## 問題の解答

## 問題の解答

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1. 問題の内容

与えられた複素数 $\alpha$ と $\beta$ に対して、$\alpha\beta$ および $\frac{\alpha}{\beta}$ を極形式で表す。ただし、偏角 $\theta$ の範囲は $0 \leq \theta < 2\pi$ とする。

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

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

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

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2. 解き方の手順

(1)

$\alpha$ と $\beta$ はすでに極形式で与えられているため、次の公式を利用する。

$z_1 = r_1(\cos \theta_1 + i \sin \theta_1), z_2 = r_2(\cos \theta_2 + i \sin \theta_2)$ のとき、

$z_1z_2 = r_1r_2(\cos (\theta_1 + \theta_2) + i \sin (\theta_1 + \theta_2))$

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

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

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

(2)

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

$\alpha$ の偏角を $\theta_\alpha$ とすると、$\cos \theta_\alpha = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$, $\sin \theta_\alpha = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$. よって、$\theta_\alpha = \frac{3}{4}\pi$

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

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

$\beta$ の偏角を $\theta_\beta$ とすると、$\cos \theta_\beta = \frac{-1}{2}$, $\sin \theta_\beta = \frac{\sqrt{3}}{2}$. よって、$\theta_\beta = \frac{2}{3}\pi$

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

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

$\frac{\alpha}{\beta} = \frac{4\sqrt{2}}{2}(\cos (\frac{3}{4}\pi - \frac{2}{3}\pi) + i \sin (\frac{3}{4}\pi - \frac{2}{3}\pi)) = 2\sqrt{2}(\cos \frac{1}{12}\pi + i \sin \frac{1}{12}\pi)$

(3)

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

$\alpha$ の偏角を $\theta_\alpha$ とすると、$\cos \theta_\alpha = \frac{-\sqrt{6}}{2\sqrt{2}} = -\frac{\sqrt{3}}{2}$, $\sin \theta_\alpha = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}$. よって、$\theta_\alpha = \frac{5}{6}\pi$

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

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

$\beta$ の偏角を $\theta_\beta$ とすると、$\cos \theta_\beta = \frac{1}{\sqrt{2}}$, $\sin \theta_\beta = \frac{1}{\sqrt{2}}$. よって、$\theta_\beta = \frac{1}{4}\pi$

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

$\alpha\beta = (2\sqrt{2}) \cdot \sqrt{2} (\cos(\frac{5}{6}\pi + \frac{1}{4}\pi) + i\sin(\frac{5}{6}\pi + \frac{1}{4}\pi)) = 4(\cos \frac{13}{12}\pi + i \sin \frac{13}{12}\pi)$

$\frac{\alpha}{\beta} = \frac{2\sqrt{2}}{\sqrt{2}}(\cos (\frac{5}{6}\pi - \frac{1}{4}\pi) + i \sin (\frac{5}{6}\pi - \frac{1}{4}\pi)) = 2(\cos \frac{7}{12}\pi + i \sin \frac{7}{12}\pi)$

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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)

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

$\frac{\alpha}{\beta} = 2\sqrt{2} (\cos \frac{1}{12}\pi + i \sin \frac{1}{12}\pi)$

(3)

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

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