$\alpha = \frac{1 + i\sqrt{3}}{2}$、$\beta = \frac{1 + i}{2}$であるとき、以下の値を求めよ。 (1) $\alpha^{10}$ (2) $\beta^{6}$ (3) $(\frac{\alpha}{\beta^{-2}})^4$

1. 問題の内容

$\alpha = \frac{1 + i\sqrt{3}}{2}$、$\beta = \frac{1 + i}{2}$であるとき、以下の値を求めよ。

(1) $\alpha^{10}$

(2) $\beta^{6}$

(3) $(\frac{\alpha}{\beta^{-2}})^4$

2. 解き方の手順

(1) $\alpha$ を極形式で表すと、

$\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$

ド・モアブルの定理より、

$\alpha^{10} = \cos \frac{10\pi}{3} + i \sin \frac{10\pi}{3}$

$= \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}$

$= -\frac{1}{2} - \frac{\sqrt{3}}{2} i$

(2) $\beta$ を極形式で表すと、

$\beta = \frac{1}{\sqrt{2}}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$

ド・モアブルの定理より、

$\beta^6 = (\frac{1}{\sqrt{2}})^6 (\cos \frac{6\pi}{4} + i \sin \frac{6\pi}{4})$

$= \frac{1}{8}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2})$

$= \frac{1}{8}(0 - i)$

$= -\frac{1}{8}i$

(3)

$\beta^{-2} = (\frac{1+i}{2})^{-2} = (\frac{2}{1+i})^2 = (\frac{2(1-i)}{2})^2 = (1-i)^2 = 1 - 2i - 1 = -2i$

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

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

ここで $-\sqrt{3}+i = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6})$ より、

$(\frac{-\sqrt{3}+i}{4})^4 = (\frac{1}{2}(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}))^4 = (\frac{1}{2})^4 (\cos \frac{20\pi}{6} + i \sin \frac{20\pi}{6})$

$= \frac{1}{16} (\cos \frac{10\pi}{3} + i \sin \frac{10\pi}{3}) = \frac{1}{16} (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) = \frac{1}{16} (-\frac{1}{2} - \frac{\sqrt{3}}{2}i)$

$= -\frac{1}{32} - \frac{\sqrt{3}}{32}i$

3. 最終的な答え

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

(2) $-\frac{1}{8}i$

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