$\alpha = 2 + 2i$, $\beta = \sqrt{3} + i$ のとき、$\alpha\beta$ と $\frac{\alpha}{\beta}$ をそれぞれ極形式で表す問題です。

1. 問題の内容

$\alpha = 2 + 2i$, $\beta = \sqrt{3} + i$ のとき、$\alpha\beta$ と $\frac{\alpha}{\beta}$ をそれぞれ極形式で表す問題です。

2. 解き方の手順

まず、$\alpha$ と $\beta$ を極形式で表します。

$\alpha = 2+2i$ について、

絶対値 $|\alpha| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$

偏角 $\arg(\alpha) = \arctan(\frac{2}{2}) = \arctan(1) = \frac{\pi}{4}$

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

$\beta = \sqrt{3}+i$ について、

絶対値 $|\beta| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$

偏角 $\arg(\beta) = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$

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

次に、$\alpha\beta$ を求めます。

$\alpha\beta = |\alpha||\beta| (\cos{(\arg(\alpha) + \arg(\beta))} + i\sin{(\arg(\alpha) + \arg(\beta))})$

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

$\alpha\beta = 4\sqrt{2}(\cos{(\frac{3\pi + 2\pi}{12})} + i\sin{(\frac{3\pi + 2\pi}{12})})$

$\alpha\beta = 4\sqrt{2}(\cos{\frac{5\pi}{12}} + i\sin{\frac{5\pi}{12}})$

最後に、$\frac{\alpha}{\beta}$ を求めます。

$\frac{\alpha}{\beta} = \frac{|\alpha|}{|\beta|} (\cos{(\arg(\alpha) - \arg(\beta))} + i\sin{(\arg(\alpha) - \arg(\beta))})$

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

$\frac{\alpha}{\beta} = \sqrt{2}(\cos{(\frac{3\pi - 2\pi}{12})} + i\sin{(\frac{3\pi - 2\pi}{12})})$

$\frac{\alpha}{\beta} = \sqrt{2}(\cos{\frac{\pi}{12}} + i\sin{\frac{\pi}{12}})$

3. 最終的な答え

$\alpha\beta = 4\sqrt{2}(\cos{\frac{5\pi}{12}} + i\sin{\frac{5\pi}{12}})$

$\frac{\alpha}{\beta} = \sqrt{2}(\cos{\frac{\pi}{12}} + i\sin{\frac{\pi}{12}})$