複素数 $\alpha$ と $\beta$ が与えられたとき、$\alpha\beta$ と $\frac{\alpha}{\beta}$ をそれぞれ極形式で表す問題です。偏角の範囲は $0 \le \theta < 2\pi$ とします。 (1) $\alpha = \cos\frac{7}{12}\pi + i\sin\frac{7}{12}\pi$, $\beta = \cos\frac{5}{12}\pi + i\sin\frac{5}{12}\pi$ (2) $\alpha = 2(\cos\frac{2}{3}\pi + i\sin\frac{2}{3}\pi)$, $\beta = 4(\cos\frac{1}{6}\pi + i\sin\frac{1}{6}\pi)$

1. 問題の内容

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

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

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

2. 解き方の手順

複素数の積と商の極形式に関する公式を利用します。

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

$\alpha\beta = r_1r_2(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2))$,

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

です。ただし、偏角が $0 \le \theta < 2\pi$ の範囲に入るように調整する必要があります。

(1)

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

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

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

$= \cos\pi + i\sin\pi$

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

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

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

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

(2)

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

$= 2\cdot4 (\cos(\frac{2}{3}\pi + \frac{1}{6}\pi) + i\sin(\frac{2}{3}\pi + \frac{1}{6}\pi))$

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

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

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

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

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

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

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

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)

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

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