複素数 $\alpha$ と $\beta$ が与えられたとき、$\alpha\beta$ と $\frac{\alpha}{\beta}$ を極形式で表す問題です。ただし、偏角 $\theta$ の範囲は $0 \le \theta < 2\pi$ とします。

1. 問題の内容

複素数 $\alpha$ と $\beta$ が与えられたとき、$\alpha\beta$ と $\frac{\alpha}{\beta}$ を極形式で表す問題です。ただし、偏角 $\theta$ の範囲は $0 \le \theta < 2\pi$ とします。

2. 解き方の手順

(1) $\alpha = -4 + 4i$, $\beta = -1 + \sqrt{3}i$ の場合

* $\alpha$ の極形式を求める:

$\alpha$ の絶対値は $|\alpha| = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

$\alpha$ の偏角は $\theta_\alpha = \arctan(\frac{4}{-4}) + \pi = \arctan(-1) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}$

したがって、$\alpha = 4\sqrt{2}(\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}))$

* $\beta$ の極形式を求める:

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

$\beta$ の偏角は $\theta_\beta = \arctan(\frac{\sqrt{3}}{-1}) + \pi = \arctan(-\sqrt{3}) + \pi = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$

したがって、$\beta = 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$

* $\alpha\beta$ を計算する:

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

$= 8\sqrt{2}(\cos(\frac{9\pi + 8\pi}{12}) + i\sin(\frac{9\pi + 8\pi}{12})) = 8\sqrt{2}(\cos(\frac{17\pi}{12}) + i\sin(\frac{17\pi}{12}))$

* $\frac{\alpha}{\beta}$ を計算する:

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

$= 2\sqrt{2}(\cos(\frac{9\pi - 8\pi}{12}) + i\sin(\frac{9\pi - 8\pi}{12})) = 2\sqrt{2}(\cos(\frac{\pi}{12}) + i\sin(\frac{\pi}{12}))$

(2) $\alpha = -\sqrt{6} + \sqrt{2}i$, $\beta = 1 + i$ の場合

* $\alpha$ の極形式を求める:

$\alpha$ の絶対値は $|\alpha| = \sqrt{(-\sqrt{6})^2 + (\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}$

$\alpha$ の偏角は $\theta_\alpha = \arctan(\frac{\sqrt{2}}{-\sqrt{6}}) + \pi = \arctan(-\frac{1}{\sqrt{3}}) + \pi = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$

したがって、$\alpha = 2\sqrt{2}(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$

* $\beta$ の極形式を求める:

$\beta$ の絶対値は $|\beta| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$

$\beta$ の偏角は $\theta_\beta = \arctan(\frac{1}{1}) = \arctan(1) = \frac{\pi}{4}$

したがって、$\beta = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$

* $\alpha\beta$ を計算する:

$\alpha\beta = (2\sqrt{2} \times \sqrt{2})(\cos(\frac{5\pi}{6} + \frac{\pi}{4}) + i\sin(\frac{5\pi}{6} + \frac{\pi}{4}))$

$= 4(\cos(\frac{10\pi + 3\pi}{12}) + i\sin(\frac{10\pi + 3\pi}{12})) = 4(\cos(\frac{13\pi}{12}) + i\sin(\frac{13\pi}{12}))$

* $\frac{\alpha}{\beta}$ を計算する:

$\frac{\alpha}{\beta} = \frac{2\sqrt{2}}{\sqrt{2}}(\cos(\frac{5\pi}{6} - \frac{\pi}{4}) + i\sin(\frac{5\pi}{6} - \frac{\pi}{4}))$

$= 2(\cos(\frac{10\pi - 3\pi}{12}) + i\sin(\frac{10\pi - 3\pi}{12})) = 2(\cos(\frac{7\pi}{12}) + i\sin(\frac{7\pi}{12}))$

3. 最終的な答え

(1) $\alpha = -4 + 4i$, $\beta = -1 + \sqrt{3}i$ の場合

$\alpha\beta = 8\sqrt{2}(\cos(\frac{17\pi}{12}) + i\sin(\frac{17\pi}{12}))$

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

(2) $\alpha = -\sqrt{6} + \sqrt{2}i$, $\beta = 1 + i$ の場合

$\alpha\beta = 4(\cos(\frac{13\pi}{12}) + i\sin(\frac{13\pi}{12}))$

$\frac{\alpha}{\beta} = 2(\cos(\frac{7\pi}{12}) + i\sin(\frac{7\pi}{12}))$