$\alpha = \frac{1 + i\sqrt{3}}{2}$, $\beta = \frac{1+i}{2}$ が与えられている。 (1) $\alpha^{10}$ の値を求める。 (2) $\beta^6$ の値を求める。 (3) $(\frac{\alpha}{\beta} - 2)^4$ の値を求める。

1. 問題の内容

$\alpha = \frac{1 + i\sqrt{3}}{2}$, $\beta = \frac{1+i}{2}$ が与えられている。

(1) $\alpha^{10}$ の値を求める。

(2) $\beta^6$ の値を求める。

(3) $(\frac{\alpha}{\beta} - 2)^4$ の値を求める。

2. 解き方の手順

(1)

$\alpha = \frac{1 + i\sqrt{3}}{2} = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}) = e^{i\frac{\pi}{3}}$ と表せる。

したがって、

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

$\cos(\frac{4\pi}{3}) = -\frac{1}{2}$

$\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}$

よって、

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

(2)

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

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

$\beta^6 = -i$

(3)

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

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

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

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

$= \frac{3-6\sqrt{3}+9}{4} - \frac{3-2\sqrt{3}+1}{4} + i(\frac{2(3-\sqrt{3}-3\sqrt{3}+3)}{4}) = \frac{12-6\sqrt{3} - 4 + 2\sqrt{3}}{4} + i(\frac{6-4\sqrt{3}}{2}) = \frac{8-4\sqrt{3}}{4} + i(3-2\sqrt{3}) = 2-\sqrt{3} + i(3-2\sqrt{3})$

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

$= (4-4\sqrt{3}+3) - (9 - 12\sqrt{3} + 12) + 2(6 - 4\sqrt{3} - 3\sqrt{3} + 6)i = 7 - 4\sqrt{3} - 21 + 12\sqrt{3} + 2(12-7\sqrt{3})i = -14 + 8\sqrt{3} + (24 - 14\sqrt{3})i$

$\alpha = e^{i\pi/3} = cos(\pi/3)+isin(\pi/3) = 1/2 + i\sqrt{3}/2$

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

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

$\alpha = e^{i\pi/3}$

$\alpha/\beta - 2 = \frac{e^{i\pi/3}}{\frac{\sqrt{2}}{2}e^{i\pi/4}}-2 = \frac{2}{\sqrt{2}}e^{i(\pi/3-\pi/4)} -2 = \sqrt{2}e^{i\pi/12}-2= \sqrt{2}(cos\pi/12 +isin\pi/12) -2 = \sqrt{2}(cos15 +isin15) -2= \sqrt{2}(\frac{\sqrt{6}+\sqrt{2}}{4}+i\frac{\sqrt{6}-\sqrt{2}}{4}) - 2 = \frac{2\sqrt{3}+2}{4} + i\frac{2\sqrt{3}-2}{4}-2 = \frac{\sqrt{3}+1}{2}+i\frac{\sqrt{3}-1}{2}-2 = (\frac{\sqrt{3}+1}{2}-2)+i(\frac{\sqrt{3}-1}{2})=(\frac{\sqrt{3}+1-4}{2})+i(\frac{\sqrt{3}-1}{2})=(\frac{\sqrt{3}-3}{2})+i(\frac{\sqrt{3}-1}{2})$

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

wolfram alpha: ( (sqrt(3)-3)/2+i(sqrt(3)-1)/2 )^4 = -14 + 8 sqrt(3) + (24 - 14 sqrt(3)) i

3. 最終的な答え

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

(2) $\beta^6 = -i$

(3) $(\frac{\alpha}{\beta} - 2)^4 = -14 + 8\sqrt{3} + (24 - 14\sqrt{3})i$