問題は以下の通りです。 (1) 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。 * ア. $\alpha = 1 + \sqrt{3}i$, $\beta = 2 + 2i$ * イ. $\alpha = -2\sqrt{3} + 2i$, $\beta = -1 + i$ (2) 次の複素数を極形式で表す。 * ア. $\frac{4+3i}{1+7i}$ * イ. $\sqrt{3} + \frac{1-i}{1+i}$ * ウ. $-4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$

1. 問題の内容

問題は以下の通りです。

(1) 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。

* ア. $\alpha = 1 + \sqrt{3}i$, $\beta = 2 + 2i$

* イ. $\alpha = -2\sqrt{3} + 2i$, $\beta = -1 + i$

(2) 次の複素数を極形式で表す。

* ア. $\frac{4+3i}{1+7i}$

* イ. $\sqrt{3} + \frac{1-i}{1+i}$

* ウ. $-4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$

2. 解き方の手順

(1) 複素数 $\alpha$ と $\beta$ を極形式で表す。$\alpha = r_1(\cos\theta_1 + i\sin\theta_1)$, $\beta = r_2(\cos\theta_2 + i\sin\theta_2)$とする。

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

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

* ア.

$\alpha = 1 + \sqrt{3}i$

$r_1 = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$

$\theta_1 = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$

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

$\beta = 2 + 2i$

$r_2 = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$

$\theta_2 = \arctan\left(\frac{2}{2}\right) = \arctan(1) = \frac{\pi}{4}$

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

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

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

* イ.

$\alpha = -2\sqrt{3} + 2i$

$r_1 = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12+4} = \sqrt{16} = 4$

$\theta_1 = \arctan\left(\frac{2}{-2\sqrt{3}}\right) + \pi = \arctan\left(-\frac{1}{\sqrt{3}}\right) + \pi = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$

$\alpha = 4\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$

$\beta = -1 + i$

$r_2 = \sqrt{(-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$

$\theta_2 = \arctan\left(\frac{1}{-1}\right) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}$

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

$\alpha\beta = 4\sqrt{2}\left(\cos\left(\frac{5\pi}{6}+\frac{3\pi}{4}\right) + i\sin\left(\frac{5\pi}{6}+\frac{3\pi}{4}\right)\right) = 4\sqrt{2}\left(\cos\frac{19\pi}{12} + i\sin\frac{19\pi}{12}\right)$

$\frac{\alpha}{\beta} = \frac{4}{\sqrt{2}}\left(\cos\left(\frac{5\pi}{6}-\frac{3\pi}{4}\right) + i\sin\left(\frac{5\pi}{6}-\frac{3\pi}{4}\right)\right) = 2\sqrt{2}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$

(2)

* ア. $\frac{4+3i}{1+7i} = \frac{(4+3i)(1-7i)}{(1+7i)(1-7i)} = \frac{4 - 28i + 3i - 21i^2}{1 - 49i^2} = \frac{4 - 25i + 21}{1 + 49} = \frac{25 - 25i}{50} = \frac{1}{2} - \frac{1}{2}i$

$r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$

$\theta = \arctan\left(\frac{-1/2}{1/2}\right) = \arctan(-1) = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}$

$\frac{4+3i}{1+7i} = \frac{\sqrt{2}}{2}\left(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}\right)$

* イ. $\sqrt{3} + \frac{1-i}{1+i} = \sqrt{3} + \frac{(1-i)(1-i)}{(1+i)(1-i)} = \sqrt{3} + \frac{1 - 2i + i^2}{1 - i^2} = \sqrt{3} + \frac{1 - 2i - 1}{1 + 1} = \sqrt{3} + \frac{-2i}{2} = \sqrt{3} - i$

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

$\theta = \arctan\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$

$\sqrt{3} + \frac{1-i}{1+i} = 2\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)$

* ウ. $-4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 4\left(-\cos\frac{\pi}{6} - i\sin\frac{\pi}{6}\right) = 4\left(\cos\left(\pi + \frac{\pi}{6}\right) + i\sin\left(\pi + \frac{\pi}{6}\right)\right) = 4\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right)$

3. 最終的な答え

(1)

* ア. $\alpha\beta = 4\sqrt{2}\left(\cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12}\right)$, $\frac{\alpha}{\beta} = \frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$

* イ. $\alpha\beta = 4\sqrt{2}\left(\cos\frac{19\pi}{12} + i\sin\frac{19\pi}{12}\right)$, $\frac{\alpha}{\beta} = 2\sqrt{2}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$

(2)

* ア. $\frac{4+3i}{1+7i} = \frac{\sqrt{2}}{2}\left(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}\right)$

* イ. $\sqrt{3} + \frac{1-i}{1+i} = 2\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)$

* ウ. $-4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 4\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right)$