489
回編集
差分
編集の要約なし
:偏角： $$\alpha$$
:絶対値： $$1$$
__TOC__
==導出==
$$\begin{align*}
\cos(x,e^{i\theta})&=\lim_{t\to\theta}\left[e^{x\cos t}\cos(x\sin t)-\frac{e^{x\cos t}\sin(x\sin t)}{\tan t}\right]\\\sin(x,e^{i\theta})&=\lim_{t\to\theta}\left[\frac{e^{x\cos t}\sin(x\sin t)}{\sin t}\right]\\
\end{align*}$$
:$$\begin{align*}
e^{xz}=&\exp(xz)\\
=&\sum_{n=0}^{\infty}\frac{(zxxz)^n}{n!}\\=&\frac{(zxxz)^0}{0!}+\frac{(zxxz)^1}{1!}+\sum_{n=2}^{n-1}\frac{(zxxz)^n}{n!}\\
=&1+z+\sum_{n=2}^{n-1}\frac{x^n}{n!}z^n\\
\end{align*}$$
この関数は $$n$$ 階微分したとき元の関数と一致する関数（'''周階原始関数'''）を構成する標準基底の元となりうる関数である。ガラパゴ三角関数は $$e^{\frac{\theta}{2\pi}}$$ が実数ではないときに周階原始関数となり、以下のように示すことが可能である。
\begin{array}{rcrcrcl}&&\textstyle\cos\left(x,e^{\frac{1}{3}\cdot2\pi i}\right)&&&=&\mathrm{exps}\left(x,3,0\right)-\mathrm{exps}\left(x,3,1\right)\\&&\textstyle\sin\left(x,e^{\frac{2}{3}\cdot2\pi i}\right)&=&-\sin\left(x,e^{\frac{1}{3}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,3,1\right)-\mathrm{exps}\left(x,3,2\right)&\\&&\textstyle-\cos\left(x,e^{\frac{2}{3}\cdot2\pi i}\right)&&&=&\mathrm{exps}\left(x,3,2\right)-\mathrm{exps}\left(x,3,0\right)\\&&\textstyle\cos\left(x,e^{\frac{2}{3}\cdot2\pi i}\right)&&&=&\mathrm{exps}\left(x,3,0\right)-\mathrm{exps}\left(x,3,2\right)\\&&\textstyle-\cos\left(x,e^{\frac{1}{3}\cdot2\pi i}\right)&&&=&\mathrm{exps}\left(x,3,1\right)-\mathrm{exps}\left(x,3,0\right)\\&&\textstyle\sin\left(x,e^{\frac{1}{3}\cdot2\pi i}\right)&=&-\sin\left(x,e^{\frac{2}{3}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,3,2\right)-\mathrm{exps}\left(x,3,1\right)\\\\\textstyle\cos(x)&=&\cos\left(x,e^{\frac{1}{4}\cdot2\pi i}\right)&=&\cos\left(x,e^{\frac{3}{4}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,4,0\right)-\mathrm{exps}\left(x,4,2\right)\\\textstyle-\sin(x)&=&-\sin\left(x,e^{\frac{1}{4}\cdot2\pi i}\right)&=&\sin\left(x,e^{\frac{3}{4}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,4,1\right)-\mathrm{exps}\left(x,4,3\right)\\\textstyle-\cos(x)&=&-\cos\left(x,e^{\frac{1}{4}\cdot2\pi i}\right)&=&-\cos\left(x,e^{\frac{3}{4}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,4,2\right)-\mathrm{exps}\left(x,4,0\right)\\\textstyle\sin(x)&=&\sin\left(x,e^{\frac{1}{4}\cdot2\pi i}\right)&=&-\sin\left(x,e^{\frac{3}{4}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,4,3\right)-\mathrm{exps}\left(x,4,1\right)&\\\\&&\textstyle\cos\left(x,e^{\frac{1}{6}\cdot2\pi i}\right)&=&\cos\left(x,e^{\frac{4}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,0\right)+\left(mathrm{exps}\left(x,6,1\right)-\mathrm{exps}\left(x,6,3\right)-\mathrm{exps}\left(x,6,4\right)\\&&\textstyle-\sin\left(x,e^{\frac{1}{6}\cdot2\pi i}\right)&=&\sin\left(x,e^{\frac{4}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,1\right)+\mathrm{exps}\left(x,6,2\right)-\mathrm{exps}\left(x,6,4\right)-\leftmathrm{exps}\left(x,6,5\right)\\&&\textstyle-\cos\left(x,e^{\frac{2}{6}\cdot2\pi i}\right)&=&-\cos\left(x,e^{\frac{5}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,2\right)+\mathrm{exps}\left(x,6,3\right)-\mathrm{exps}\left(x,6,5\right)-\mathrm{exps}\left(x,6,0\right)\\&&\textstyle-\cos\left(x,e^{\frac{1}{6}\cdot2\pi i}\right)&=&-\cos\left(x,e^{\frac{4}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,3\right)+\mathrm{exps}\left(x,6,4\right)-\left(mathrm{exps}\left(x,6,0\right)-\mathrm{exps}\left(x,6,1\right)\\&&\textstyle\sin\left(x,e^{\frac{1}{6}\cdot2\pi i}\right)&=&\sin\left(x,e^{\frac{2}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,4\right)+\mathrm{exps}\left(x,6,5\right)-\mathrm{exps}\left(x,6,1\right)-\mathrm{exps}\left(x,6,2\right)\\&&\textstyle\cos\left(x,e^{\frac{2}{6}\cdot2\pi i}\right)&=&\cos\left(x,e^{\frac{4}{6}\cdot2\pi i}\right)&=&\mathrm{exps}\left(x,6,5\right)+\mathrm{exps}\left(x,6,0\right)-\mathrm{exps}\left(x,6,2\right)-\mathrm{exps}\left(x,6,3\right)\\
\end{array}