双曲線関数の実部と虚部

命題
\(x,y\in\mathbb{R}\)に対して，以下が成り立つ: \[\begin{aligned} & \sinh(x+\mathrm{i}y) = \sinh x \cos y + \mathrm{i}\cosh x \sin y,\\ & \cosh(x+\mathrm{i}y) = \cosh x \cos y + \mathrm{i}\sinh x \sin y,\\ & \tanh(x+\mathrm{i}y) = \frac{\sinh x\cosh x + \mathrm{i}\sin y \cos y}{1 + \sinh^2 x - \sin^2 y}. \end{aligned}\]

証明
\[\begin{aligned} \sinh(x+\mathrm{i}y) &= \frac{1}{2}\mathrm{e}^{x+\mathrm{i}y} - \frac{1}{2}\mathrm{e}^{-x-\mathrm{i}y}\\ &= \frac{1}{2}\mathrm{e}^x(\cos y+\mathrm{i}\sin y) - \frac{1}{2}\mathrm{e}^{-x}(\cos y-\mathrm{i}\sin y)\\ &= \frac{1}{2}(\mathrm{e}^x-\mathrm{e}^{-x})\cos y+ \frac{1}{2}(\mathrm{e}^x+\mathrm{e}^{-x})\sin y\\ &= \sinh x\cos y+ \mathrm{i}\cosh x\sin y \end{aligned}\] \[\begin{aligned} \cosh(x+\mathrm{i}y) &= \frac{1}{2}\mathrm{e}^{x+\mathrm{i}y} + \frac{1}{2}\mathrm{e}^{-x-\mathrm{i}y}\\ &= \frac{1}{2}\mathrm{e}^x(\cos y+\mathrm{i}\sin y) + \frac{1}{2}\mathrm{e}^{-x}(\cos y-\mathrm{i}\sin y)\\ &= \frac{1}{2}(\mathrm{e}^x+\mathrm{e}^{-x})\cos y+ \frac{1}{2}(\mathrm{e}^x-\mathrm{e}^{-x})\sin y\\ &= \cosh x\cos y+ \mathrm{i}\sinh x\sin y \end{aligned}\] \[\begin{aligned} \tanh(x+\mathrm{i}y) &= \frac{\sinh(x+\mathrm{i}y)}{\cosh(x+\mathrm{i}y)}\\ &= \frac{\sinh x\cos y+ \mathrm{i}\cosh x\sin y}{\cosh x\cos y+ \mathrm{i}\sinh x\sin y}\\ &= \frac{\sinh x\cos y+ \mathrm{i}\cosh x\sin y}{\cosh x\cos y+ \mathrm{i}\sinh x\sin y} ~\frac{\cosh x\cos y- \mathrm{i}\sinh x\sin y}{\cosh x\cos y- \mathrm{i}\sinh x\sin y}\\ \end{aligned}\] ここで，分母・分子について \[\begin{aligned} \text{分子} &= (\sinh x\cos y+ \mathrm{i}\cosh x\sin y)(\cosh x\cos y- \mathrm{i}\sinh x\sin y)\\ &= \sinh x\cosh x(\cos y^2+\sin y^2) + \mathrm{i}(\cosh^2x-\sinh^2x)\sin y\cos y\\ &= \sinh x\cosh x+ \mathrm{i}\sin y\cos y, \end{aligned}\] \[\begin{aligned} \text{分母} &= (\cosh x\cos y+ \mathrm{i}\sinh x\sin y)(\cosh x\cos y- \mathrm{i}\sinh x\sin y)\\ &= \cosh x^2\cos y^2 + \sinh^2x\sin^2y\\ &= (1+\sinh^2x)(1-\sin^2y) + \sinh^2x\sin^2y\\ &= 1 + \sinh^2x - \sin^2y \end{aligned}\] が成り立つ． ゆえに， \[\begin{aligned} \tanh(x+\mathrm{i}y) = \frac{\sinh x\cosh x+ \mathrm{i}\sin y\cos y}{1 + \sinh^2x - \sin^2y} \end{aligned}\]