与えられた複素数の累乗を計算する問題です。具体的には、以下の4つの問題を解きます。 (1) $(1 + \sqrt{3}i)^6$ (2) $(\frac{\sqrt{3}}{2} - \frac{1}{2}i)^7$ (3) $(1 - i)^{-10}$ (4) $(\frac{3 + \sqrt{3}i}{2})^{-4}$

1. 問題の内容

与えられた複素数の累乗を計算する問題です。具体的には、以下の4つの問題を解きます。

(1) $(1 + \sqrt{3}i)^6$

(2) $(\frac{\sqrt{3}}{2} - \frac{1}{2}i)^7$

(3) $(1 - i)^{-10}$

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

2. 解き方の手順

(1) $(1 + \sqrt{3}i)^6$

$1 + \sqrt{3}i$を極形式で表すと、

$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$

$\theta = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}$

よって、$1 + \sqrt{3}i = 2(\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}})$

ド・モアブルの定理より、

$(1 + \sqrt{3}i)^6 = [2(\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}})]^6 = 2^6(\cos{2\pi} + i\sin{2\pi}) = 64(1 + 0i) = 64$

(2) $(\frac{\sqrt{3}}{2} - \frac{1}{2}i)^7$

$\frac{\sqrt{3}}{2} - \frac{1}{2}i$を極形式で表すと、

$r = \sqrt{(\frac{\sqrt{3}}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1$

$\theta = \arctan{\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}} = \arctan{-\frac{1}{\sqrt{3}}} = -\frac{\pi}{6}$

よって、$\frac{\sqrt{3}}{2} - \frac{1}{2}i = \cos{(-\frac{\pi}{6})} + i\sin{(-\frac{\pi}{6})}$

ド・モアブルの定理より、

$(\frac{\sqrt{3}}{2} - \frac{1}{2}i)^7 = [\cos{(-\frac{\pi}{6})} + i\sin{(-\frac{\pi}{6})}]^7 = \cos{(-\frac{7\pi}{6})} + i\sin{(-\frac{7\pi}{6})} = \cos{(\frac{5\pi}{6})} - i\sin{(\frac{5\pi}{6})} = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$

(3) $(1 - i)^{-10}$

$1 - i$を極形式で表すと、

$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

$\theta = \arctan{\frac{-1}{1}} = -\frac{\pi}{4}$

よって、$1 - i = \sqrt{2}(\cos{(-\frac{\pi}{4})} + i\sin{(-\frac{\pi}{4})})$

$(1 - i)^{-10} = [\sqrt{2}(\cos{(-\frac{\pi}{4})} + i\sin{(-\frac{\pi}{4})})]^{-10} = (\sqrt{2})^{-10}(\cos{(\frac{10\pi}{4})} + i\sin{(\frac{10\pi}{4})}) = 2^{-5}(\cos{(\frac{5\pi}{2})} + i\sin{(\frac{5\pi}{2})}) = \frac{1}{32}(0 + i) = \frac{1}{32}i$

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

$\frac{3 + \sqrt{3}i}{2}$を極形式で表すと、

$r = \sqrt{(\frac{3}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$

$\theta = \arctan{\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}} = \arctan{\frac{\sqrt{3}}{3}} = \frac{\pi}{6}$

よって、$\frac{3 + \sqrt{3}i}{2} = \sqrt{3}(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}})$

$(\frac{3 + \sqrt{3}i}{2})^{-4} = [\sqrt{3}(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}})]^{-4} = (\sqrt{3})^{-4}(\cos{(-\frac{4\pi}{6})} + i\sin{(-\frac{4\pi}{6})}) = (\frac{1}{\sqrt{3}})^4 (\cos{(-\frac{2\pi}{3})} + i\sin{(-\frac{2\pi}{3})}) = \frac{1}{9}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -\frac{1}{18} - \frac{\sqrt{3}}{18}i$

3. 最終的な答え

(1) 64

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

(3) $\frac{1}{32}i$

(4) $-\frac{1}{18} - \frac{\sqrt{3}}{18}i$