与えられた数のべき根をすべて求め、複素平面上に図示する。 (1) $\sqrt[3]{64}$ (2) $\sqrt[3]{-8}$ (3) $\sqrt[6]{i}$ (4) $\sqrt[4]{-\sqrt{3}+3i}$

1. 問題の内容

与えられた数のべき根をすべて求め、複素平面上に図示する。

(1) $\sqrt[3]{64}$

(2) $\sqrt[3]{-8}$

(3) $\sqrt[6]{i}$

(4) $\sqrt[4]{-\sqrt{3}+3i}$

2. 解き方の手順

べき根を求めるには、複素数を極形式で表し、ド・モアブルの定理を用いる。

$z = r(\cos\theta + i\sin\theta)$のとき、$\sqrt[n]{z} = \sqrt[n]{r}(\cos(\frac{\theta+2k\pi}{n})+i\sin(\frac{\theta+2k\pi}{n}))$ ($k=0, 1, ..., n-1$)

(1) $\sqrt[3]{64}$

$64 = 64(\cos 0 + i\sin 0)$

$\sqrt[3]{64} = \sqrt[3]{64}(\cos(\frac{0+2k\pi}{3}) + i\sin(\frac{0+2k\pi}{3})) = 4(\cos(\frac{2k\pi}{3}) + i\sin(\frac{2k\pi}{3}))$

$k=0: 4(\cos 0 + i\sin 0) = 4$

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

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

(2) $\sqrt[3]{-8}$

$-8 = 8(\cos \pi + i\sin \pi)$

$\sqrt[3]{-8} = \sqrt[3]{8}(\cos(\frac{\pi+2k\pi}{3}) + i\sin(\frac{\pi+2k\pi}{3})) = 2(\cos(\frac{(2k+1)\pi}{3}) + i\sin(\frac{(2k+1)\pi}{3}))$

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

$k=1: 2(\cos \pi + i\sin \pi) = 2(-1+0) = -2$

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

(3) $\sqrt[6]{i}$

$i = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2}$

$\sqrt[6]{i} = \cos(\frac{\pi/2+2k\pi}{6}) + i\sin(\frac{\pi/2+2k\pi}{6}) = \cos(\frac{(4k+1)\pi}{12}) + i\sin(\frac{(4k+1)\pi}{12})$

$k=0: \cos \frac{\pi}{12} + i\sin \frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4} + i\frac{\sqrt{6}-\sqrt{2}}{4}$

$k=1: \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4} + i\frac{\sqrt{6}+\sqrt{2}}{4}$

$k=2: \cos \frac{9\pi}{12} + i\sin \frac{9\pi}{12} = \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$

$k=3: \cos \frac{13\pi}{12} + i\sin \frac{13\pi}{12} = \cos (\pi + \frac{\pi}{12}) + i\sin (\pi + \frac{\pi}{12}) = -\frac{\sqrt{6}+\sqrt{2}}{4} - i\frac{\sqrt{6}-\sqrt{2}}{4}$

$k=4: \cos \frac{17\pi}{12} + i\sin \frac{17\pi}{12} = \cos (\pi + \frac{5\pi}{12}) + i\sin (\pi + \frac{5\pi}{12}) = -\frac{\sqrt{6}-\sqrt{2}}{4} - i\frac{\sqrt{6}+\sqrt{2}}{4}$

$k=5: \cos \frac{21\pi}{12} + i\sin \frac{21\pi}{12} = \cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

(4) $\sqrt[4]{-\sqrt{3}+3i}$

$-\sqrt{3} + 3i = r(\cos\theta + i\sin\theta)$

$r = \sqrt{(-\sqrt{3})^2 + 3^2} = \sqrt{3+9} = \sqrt{12} = 2\sqrt{3}$

$\cos\theta = \frac{-\sqrt{3}}{2\sqrt{3}} = -\frac{1}{2}$

$\sin\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$

$\theta = \frac{2\pi}{3}$

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

$\sqrt[4]{-\sqrt{3}+3i} = \sqrt[4]{2\sqrt{3}}(\cos(\frac{2\pi/3+2k\pi}{4}) + i\sin(\frac{2\pi/3+2k\pi}{4})) = (2\sqrt{3})^{1/4}(\cos(\frac{(3k+1)\pi}{6}) + i\sin(\frac{(3k+1)\pi}{6}))$

$(2\sqrt{3})^{1/4} = (2 \cdot 3^{1/2})^{1/4} = 2^{1/4} \cdot 3^{1/8}$

$k=0: (2\sqrt{3})^{1/4}(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}) = 2^{1/4}3^{1/8}(\frac{\sqrt{3}}{2} + i\frac{1}{2})$

$k=1: (2\sqrt{3})^{1/4}(\cos \frac{4\pi}{6} + i\sin \frac{4\pi}{6}) = 2^{1/4}3^{1/8}(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}) = 2^{1/4}3^{1/8}(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$

$k=2: (2\sqrt{3})^{1/4}(\cos \frac{7\pi}{6} + i\sin \frac{7\pi}{6}) = 2^{1/4}3^{1/8}(-\frac{\sqrt{3}}{2} - i\frac{1}{2})$

$k=3: (2\sqrt{3})^{1/4}(\cos \frac{10\pi}{6} + i\sin \frac{10\pi}{6}) = 2^{1/4}3^{1/8}(\cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3}) = 2^{1/4}3^{1/8}(\frac{1}{2} - i\frac{\sqrt{3}}{2})$

3. 最終的な答え

(1) $4, -2+2\sqrt{3}i, -2-2\sqrt{3}i$

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

(3) $\frac{\sqrt{6}+\sqrt{2}}{4} + i\frac{\sqrt{6}-\sqrt{2}}{4}, \frac{\sqrt{6}-\sqrt{2}}{4} + i\frac{\sqrt{6}+\sqrt{2}}{4}, -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}, -\frac{\sqrt{6}+\sqrt{2}}{4} - i\frac{\sqrt{6}-\sqrt{2}}{4}, -\frac{\sqrt{6}-\sqrt{2}}{4} - i\frac{\sqrt{6}+\sqrt{2}}{4}, \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

(4) $2^{1/4}3^{1/8}(\frac{\sqrt{3}}{2} + i\frac{1}{2}), 2^{1/4}3^{1/8}(-\frac{1}{2} + i\frac{\sqrt{3}}{2}), 2^{1/4}3^{1/8}(-\frac{\sqrt{3}}{2} - i\frac{1}{2}), 2^{1/4}3^{1/8}(\frac{1}{2} - i\frac{\sqrt{3}}{2})$