複素三角関数の公式

複素三角関数は次の式で表されます。$$\sin z=\frac{e^{iz}-e^{-iz}}{2i},  \cos z=\frac{e^{iz}+e^{-iz}}{2},  \tan z=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$
複素三角関数について次式が成り立ちます。
\begin{alignat}{2}
&(1) \sin (-z)=-\sin z\\
&(2) \cos(-z)=\cos z\\
&(3) \tan(-z)=-\tan z\\
&(4) \sin (z+2nπ)=\sin z\\
&(5) \cos (z+2nπ)=\cos z\\
&(6) \tan (z+nπ)=\tan z\\
&(7) \frac{\sin z}{\cos z}=\tan x\\
&(8) \sin^2 z+\cos^2 z=1\\
&(9) 1+\tan^2 z=\frac{1}{\cos^2 z}\\
&(10) \sin (z_1 \pm z_2)=\sin z_1 \cos z_2 \pm \cos z_1 \sin z_2\\
&(11) \cos (z_1 \pm z_2)=\cos z_1 \cos z_2 \mp \sin z_1 \sin z_2\\
&(12) \tan (z_1 \pm z_2)=\frac{\tan z_1 \pm \tan z_2}{1 \pm \tan z_1 \tan z_2}\\
&(13) \sin z = \sin (x+iy)=\sin x \cosh y+i \cos x \sinh y\\
&(14) \cos z= \cos (x+iy)=\cos x \cosh y-i \sin x \sinh y
\end{alignat}







<証明>
\begin{alignat}{2}
&(1) \sin (-z)=\frac{e^{-iz}-e^{iz}}{2i}=-\frac{e^{iz}-e^{-iz}}{2i}=\sin z\\
&(2) \cos (-z)=\frac{e^{-iz}+e^{iz}}{2}=\frac{e^{iz}+e^{-iz}}{2}=\cos z\\
&(3) \tan (-z)=\frac{e^{-iz}-e^{iz}}{i(e^{-iz}+e^{iz})}=-\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=-\tan z\\
&\\
&\\
&(4) \sin (z+2nπ)=\frac{e^{i(z+2nπ)}-e^{-i(z+2nπ)}}{2i}=\frac{e^{iz} \cdot e^{2nπi}-e^{-iz} \cdot e^{-2nπi}}{2i}=\frac{e^{iz}-e^{-iz}}{2i}=\sin z\\
&(5) \cos (z+2nπ)=\frac{e^{i(z+2nπ)}+e^{-i(z+2nπ)}}{2}=\frac{e^{iz} \cdot e^{2nπi}+e^{-iz} \cdot e^{-2nπi}}{2}=\frac{e^{iz}+e^{-iz}}{2}=\cos z\\
&(6) \tan (z+2nπ)=\frac{e^{i(z+nπ)}-e^{-i(z+nπ)}}{i(e^{i(z+nπ)}+e^{-i(z+nπ)})}=\frac{e^{iz} \cdot e^{nπi}-e^{-iz} \cdot e^{-nπi}}{i(e^{iz} \cdot e^{nπi}+e^{-iz} \cdot e^{-nπi})}\\
&              =\frac{e^{iz} \cdot e^{2nπi}-e^{-iz}}{i(e^{iz} \cdot e^{2nπi}+e^{-iz})}=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=\tan z\\
&\\
&\\
&(7) \frac{\sin z}{\cos z}=\frac{e^{iz}-e^{-iz}}{2i} \cdot \frac{2}{e^{iz}+e^{-iz}}=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=\tan z\\
&(8) \sin^2 z+ \cos^2 z=\left(\frac{e^{iz}-e^{-iz}}{2i}\right)^2+\left(\frac{e^{iz}+e^{-iz}}{2}\right)^2\\
&               =-\frac{e^{2iz}-2+e^{-2iz}}{4}+\frac{e^{2iz}+2+e^{-2iz}}{4}=1\\
&(9) 1+\tan^2 z=1+\left\{\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}\right\}=\frac{(e^{iz}+e^{-iz})^2-(e^{iz}-e^{-iz})^2}{(e^{iz}+e^{-iz})^2}\\
&            =\frac{e^{2iz}+2+e^{-2iz}-(e^{2iz}-2+e^{-2iz})}{(e^{iz}+e^{-iz})^2}=\frac{4}{(e^{iz}+e^{-iz})^2}\\
&            =\left(\frac{2}{e^{iz}+e^{-iz}}\right)^2=\frac{1}{\cos^2 z}
&\\
&\\
&\\
&(10) \sin z_1 \cos z_2+ \cos z_1 \sin z_2\\
&=\frac{e^{iz_1}-e^{-iz_1}}{2i} \cdot \frac{e^{iz_2}+e^{-iz_2}}{2}+\frac{e^{iz_1}+e^{-iz_1}}{2} \cdot\frac{e^{iz_2}-e^{-iz_2}}{2i}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}+e^{i(z_1+z_2)}-e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}}{4i}\\
&=\frac{2e^{i(z_1+z_2)}-2e^{-i(z_1+z_2)}}{4i}=\frac{e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}}{2i}=\sin (z_1+z_2)\\
&\\
&\\
&\sin z_1 \cos z_2- \cos z_1 \sin z_2\\
&=\frac{e^{iz_1}-e^{-iz_1}}{2i} \cdot \frac{e^{iz_2}+e^{-iz_2}}{2}-\frac{e^{iz_1}+e^{-iz_1}}{2} \cdot\frac{e^{iz_2}-e^{-iz_2}}{2i}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}-e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}}{4i}\\
&=\frac{2e^{i(z_1-z_2)}-2e^{-i(z_1-z_2)}}{4i}=\frac{e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}}{2i}=\sin (z_1-z_2)\\
&\\
&\\
&(11) \cos z_1 \cos z_2- \sin z_1 \sin z_2\\
&=\frac{e^{iz_1}+e^{-iz_1}}{2} \cdot \frac{e^{iz_2}+e^{-iz_2}}{2}-\frac{e^{iz_1}-e^{-iz_1}}{2i} \cdot\frac{e^{iz_2}-e^{-iz_2}}{2i}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}+e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}}{4}\\
&=\frac{2e^{i(z_1+z_2)}+2e^{-i(z_1+z_2)}}{4}=\frac{e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}}{2}=\cos (z_1+z_2)\\
&\\
&\\
&\cos z_1 \cos z_2+ \sin z_1 \sin z_2\\
&=\frac{e^{iz_1}+e^{-iz_1}}{2} \cdot \frac{e^{iz_2}+e^{-iz_2}}{2}+\frac{e^{iz_1}-e^{-iz_1}}{2i} \cdot\frac{e^{iz_2}-e^{-iz_2}}{2i}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}-\{e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}\}}{4}\\
&=\frac{2e^{i(z_1-z_2)}+2e^{-i(z_1-z_2)}}{4}=\frac{e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}}{2}=\cos (z_1-z_2)\\
&\\
&\\
&(12) \frac{\tan z_1 + \tan z_2}{1-\tan z_1 \tan z_2}=\frac{\frac{e^{iz_1}-e^{-iz_1}}{i(e^{iz_1}+e^{-iz_1})}+\frac{e^{iz_2}-e^{-iz_2}}{i(e^{iz_2}+e^{-iz_2})}}{1-\frac{e^{iz_1}-e^{-iz_1}}{i(e^{iz_1}+e^{-iz_1})} \cdot \frac{e^{iz_2}-e^{-iz_2}}{i(e^{iz_2}+e^{-iz_2})}}\\
&=\frac{(e^{iz_1}-e^{-iz_1})(e^{iz_2}+e^{-iz_2})+(e^{iz_1}+e^{-iz_1})(e^{iz_2}-e^{-iz_2})}{i(e^{iz_1}+e^{-iz_1})(e^{iz_2}+e^{-iz_2})+i(e^{iz_1}-e^{-iz_1})(e^{iz_2}-e^{-iz_2})}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}+e^{i(z_1+z_2)}-e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}}{i\{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}+e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}\}}\\
&=\frac{2e^{i(z_1+z_2)}-2e^{-i(z_1+z_2)}}{i\{2e^{i(z_1+z_2)}+2e^{-i(z_1+z_2)}\}}\\
&=\frac{e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}}{i\{e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}\}}=\tan (z_1+z_2)\\
&\\
&\\
&\frac{\tan z_1 – \tan z_2}{1+\tan z_1 \tan z_2}=\frac{\frac{e^{iz_1}-e^{-iz_1}}{i(e^{iz_1}+e^{-iz_1})}-\frac{e^{iz_2}-e^{-iz_2}}{i(e^{iz_2}+e^{-iz_2})}}{1+\frac{e^{iz_1}-e^{-iz_1}}{i(e^{iz_1}+e^{-iz_1})} \cdot \frac{e^{iz_2}-e^{-iz_2}}{i(e^{iz_2}+e^{-iz_2})}}\\
&=\frac{(e^{iz_1}-e^{-iz_1})(e^{iz_2}+e^{-iz_2})-(e^{iz_1}+e^{-iz_1})(e^{iz_2}-e^{-iz_2})}{i(e^{iz_1}+e^{-iz_1})(e^{iz_2}+e^{-iz_2})-i(e^{iz_1}-e^{-iz_1})(e^{iz_2}-e^{-iz_2})}\\
&=\frac{e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}-\{e^{i(z_1+z_2)}-e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}\}}{i[e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}-\{e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}\}]}\\
&=\frac{2e^{i(z_1-z_2)}-2e^{-i(z_1-z_2)}}{i\{2e^{i(z_1-z_2)}+2e^{-i(z_1-z_2)}\}}\\
&=\frac{e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}}{i\{e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}\}}=\tan (z_1-z_2)\\
&\\
&\\
&(13) \sin z= \sin (x+iy)=\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}=\frac{e^{ix} \cdot e^{-y}-e^{-ix} \cdot e^y}{2i}\\
&        =\frac{e^{-y}}{2i}(\cos x+i \sin x)-\frac{e^y}{2i}(\cos x- i \sin x)\\
&        =\frac{e^{-y} \cos x}{2i}+\frac{e^{-y} \sin x}{2}-\frac{e^y \cos x}{2i}+\frac{e^y \sin x}{2}\\
&        =\frac{e^{-y}-e^y}{2i} \cos x+\frac{e^y+e^{-y}}{2} \sin x\\
&        =\frac{e^y+e^{-y}}{2} \sin x+i \cdot \frac{e^y-e^{-y}}{2}\cos x\\
&        =\sin x \cosh y +i \cos x\sinh y
&\\
&\\
&\\
&(14) \cos z= \cos (x+iy)=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2}=\frac{e^{ix} \cdot e^{-y}+e^{-ix} \cdot e^y}{2}\\
&        =\frac{e^{-y}}{2}(\cos x+i \sin x)+\frac{e^y}{2}(\cos x- i \sin x)\\
&        =\frac{e^{-y} \cos x}{2}+\frac{ie^{-y} \sin x}{2}+\frac{e^y \cos x}{2}-\frac{ie^y \sin x}{2}\\
&        =\frac{e^y+e^{-y}}{2} \cos x-i \cdot \frac{e^y-e^{-y}}{2} \sin x\\
&        =\cos x \cosh y -i \sin x\sinh y
\end{alignat}

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