2つの複素数$\alpha$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$\theta$の範囲は$0 \leq \theta < 2\pi$です。

1. 問題の内容

2つの複素数$\alpha$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$\theta$の範囲は$0 \leq \theta < 2\pi$です。

2. 解き方の手順

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

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

$\alpha\beta = 8 \left[ \cos\left(\frac{4\pi}{6} + \frac{\pi}{6}\right) + i\sin\left(\frac{4\pi}{6} + \frac{\pi}{6}\right) \right]$

$\alpha\beta = 8 \left[ \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right]$

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

$\frac{\alpha}{\beta} = \frac{1}{2} \left[ \cos\left(\frac{4\pi}{6} - \frac{\pi}{6}\right) + i\sin\left(\frac{4\pi}{6} - \frac{\pi}{6}\right) \right]$

$\frac{\alpha}{\beta} = \frac{1}{2} \left[ \cos\frac{3\pi}{6} + i\sin\frac{3\pi}{6} \right]$

$\frac{\alpha}{\beta} = \frac{1}{2} \left[ \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right]$

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

$\alpha$の極形式を求める。

$r = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$

$\tan \theta = \frac{6}{6} = 1$, よって $\theta = \frac{\pi}{4}$

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

$\beta$の極形式を求める。

$r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$

$\tan \theta = \frac{1}{\sqrt{3}}$, よって $\theta = \frac{\pi}{6}$

$\beta = 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$

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

$\alpha\beta = 12\sqrt{2} \left[ \cos\left(\frac{3\pi}{12} + \frac{2\pi}{12}\right) + i\sin\left(\frac{3\pi}{12} + \frac{2\pi}{12}\right) \right]$

$\alpha\beta = 12\sqrt{2} \left[ \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} \right]$

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

$\frac{\alpha}{\beta} = 3\sqrt{2} \left[ \cos\left(\frac{3\pi}{12} - \frac{2\pi}{12}\right) + i\sin\left(\frac{3\pi}{12} - \frac{2\pi}{12}\right) \right]$

$\frac{\alpha}{\beta} = 3\sqrt{2} \left[ \cos\frac{\pi}{12} + i\sin\frac{\pi}{12} \right]$

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

$\alpha$の極形式を求める。

$r = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$

$\tan \theta = \frac{-2\sqrt{3}}{-2} = \sqrt{3}$

$\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

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

$\beta$の極形式を求める。

$r = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$

$\tan \theta = \frac{1}{-1} = -1$

$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

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

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

$\alpha\beta = 4\sqrt{2} \left[ \cos\left(\frac{16\pi}{12} + \frac{9\pi}{12}\right) + i\sin\left(\frac{16\pi}{12} + \frac{9\pi}{12}\right) \right]$

$\alpha\beta = 4\sqrt{2} \left[ \cos\frac{25\pi}{12} + i\sin\frac{25\pi}{12} \right]$

$\frac{25\pi}{12} = 2\pi + \frac{\pi}{12}$. よって $\frac{25\pi}{12}$ と $\frac{\pi}{12}$ は同じ角を表す。

$\alpha\beta = 4\sqrt{2} \left[ \cos\frac{\pi}{12} + i\sin\frac{\pi}{12} \right]$

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

$\frac{\alpha}{\beta} = 2\sqrt{2} \left[ \cos\left(\frac{16\pi}{12} - \frac{9\pi}{12}\right) + i\sin\left(\frac{16\pi}{12} - \frac{9\pi}{12}\right) \right]$

$\frac{\alpha}{\beta} = 2\sqrt{2} \left[ \cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12} \right]$

3. 最終的な答え

(1)

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

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

(2)

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

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

(3)

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

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