1±itan(z)など
1±itan(z)など
(1)
\[ 1\pm i\tan z=\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{\pm2i\Re z}+e^{\mp2\Im z}\right) \](2)
\[ 1\pm i\tan^{-1}z=\frac{1}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(-e^{\mp2i\Re z}+e^{\pm2\Im z}\right) \](3)
\[ 1\pm\tanh z=\frac{1}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)}\left(e^{\pm2\Re z}+e^{\pm2i\Im z}\right) \](4)
\[ 1\pm\tanh^{-1}z=\frac{1}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)}\left(e^{\pm2\Re z}-e^{\pm2i\Im z}\right) \](1)
\begin{align*} 1\pm i\tan z & =1\pm i\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)\right)\\ & =\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)\mp\sinh\left(2\Im z\right)\pm i\sin\left(2\Re z\right)\right)\\ & =\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{\pm2i\Re z}+e^{\mp2\Im z}\right) \end{align*}(1)-2
\begin{align*} 1\pm i\tan z & =\frac{1}{\cos z}\left(\cos z\pm i\sin z\right)\\ & =\frac{e^{\pm iz}}{\cos z}\\ & =e^{\pm iz}\frac{2\cos\overline{z}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =e^{\pm iz}\frac{e^{i\overline{z}}+e^{-i\overline{z}}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{e^{i\left(\overline{z}\pm z\right)}+e^{-i\left(\overline{z}\mp z\right)}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{e^{i\left(2H\left(\pm1\right)\Re z-2iH\left(\mp1\right)\Im z\right)}+e^{-i\left(2H\left(\mp1\right)\Re z-2iH\left(\pm1\right)\Im z\right)}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{e^{2iH\left(\pm1\right)\Re z}e^{2H\left(\mp1\right)\Im z}+e^{-2iH\left(\mp1\right)\Re z}e^{-2H\left(\pm1\right)\Im z}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{e^{2i0\Re z}e^{2\left(\mp1\right)\Im z}+e^{-2i\left(\mp1\right)\Re z}e^{-2\cdot0\Im z}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{e^{\pm2i\Re z}+e^{\mp2\Im z}}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \end{align*}(2)
\begin{align*} 1\pm i\tan^{-1}z & =1\pm i\frac{\sin\left(2\Re z\right)-i\sinh\left(2\Im z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =1+\frac{\pm\sinh\left(2\Im z\right)\pm i\sin\left(2\Re z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{1}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left\{ -\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)\pm\sinh\left(2\Im z\right)\pm i\sin\left(2\Re z\right)\right\} \\ & =\frac{1}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(-e^{\mp2i\Re z}+e^{\pm2\Im z}\right) \end{align*}(2)-2
\begin{align*} 1\pm i\tan^{-1}z & =\frac{1}{\sin z}\left(\sin z\pm i\cos z\right)\\ & =\frac{\pm ie^{\mp iz}}{\sin z}\\ & =\pm ie^{\mp iz}\frac{2\sin\overline{z}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\pm e^{\mp iz}\frac{e^{i\overline{z}}-e^{-i\overline{z}}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\pm\frac{e^{i\left(\overline{z}\mp z\right)}-e^{-i\left(\overline{z}\pm z\right)}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\pm\frac{e^{i\left(2H\left(\mp1\right)\Re z-2iH\left(\pm1\right)\Im z\right)}-e^{-i\left(2H\left(\pm1\right)\Re z-2iH\left(\mp1\right)\Im z\right)}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\pm\frac{e^{2iH_{0}\left(\mp1\right)\Re z}e^{2H_{0}\left(\pm1\right)\Im z}-e^{-2iH_{0}\left(\pm1\right)\Re z}e^{-2H_{0}\left(\mp1\right)\Im z}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\pm\frac{\mp\left(e^{\mp2i\Re z}e^{2\cdot0\Im z}-e^{2i\cdot0\Re z}e^{\pm2\Im z}\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{-e^{\mp2i\Re z}+e^{\pm2\Im z}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =\frac{-e^{\mp2i\Re z}+e^{\pm2\Im z}}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \end{align*}(3)
\begin{align*} 1\pm\tanh z & =1\mp i\tan\left(iz\right)\\ & =\frac{1}{\cos\left(2\Re\left(iz\right)\right)+\cosh\left(2\Im\left(iz\right)\right)}\left(e^{\mp2i\Re\left(iz\right)}+e^{\pm2\Im\left(iz\right)}\right)\\ & =\frac{1}{\cos\left(-2\Im z\right)+\cosh\left(2\Re z\right)}\left(e^{\pm2i\Im z}+e^{\pm2\Re z}\right)\\ & =\frac{1}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)}\left(e^{\pm2\Re z}+e^{\pm2i\Im z}\right) \end{align*}(4)
\begin{align*} 1\pm\tanh^{-1}z & =1\pm i\tan^{-1}\left(iz\right)\\ & =\frac{1}{-\cos\left(2\Re\left(iz\right)\right)+\cosh\left(2\Im\left(iz\right)\right)}\left(-e^{\mp2i\Re\left(iz\right)}+e^{\pm2\Im\left(iz\right)}\right)\\ & =\frac{1}{-\cos\left(-2\Im z\right)+\cosh\left(2\Re z\right)}\left(-e^{\pm2i\Im z}+e^{\pm2\Re z}\right)\\ & =\frac{1}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)}\left(e^{\pm2\Re z}-e^{\pm2i\Im z}\right) \end{align*}ページ情報
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逆三角関数と逆双曲線関数の級数表示
\[
\sin^{\bullet}x=\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\qquad,(|x|\leq1)
\]
三角関数と双曲線関数の実部と虚部
\[
\tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}
\]
ピタゴラスの基本三角関数公式
\[
\cos^{2}x+\sin^{2}x=1
\]
逆三角関数の三角関数と逆双曲線関数の双曲線関数
\[
\sin\Cos^{\bullet}z=\sqrt{1-z^{2}}
\]