複素数 $z$ が $z + \frac{1}{z} = \sqrt{2}$ を満たすとき、以下の問いに答える。 (1) $z$ を極形式で表す。 (2) $\frac{1}{z^{10}} + z^{10}$ と $\frac{1}{z^{99}} + z^{99}$ の値を求める。

1. 問題の内容

複素数 $z$ が $z + \frac{1}{z} = \sqrt{2}$ を満たすとき、以下の問いに答える。

(1) $z$ を極形式で表す。

(2) $\frac{1}{z^{10}} + z^{10}$ と $\frac{1}{z^{99}} + z^{99}$ の値を求める。

2. 解き方の手順

(1) $z + \frac{1}{z} = \sqrt{2}$ より、両辺に $z$ をかけて整理すると

$z^2 - \sqrt{2}z + 1 = 0$

この2次方程式を解くと

$z = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm \sqrt{2}i}{2} = \frac{1 \pm i}{\sqrt{2}}$

$z = \frac{1 + i}{\sqrt{2}}$ のとき、$|z| = \sqrt{(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} = 1$

$\arg z = \frac{\pi}{4}$

よって $z = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = e^{i\frac{\pi}{4}}$

$z = \frac{1 - i}{\sqrt{2}}$ のとき、$|z| = \sqrt{(\frac{1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2} = 1$

$\arg z = -\frac{\pi}{4}$

よって $z = \cos (-\frac{\pi}{4}) + i \sin (-\frac{\pi}{4}) = e^{-i\frac{\pi}{4}}$

(2) ド・モアブルの定理より

$z^{10} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{10} = \cos \frac{10\pi}{4} + i \sin \frac{10\pi}{4} = \cos \frac{5\pi}{2} + i \sin \frac{5\pi}{2} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = i$

$\frac{1}{z^{10}} = \frac{1}{i} = -i$

よって $\frac{1}{z^{10}} + z^{10} = -i + i = 0$

$z^{99} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{99} = \cos \frac{99\pi}{4} + i \sin \frac{99\pi}{4} = \cos (\frac{3\pi}{4} + 24\pi) + i \sin (\frac{3\pi}{4} + 24\pi) = \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

$\frac{1}{z^{99}} = \frac{1}{-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i} = \frac{-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i}{\frac{1}{2} + \frac{1}{2}} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i$

よって $\frac{1}{z^{99}} + z^{99} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i) + (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) = -\sqrt{2}$

同様に $z = e^{-i\frac{\pi}{4}}$ のとき

$z^{10} = (\cos (-\frac{\pi}{4}) + i \sin (-\frac{\pi}{4}))^{10} = \cos (-\frac{10\pi}{4}) + i \sin (-\frac{10\pi}{4}) = \cos (-\frac{5\pi}{2}) + i \sin (-\frac{5\pi}{2}) = \cos (-\frac{\pi}{2}) + i \sin (-\frac{\pi}{2}) = -i$

$\frac{1}{z^{10}} = \frac{1}{-i} = i$

よって $\frac{1}{z^{10}} + z^{10} = i - i = 0$

$z^{99} = (\cos (-\frac{\pi}{4}) + i \sin (-\frac{\pi}{4}))^{99} = \cos (-\frac{99\pi}{4}) + i \sin (-\frac{99\pi}{4}) = \cos (-\frac{3\pi}{4} - 24\pi) + i \sin (-\frac{3\pi}{4} - 24\pi) = \cos (-\frac{3\pi}{4}) + i \sin (-\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i$

$\frac{1}{z^{99}} = \frac{1}{-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i} = \frac{-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i}{\frac{1}{2} + \frac{1}{2}} = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

よって $\frac{1}{z^{99}} + z^{99} = (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) + (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i) = -\sqrt{2}$

3. 最終的な答え

(1) $z = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4}$ または $z = \cos (-\frac{\pi}{4}) + i \sin (-\frac{\pi}{4})$

(2) $\frac{1}{z^{10}} + z^{10} = 0$

$\frac{1}{z^{99}} + z^{99} = -\sqrt{2}$