$$

$$\begin{eqnarray} \sin\left(z\right)&=&\frac{ e^{iz}-e^{-iz} }{2i} \\&=&\frac{ X-X^{-1} }{2i}\;\cdots\;X=e^{iz} \\\frac{ X-X^{-1} }{2i}&=&2 \\X-X^{-1}&=&2\cdot2i \\X-X^{-1}&=&4i \\X-4i-X^{-1}&=&0 \\X\cdot\left(X-4i-X^{-1}\right)&=&X\cdot0 \\X^2-4iX-1&=&0 \\X&=&\frac{-(-4i)\pm\sqrt{(-4i)^2-4\cdot1\cdot(-1)}}{2\cdot1} \\&=&\frac{4i\pm\sqrt{-12}}{2} \\&=&\frac{4i\pm2\sqrt{3}i}{2} \\&=&2i \pm \sqrt{3}i \\X=e^{iz}&=&2i \pm \sqrt{3}i \\&=&\left(2\pm \sqrt{3}\right)i \\iz&=&\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \\&=&-i\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \end{eqnarray}$$

$$\begin{eqnarray} \frac{ e^{iz}-e^{-iz} }{2i} &=& \frac{ e^{iz}-e^{i(-z)} }{2i} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) -\left(\cos{\left(-z\right)}+i\sin{\left(-z\right)}\right) }{2i} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) -\left(\cos{\left(z\right)}-i\sin{\left(z\right)}\right) }{2i} \\&=& \frac{ \cancel{\cos{\left(z\right)}}+i\sin{\left(z\right)} \cancel{-\cos{\left(z\right)}}+i\sin{\left(z\right)} }{2i} \\&=& \frac{ \cancel{2i}\sin{\left(z\right)} }{\cancel{2i}} \\&=&\sin{\left(z\right)} \end{eqnarray}$$

メビウス変換のその逆変換

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$\beta$ $$ \begin{eqnarray} \beta&=&\frac{-\delta b}{a}=\frac{- b}{a}\delta \\&=&\frac{- b}{\cancel{a}}\frac{\cancel{a}}{ad-bc} \\&=&\frac{- b}{ad-bc} \end{eqnarray}$$

$$ \begin{eqnarray} \phi^{-1}\left(z\right)&=&\frac{z \alpha + \beta}{ z \gamma + \delta} \\&=&\frac{z \frac{d}{\cancel{ad-bc}} + \frac{- b}{\cancel{ad-bc}}}{ z \frac{-c}{\cancel{ad-bc}} + \frac{a}{\cancel{ad-bc}}} \\&=&\frac{z\,d+(-b)}{z\,(-c)+a} \end{eqnarray}$$

確認

$$ \begin{eqnarray} \phi^{-1}\left(\phi\left(z\right)\right)&=&\phi^{-1}\left(\frac{z\,a+b}{z\,c+d}\right) \\&=&\frac{\frac{z\,a+b}{z\,c+d}\,d+(-b)} {\frac{z\,a+b}{z\,c+d}\,(-c)+a} \\&=&\frac{\left(z\,a+b\right)\,d+(-b)\left(z\,c+d\right)} {\left(z\,a+b\right)\,(-c)+a\left(z\,c+d\right)} \\&=&\frac{z\,ad\cancel{+bd}-z\,bc\cancel{-bd}} {\cancel{-z\,ac}-bc\cancel{+z\,ac}+ad} \\&=&\frac{z\cancel{\left(ad-bc\right)}}{\cancel{ad-bc}} \\&=&z \end{eqnarray}$$ $$ \begin{eqnarray} \phi\left(\phi^{-1}\left(z\right)\right)&=&\phi\left(\frac{z\,d+(-b)}{z\,(-c)+a}\right) \\&=&\frac{\frac{z\,d+(-b)}{z\,(-c)+a} a + b}{ \frac{z\,d+(-b)}{z\,(-c)+a} c + d} \\&=&\frac{\left(z\,d+(-b)\right) a + b\left(z\,(-c)+a\right)} { \left(z\,d+(-b)\right) c+d\left(z\,(-c)+a\right)} \\&=&\frac{z\,ad\cancel{-ab} -z\,bc \cancel{+ab}} {\cancel{z\,cd}-bc\cancel{-z\,cd}+ad} \\&=&\frac{z\,\cancel{\left(ad-bc\right)}} {\cancel{ad-bc}} \\&=&z \end{eqnarray}$$