<s>視えたら自明かもしれない</s>
==観察2==
$$
\begin{align*}
& e^{i(x+y)} \\
&= e^{ix}e^{iy} \\
&= (\cos{x}+i\sin{x})(\cos{y}+i\sin{y}) \\
&= (\cos{x}\cos{y}-\sin{x}\sin{y}) + i(\sin{x}\cos{y} + \cos{x}\sin{y})
\end{align*}
$$
==主定理==
$$
\begin{align*}
& e^{z(x+y)} \\
&= e^{zx}e^{zy} \\
&= (\cos(x,z)+z\sin(x,z))(\cos(y,z)+z\sin(y,z)) \\
&= (\cos(x,z)\cos(y,z)) + z(\sin(x,z)\cos(y,z) + \cos(x,z)\sin(y,z)) + z^2\sin(x,z)\sin(y,z) \\
&= (\cos(x,z)\cos(y,z)) + z(\sin(x,z)\cos(y,z) + \cos(x,z)\sin(y,z)) + (2\mathrm{Re}(z)-1)\sin(x,z)\sin(y,z) \\
&= (\cos(x,z)\cos(y,z) - \sin(x,z)\sin(y,z)) + z(\sin(x,z)\cos(y,z) + \cos(x,z)\sin(y,z) + 2\mathrm{Re}(z)\sin(x,z)\sin(y,z))
\end{align*}
$$
==結論==
コスモスがきれいに咲くのは$$ z = \pm i $$の特殊な場合のみであることが分かった。