1.次の計算をしなさい。
(1)\((1+\sqrt{3}i)^5\)
\(\displaystyle =\left\{2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\right\}^5\)
\(\displaystyle =32\left(\cos\frac{5}{3}\pi+i\sin\frac{5}{3}\pi\right)\)
\(\displaystyle =32\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\)
\(=16-16\sqrt{3}i\)
(2)\((1-\sqrt{3}i)^6\)
\(\displaystyle =\left\{2\left(\cos\frac{5}{3}\pi+i\sin\frac{5}{3}\pi\right)\right\}^6\)
\(\displaystyle =64\left(\cos\frac{30}{3}\pi+i\sin\frac{30}{3}\pi\right)\)
\(\displaystyle =64(1-0i)\)
\(=64\)
(3)\((1+i)^8\)
\(\displaystyle =\left\{\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right\}^8\)
\(\displaystyle =16\left(\cos2\pi+i\sin2\pi\right)\)
\(\displaystyle =64(1+0i)\)
\(=16\)
(4)\((-1+i)^{-4}\)
\(\displaystyle =\left\{\sqrt{2}\left(\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\right)\right\}^{-4}\)
\(\displaystyle =\frac{1}{4}\{\cos(-3\pi)+i\sin(-3\pi)\}\)
\(\displaystyle =\frac{1}{4}(-1+0i)\)
\(\displaystyle =-\frac{1}{4}\)
(5)\((1-\sqrt{3}i)^{-6}\)
\(\displaystyle =\left\{2\left(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})\right)\right\}^{-6}\)
\(\displaystyle =\frac{1}{64}(\cos2\pi+i\sin2\pi)\)
\(\displaystyle =\frac{1}{64}(1+0i)\)
\(\displaystyle =\frac{1}{64}\)
2.次の方程式を解きなさい。
(1)\(z^6=1\)
\(r^6(\cos6\theta+i\sin6\theta)=1(\cos0+i\sin0)\)
\(r^6=1,6\theta=2k\pi\)
\(\displaystyle r=1,\theta=\frac{1}{3}k\pi\)
\(k=0,1,2,3,4,5\)のとき、
\(\displaystyle \theta=0,\frac{1}{3}\pi,\frac{2}{3}\pi,\pi,\frac{4}{3}\pi,\frac{5}{3}\pi\)なので、求める解は
\(z=\cos0+i\sin0\)
\(\displaystyle z=\cos\frac{1}{3}\pi+i\sin\frac{1}{3}\pi\)
\(\displaystyle z=\cos\frac{2}{3}\pi+i\sin\frac{2}{3}\pi\)
\(\displaystyle z=\cos\pi+i\sin\pi\)
\(\displaystyle z=\cos\frac{4}{3}\pi+i\sin\frac{4}{3}\pi\)
\(\displaystyle z=\cos\frac{5}{3}\pi+i\sin\frac{5}{3}\pi\)
よって、
\(\displaystyle z=1,\frac{1+\sqrt{3}i}{2},\frac{-1+\sqrt{3}i}{2}\)
\(\displaystyle \ \ \ ,-1,\frac{-1-\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2}\)
(2)\(z^2=i\)
\(\displaystyle r^2(\cos2\theta+i\sin2\theta)=1\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)
\(\displaystyle r^2=1,2\theta=\frac{\pi}{2}+2k\pi\)
\(\displaystyle r=1,\theta=\frac{\pi}{4}+k\pi\)
\(k=0,1\)のとき、
\(\displaystyle \theta=\frac{1}{4}\pi,\frac{5}{4}\pi\)なので、求める解は
\(\displaystyle z=\cos\frac{1}{4}\pi+i\sin\frac{1}{4}\pi\)
\(\displaystyle z=\cos\frac{5}{4}\pi+i\sin\frac{5}{4}\pi\)
よって、
\(\displaystyle z=\frac{1+i}{\sqrt{2}},\frac{-1-i}{\sqrt{2}}\)
(3)\(z^3=8i\)
\(\displaystyle r^3(\cos3\theta+i\sin3\theta)=8\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)
\(\displaystyle r^3=8,3\theta=\frac{\pi}{2}+2k\pi\)
\(\displaystyle r=2,\theta=\frac{\pi}{6}+\frac{2}{3}k\pi\)
\(k=0,1,2\)のとき、
\(\displaystyle \theta=\frac{1}{6}\pi,\frac{5}{6}\pi,\frac{3}{2}\pi\)なので、求める解は
\(\displaystyle z=2\left(\cos\frac{1}{6}\pi+i\sin\frac{1}{6}\pi\right)\)
\(\displaystyle z=2\left(\cos\frac{5}{6}\pi+i\sin\frac{5}{6}\pi\right)\)
\(\displaystyle z=2\left(\cos\frac{3}{2}\pi+i\sin\frac{3}{2}\pi\right)\)
よって、
\(\displaystyle z=\sqrt{3}+i,-\sqrt{3}+i,-2i\)
(4)\(z^4=-4\)
\(\displaystyle r^4(\cos4\theta+i\sin4\theta)=4(\cos\pi+i\sin\pi)\)
\(\displaystyle r^4=4,4\theta=\pi+2k\pi\)
\(\displaystyle r=\sqrt{2},\theta=\frac{1+2k}{4}\pi\)
\(k=0,1,2,3\)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
\(\displaystyle z=\sqrt{2}\left(\cos\frac{1}{4}\pi+i\sin\frac{1}{4}\pi\right)\)
\(\displaystyle z=\sqrt{2}\left(\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\right)\)
\(\displaystyle z=\sqrt{2}\left(\cos\frac{5}{4}\pi+i\sin\frac{5}{4}\pi\right)\)
\(\displaystyle z=\sqrt{2}\left(\cos\frac{7}{4}\pi+i\sin\frac{7}{4}\pi\right)\)
よって、
\(\displaystyle z=1+i,-1+i,-1-i,1-i\)
(5)\(z^2=1+\sqrt{3}i\)
\(\displaystyle r^2(\cos2\theta+i\sin2\theta)=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)
\(\displaystyle r^2=2,2\theta=\frac{\pi}{3}+2k\pi\)
\(\displaystyle r=\sqrt{2},\theta=\frac{1+6k}{6}\pi\)
\(k=0,1\)のとき、
\(\displaystyle \theta=\frac{1}{6}\pi,\frac{7}{6}\pi\)なので、求める解は
\(\displaystyle z=\sqrt{2}\left(\cos\frac{1}{6}\pi+i\sin\frac{1}{6}\pi\right)\)
\(\displaystyle z=\sqrt{2}\left(\cos\frac{7}{6}\pi+i\sin\frac{7}{6}\pi\right)\)
よって、
\(\displaystyle z=\frac{\sqrt{6}+\sqrt{2}i}{2},\frac{-\sqrt{6}-\sqrt{2}i}{2}\)
