はい、承知いたしました。問題の複素数 $\alpha, \beta$ について、$\alpha\beta$ と $\frac{\alpha}{\beta}$ をそれぞれ極形式で表します。偏角の範囲は $0 \le \theta < 2\pi$ とします。

はい、承知いたしました。問題の複素数 $\alpha, \beta$ について、$\alpha\beta$ と $\frac{\alpha}{\beta}$ をそれぞれ極形式で表します。偏角の範囲は $0 \le \theta < 2\pi$ とします。

(1) $\alpha = \cos{\frac{7}{12}\pi} + i\sin{\frac{7}{12}\pi}, \quad \beta = \cos{\frac{5}{12}\pi} + i\sin{\frac{5}{12}\pi}$

2. 解き方の手順

複素数 $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ と $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ について、積と商は次のように計算できます。

* $z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))$

* $\frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))$

(1) の場合:

* $\alpha\beta = 1 \cdot 1 (\cos(\frac{7}{12}\pi + \frac{5}{12}\pi) + i\sin(\frac{7}{12}\pi + \frac{5}{12}\pi)) = \cos{\pi} + i\sin{\pi}$

* $\frac{\alpha}{\beta} = \frac{1}{1} (\cos(\frac{7}{12}\pi - \frac{5}{12}\pi) + i\sin(\frac{7}{12}\pi - \frac{5}{12}\pi)) = \cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}$

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

まずは $\alpha$ と $\beta$ を極形式で表します。

* $\alpha = -4 + 4i = 4\sqrt{2}(\cos\theta + i\sin\theta)$。ここで、$\cos\theta = -\frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}}$ なので、$\theta = \frac{3}{4}\pi$。したがって、$\alpha = 4\sqrt{2}(\cos{\frac{3}{4}\pi} + i\sin{\frac{3}{4}\pi})$。

* $\beta = -1 + \sqrt{3}i = 2(\cos\phi + i\sin\phi)$。ここで、$\cos\phi = -\frac{1}{2}, \sin\phi = \frac{\sqrt{3}}{2}$ なので、$\phi = \frac{2}{3}\pi$。したがって、$\beta = 2(\cos{\frac{2}{3}\pi} + i\sin{\frac{2}{3}\pi})$。

* $\alpha\beta = 4\sqrt{2} \cdot 2 (\cos(\frac{3}{4}\pi + \frac{2}{3}\pi) + i\sin(\frac{3}{4}\pi + \frac{2}{3}\pi)) = 8\sqrt{2}(\cos{\frac{17}{12}\pi} + i\sin{\frac{17}{12}\pi})$

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

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

まずは $\alpha$ と $\beta$ を極形式で表します。

* $\alpha = -\sqrt{6} + \sqrt{2}i = r(\cos\theta + i\sin\theta)$。$r = \sqrt{(-\sqrt{6})^2 + (\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}$。$\cos\theta = \frac{-\sqrt{6}}{2\sqrt{2}} = -\frac{\sqrt{3}}{2}, \sin\theta = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}$。したがって、$\theta = \frac{5\pi}{6}$。$\alpha = 2\sqrt{2}(\cos{\frac{5\pi}{6}} + i\sin{\frac{5\pi}{6}})$。

* $\beta = 1 + i = \sqrt{2}(\cos\phi + i\sin\phi)$。$\cos\phi = \frac{1}{\sqrt{2}}, \sin\phi = \frac{1}{\sqrt{2}}$。したがって、$\phi = \frac{\pi}{4}$。$\beta = \sqrt{2}(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}})$。

* $\alpha\beta = 2\sqrt{2} \cdot \sqrt{2} (\cos(\frac{5\pi}{6} + \frac{\pi}{4}) + i\sin(\frac{5\pi}{6} + \frac{\pi}{4})) = 4(\cos{\frac{13\pi}{12}} + i\sin{\frac{13\pi}{12}})$

* $\frac{\alpha}{\beta} = \frac{2\sqrt{2}}{\sqrt{2}} (\cos(\frac{5\pi}{6} - \frac{\pi}{4}) + i\sin(\frac{5\pi}{6} - \frac{\pi}{4})) = 2(\cos{\frac{7\pi}{12}} + i\sin{\frac{7\pi}{12}})$

3. 最終的な答え

(1)

* $\alpha\beta = \cos{\pi} + i\sin{\pi}$

* $\frac{\alpha}{\beta} = \cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}$

(2)

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

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

(3)

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

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