【高校数学C】3-1-3 ド・モアブルの定理｜要点まとめ

\(\displaystyle =\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)^3\)
\(\displaystyle =\cos\left(3・\frac{\pi}{3}\right)+i\sin\left(3・\frac{\pi}{3}\right)\)
\(=\cos\pi+i\sin\pi\)
\(=-1\)

\(\displaystyle =\left\{\sqrt{2}\left(\cos\frac{7}{4}\pi+i\sin\frac{7}{4}\pi\right)\right\}^6\)
\(\displaystyle =8・\left\{\cos\left(6・\frac{7}{4}\pi\right)+i\sin\left(6・\frac{7}{4}\pi\right)\right\}\)
\(\displaystyle =8\left(\cos\frac{21\pi}{2}+i\sin\frac{21\pi}{2}\right)\)
\(\displaystyle =8\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)
\(=8i\)

\(r^3(\cos3\theta+i\sin3\theta)=1(\cos0+i\sin0)\)
\(r^3=1,3\theta=2k\pi\)
\(\displaystyle r=1,\theta=\frac{2}{3}k\pi\)
\(k=0,1,2\)のとき、\(\displaystyle \theta=0,\frac{2}{3}\pi,\frac{4}{3}\pi\)なので、求める解は
\(z=\cos0+i\sin0\)
\(\displaystyle z=\cos\frac{2}{3}\pi+i\sin\frac{2}{3}\pi\)
\(\displaystyle z=\cos\frac{4}{3}\pi+i\sin\frac{4}{3}\pi\)
よって、
\(\displaystyle z=1,\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}\)

\(r^4(\cos4\theta+i\sin4\theta)=1(\cos\pi+i\sin\pi)\)
\(r^4=1,4\theta=\pi+2k\pi\)
\(\displaystyle r=1,\theta=\frac{\pi}{4}+\frac{k}{2}\pi\)
\(k=0,1,2,3\)のとき、\(\displaystyle \theta=\frac{1}{4}\pi,\frac{3}{4}\pi,\frac{5}{4}\pi,\frac{7}{4}\pi\)なので、求める解は
\(\displaystyle z=\cos\frac{1}{4}\pi+i\sin\frac{1}{4}\pi\)
\(\displaystyle z=\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\)
\(\displaystyle z=\cos\frac{5}{4}\pi+i\sin\frac{5}{4}\pi\)
\(\displaystyle z=\cos\frac{7}{4}\pi+i\sin\frac{7}{4}\pi\)
よって、
\(\displaystyle z=\frac{\sqrt{2}+\sqrt{2}i}{2},\frac{-\sqrt{2}+\sqrt{2}i}{2}\)
\(\displaystyle \ \ \ ,\frac{-\sqrt{2}-\sqrt{2}i}{2},\frac{\sqrt{2}-\sqrt{2}i}{2}\)