問題8(1): 複素数 $z = 1 + i$ と $w = \sqrt{3} + i$ に対して、$zw$, $\frac{z}{w}$, $\frac{w}{z}$ を極形式で表してください。 問題9(1): $(\sqrt{3} + i)^6$ をド・モアブルの公式を用いて計算してください。

以下に問題8の(1)と問題9の(1)の解答を示します。

1. 問題の内容

問題8(1): 複素数 $z = 1 + i$ と $w = \sqrt{3} + i$ に対して、$zw$, $\frac{z}{w}$, $\frac{w}{z}$ を極形式で表してください。

問題9(1): $(\sqrt{3} + i)^6$ をド・モアブルの公式を用いて計算してください。

2. 解き方の手順

問題8(1):

まず、$z$と$w$を極形式で表します。

$z = 1 + i$ の絶対値 $|z|$ は $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$

$z$ の偏角 $\arg(z)$ は $\arg(z) = \arctan(\frac{1}{1}) = \arctan(1) = \frac{\pi}{4}$

よって、$z = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$

$w = \sqrt{3} + i$ の絶対値 $|w|$ は $|w| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$

$w$ の偏角 $\arg(w)$ は $\arg(w) = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$

よって、$w = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$

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

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

$\frac{z}{w} = \frac{\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))}{2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))} = \frac{\sqrt{2}}{2}(\cos(\frac{\pi}{4} - \frac{\pi}{6}) + i\sin(\frac{\pi}{4} - \frac{\pi}{6})) = \frac{\sqrt{2}}{2}(\cos(\frac{3\pi - 2\pi}{12}) + i\sin(\frac{3\pi - 2\pi}{12}))$

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

$\frac{w}{z} = \frac{2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))}{\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))} = \frac{2}{\sqrt{2}}(\cos(\frac{\pi}{6} - \frac{\pi}{4}) + i\sin(\frac{\pi}{6} - \frac{\pi}{4})) = \sqrt{2}(\cos(\frac{2\pi - 3\pi}{12}) + i\sin(\frac{2\pi - 3\pi}{12}))$

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

問題9(1):

まず、$\sqrt{3} + i$ を極形式で表します。

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

$\arg(\sqrt{3} + i) = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$

よって、$\sqrt{3} + i = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$

ド・モアブルの公式より、$(\sqrt{3} + i)^6 = [2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))]^6 = 2^6(\cos(6 \cdot \frac{\pi}{6}) + i\sin(6 \cdot \frac{\pi}{6})) = 64(\cos(\pi) + i\sin(\pi))$

$= 64(-1 + i \cdot 0) = -64$

3. 最終的な答え

問題8(1):

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

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

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

問題9(1):

$(\sqrt{3} + i)^6 = -64$