与えられた複素数の方程式を解きます。具体的には以下の4つ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

与えられた複素数の方程式を解きます。具体的には以下の4つ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) $z^3 = 27$ の場合：

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

$z = \sqrt[3]{27}(\cos \frac{2\pi k}{3} + i\sin \frac{2\pi k}{3}) = 3(\cos \frac{2\pi k}{3} + i\sin \frac{2\pi k}{3})$ (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) $z^6 = -1$ の場合：

$z^6 = -1 = \cos(\pi + 2\pi k) + i\sin(\pi + 2\pi k)$

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

k = 0 のとき、 $z = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$

k = 1 のとき、 $z = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = i$

k = 2 のとき、 $z = \cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$

k = 3 のとき、 $z = \cos \frac{7\pi}{6} + i\sin \frac{7\pi}{6} = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$

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

k = 5 のとき、 $z = \cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6} = \frac{\sqrt{3}}{2} - \frac{1}{2}i$

(3) $z^3 = -8i$ の場合：

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

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

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

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

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

(4) $z^4 = -32(1 + \sqrt{3}i)$ の場合：

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

$z^4 = 64(\cos (\frac{4\pi}{3} + 2\pi k) + i\sin (\frac{4\pi}{3} + 2\pi k))$

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

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

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

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

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

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

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

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

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