次の方程式を解きます。 (1) $z^3 = 27$ (2) $z^6 = -1$ (3) $z^3 = -8i$ (4) $z^4 = -32(1 + \sqrt{3}i)$

1. 問題の内容

次の方程式を解きます。

(1) $z^3 = 27$

(2) $z^6 = -1$

(3) $z^3 = -8i$

(4) $z^4 = -32(1 + \sqrt{3}i)$

2. 解き方の手順

(1) $z^3 = 27$

$z^3 = 27 = 27(\cos 0 + i\sin 0)$

$z = \sqrt[3]{27} (\cos \frac{0 + 2k\pi}{3} + i\sin \frac{0 + 2k\pi}{3}) = 3 (\cos \frac{2k\pi}{3} + i\sin \frac{2k\pi}{3})$, $k = 0, 1, 2$.

$k=0$ のとき $z = 3 (\cos 0 + i\sin 0) = 3$

$k=1$ のとき $z = 3 (\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}) = 3(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -\frac{3}{2} + i\frac{3\sqrt{3}}{2}$

$k=2$ のとき $z = 3 (\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}) = 3(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -\frac{3}{2} - i\frac{3\sqrt{3}}{2}$

(2) $z^6 = -1$

$z^6 = -1 = \cos \pi + i\sin \pi$

$z = \cos (\frac{\pi + 2k\pi}{6}) + i \sin (\frac{\pi + 2k\pi}{6})$, $k = 0, 1, 2, 3, 4, 5$.

$k=0: z = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$

$k=1: z = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = i$

$k=2: z = \cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2}$

$k=3: z = \cos \frac{7\pi}{6} + i\sin \frac{7\pi}{6} = -\frac{\sqrt{3}}{2} - \frac{i}{2}$

$k=4: z = \cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2} = -i$

$k=5: z = \cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6} = \frac{\sqrt{3}}{2} - \frac{i}{2}$

(3) $z^3 = -8i$

$z^3 = -8i = 8(\cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2})$

$z = \sqrt[3]{8} (\cos (\frac{\frac{3\pi}{2} + 2k\pi}{3}) + i\sin (\frac{\frac{3\pi}{2} + 2k\pi}{3})) = 2(\cos (\frac{3\pi + 4k\pi}{6}) + i\sin (\frac{3\pi + 4k\pi}{6}))$, $k=0, 1, 2$.

$k=0: z = 2 (\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}) = 2i$

$k=1: z = 2 (\cos \frac{7\pi}{6} + i\sin \frac{7\pi}{6}) = 2(-\frac{\sqrt{3}}{2} - \frac{i}{2}) = -\sqrt{3} - i$

$k=2: z = 2 (\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}) = 2(\frac{\sqrt{3}}{2} - \frac{i}{2}) = \sqrt{3} - i$

(4) $z^4 = -32(1 + \sqrt{3}i)$

$z^4 = -32(1 + \sqrt{3}i) = 64(-\frac{1}{2} - \frac{\sqrt{3}}{2}i) = 64(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3})$

$z = \sqrt[4]{64} (\cos (\frac{\frac{4\pi}{3} + 2k\pi}{4}) + i\sin (\frac{\frac{4\pi}{3} + 2k\pi}{4})) = 2\sqrt{2} (\cos (\frac{4\pi + 6k\pi}{12}) + i\sin (\frac{4\pi + 6k\pi}{12})) = 2\sqrt{2} (\cos (\frac{2\pi + 3k\pi}{6}) + i\sin (\frac{2\pi + 3k\pi}{6}))$, $k=0, 1, 2, 3$.

$k=0: z = 2\sqrt{2} (\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2\sqrt{2} (\frac{1}{2} + i\frac{\sqrt{3}}{2}) = \sqrt{2} + i\sqrt{6}$

$k=1: z = 2\sqrt{2} (\cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6}) = 2\sqrt{2} (-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{6} + i\sqrt{2}$

$k=2: z = 2\sqrt{2} (\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}) = 2\sqrt{2} (-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -\sqrt{2} - i\sqrt{6}$

$k=3: z = 2\sqrt{2} (\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}) = 2\sqrt{2} (\frac{\sqrt{3}}{2} - i\frac{1}{2}) = \sqrt{6} - i\sqrt{2}$

3. 最終的な答え

(1) $z = 3, -\frac{3}{2} + i\frac{3\sqrt{3}}{2}, -\frac{3}{2} - i\frac{3\sqrt{3}}{2}$

(2) $z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i, -\frac{\sqrt{3}}{2} + \frac{i}{2}, -\frac{\sqrt{3}}{2} - \frac{i}{2}, -i, \frac{\sqrt{3}}{2} - \frac{i}{2}$

(3) $z = 2i, -\sqrt{3} - i, \sqrt{3} - i$

(4) $z = \sqrt{2} + i\sqrt{6}, -\sqrt{6} + i\sqrt{2}, -\sqrt{2} - i\sqrt{6}, \sqrt{6} - i\sqrt{2}$