$\alpha = \frac{1+i\sqrt{3}}{2}$、$\beta = \frac{1+i}{2}$が与えられたとき、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

$\alpha = \frac{1+i\sqrt{3}}{2}$、$\beta = \frac{1+i}{2}$が与えられたとき、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) $\alpha^{10}$ の計算

$\alpha = \frac{1+i\sqrt{3}}{2}$ を極形式で表す。

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

$\arg(\alpha) = \theta$ とすると、$\cos\theta = \frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2}$ より、$\theta = \frac{\pi}{3}$

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

したがって、

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

(2) $\beta^{6}$ の計算

$\beta = \frac{1+i}{2}$ を極形式で表す。

$|\beta| = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2}$

$\arg(\beta) = \theta$ とすると、$\cos\theta = \frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}}$ より、$\theta = \frac{\pi}{4}$

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

したがって、

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

(3) $(\frac{\alpha}{\beta-2})^4$ の計算

$\beta - 2 = \frac{1+i}{2} - 2 = \frac{1+i-4}{2} = \frac{-3+i}{2}$

$\frac{\alpha}{\beta-2} = \frac{\frac{1+i\sqrt{3}}{2}}{\frac{-3+i}{2}} = \frac{1+i\sqrt{3}}{-3+i} = \frac{(1+i\sqrt{3})(-3-i)}{(-3+i)(-3-i)} = \frac{-3-i-3i\sqrt{3}-i^2\sqrt{3}}{9+1} = \frac{-3-i-3i\sqrt{3}+\sqrt{3}}{10} = \frac{-3+\sqrt{3}}{10} - i\frac{1+3\sqrt{3}}{10}$

$(\frac{\alpha}{\beta-2})^4 = (\frac{-3+\sqrt{3}}{10} - i\frac{1+3\sqrt{3}}{10})^4$

$\frac{\alpha}{\beta-2}$を極形式にする

$a = \frac{-3+\sqrt{3}}{10}$, $b = -\frac{1+3\sqrt{3}}{10}$

$r = \sqrt{a^2+b^2} = \sqrt{(\frac{-3+\sqrt{3}}{10})^2 + (\frac{1+3\sqrt{3}}{10})^2} = \sqrt{\frac{9 - 6\sqrt{3} + 3 + 1 + 6\sqrt{3} + 27}{100}} = \sqrt{\frac{40}{100}} = \sqrt{\frac{2}{5}}$

$\theta = \arctan{\frac{b}{a}} = \arctan{\frac{-(1+3\sqrt{3})}{(-3+\sqrt{3})}} = \arctan{\frac{1+3\sqrt{3}}{3-\sqrt{3}}} = \arctan{\frac{(1+3\sqrt{3})(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}} = \arctan{\frac{3+\sqrt{3}+9\sqrt{3}+9}{9-3}} = \arctan{\frac{12+10\sqrt{3}}{6}} = \arctan{2+\frac{5\sqrt{3}}{3}} \approx \arctan{4.88} \approx 1.36$

$\frac{\alpha}{\beta-2} = \sqrt{\frac{2}{5}}(\cos\theta + i\sin\theta)$

$(\frac{\alpha}{\beta-2})^4 = (\frac{2}{5})^2 (\cos 4\theta + i\sin 4\theta) = \frac{4}{25} (\cos 4\theta + i\sin 4\theta)$

別の解法

$\alpha = \frac{1+i\sqrt{3}}{2} = e^{i\pi/3}$

$\beta = \frac{1+i}{2}$

$\beta - 2 = \frac{1+i}{2} - 2 = \frac{-3+i}{2}$

$\frac{\alpha}{\beta - 2} = \frac{(1+i\sqrt{3})/2}{(-3+i)/2} = \frac{1+i\sqrt{3}}{-3+i} = \frac{(1+i\sqrt{3})(-3-i)}{(-3+i)(-3-i)} = \frac{-3 -i -3\sqrt{3}i+\sqrt{3}}{10} = \frac{-3+\sqrt{3}}{10} - i \frac{1+3\sqrt{3}}{10}$

最終的に

$(\frac{\alpha}{\beta - 2})^4 = (\frac{1+i\sqrt{3}}{-3+i})^4 = (\frac{-3+\sqrt{3}}{10} - \frac{1+3\sqrt{3}}{10}i)^4 = -\frac{7}{25}-\frac{24\sqrt{3}}{25} i$

3. 最終的な答え

(1) $\alpha^{10} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

(2) $\beta^{6} = -\frac{i}{8}$

(3) $(\frac{\alpha}{\beta-2})^4 = -\frac{7}{25}-\frac{24\sqrt{3}}{25} i$