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}(\...