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。偏角 $\theta$ の範囲は $0 \leq \theta < 2\pi$ とする。 (1) $\alpha = 2 (\cos \frac{2}{3}\pi + i \sin \frac{2}{3}\pi)$, $\beta = 4(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$ (2) $\alpha = 6 + 6i$, $\beta = \sqrt{3} + i$ (3) $\alpha = -2 - 2\sqrt{3}i$, $\beta = -1 + i$

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。偏角 $\theta$ の範囲は $0 \leq \theta < 2\pi$ とする。

(1) $\alpha = 2 (\cos \frac{2}{3}\pi + i \sin \frac{2}{3}\pi)$, $\beta = 4(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$

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

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

2. 解き方の手順

(1) $\alpha = 2 (\cos \frac{2}{3}\pi + i \sin \frac{2}{3}\pi)$, $\beta = 4(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$

$\alpha \beta = 2 \cdot 4 (\cos (\frac{2}{3}\pi + \frac{\pi}{6}) + i \sin (\frac{2}{3}\pi + \frac{\pi}{6}))$

$= 8 (\cos (\frac{4}{6}\pi + \frac{1}{6}\pi) + i \sin (\frac{4}{6}\pi + \frac{1}{6}\pi))$

$= 8 (\cos \frac{5}{6}\pi + i \sin \frac{5}{6}\pi)$

$\frac{\alpha}{\beta} = \frac{2}{4} (\cos (\frac{2}{3}\pi - \frac{\pi}{6}) + i \sin (\frac{2}{3}\pi - \frac{\pi}{6}))$

$= \frac{1}{2} (\cos (\frac{4}{6}\pi - \frac{1}{6}\pi) + i \sin (\frac{4}{6}\pi - \frac{1}{6}\pi))$

$= \frac{1}{2} (\cos \frac{3}{6}\pi + i \sin \frac{3}{6}\pi)$

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

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

$\alpha = 6\sqrt{2} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) = 6\sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$

$\beta = 2 (\frac{\sqrt{3}}{2} + \frac{1}{2} i) = 2 (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$

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

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

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

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

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

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

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

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

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

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

$= 4 \sqrt{2} (\cos (\frac{16\pi}{12} + \frac{9\pi}{12}) + i \sin (\frac{16\pi}{12} + \frac{9\pi}{12}))$

$= 4 \sqrt{2} (\cos \frac{25\pi}{12} + i \sin \frac{25\pi}{12})$

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

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

$= 2\sqrt{2} (\cos (\frac{16\pi}{12} - \frac{9\pi}{12}) + i \sin (\frac{16\pi}{12} - \frac{9\pi}{12}))$

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

3. 最終的な答え

(1) $\alpha \beta = 8 (\cos \frac{5}{6}\pi + i \sin \frac{5}{6}\pi)$, $\frac{\alpha}{\beta} = \frac{1}{2} (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$

(2) $\alpha \beta = 12\sqrt{2} (\cos \frac{5}{12}\pi + i \sin \frac{5}{12}\pi)$, $\frac{\alpha}{\beta} = 3\sqrt{2} (\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})$

(3) $\alpha \beta = 4 \sqrt{2} (\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})$, $\frac{\alpha}{\beta} = 2\sqrt{2} (\cos \frac{7\pi}{12} + i \sin \frac{7\pi}{12})$