複素数 $z$ が $z + \frac{1}{z} = \sqrt{2}$ を満たすとき、次の問いに答えます。 (1) $z$ を極形式で表しなさい。ただし、偏角 $\theta$ は $-\pi \le \theta < \pi$ とします。 (2) $z^{18} + \frac{1}{z^{12}}$ のとり得る値を求めなさい。

1. 問題の内容

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

(1) $z$ を極形式で表しなさい。ただし、偏角 $\theta$ は $-\pi \le \theta < \pi$ とします。

(2) $z^{18} + \frac{1}{z^{12}}$ のとり得る値を求めなさい。

2. 解き方の手順

(1)

$z + \frac{1}{z} = \sqrt{2}$ を $z$ について解きます。両辺に $z$ をかけると、

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

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

これを解の公式を用いて解くと、

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

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

偏角は $\theta = \arctan(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}) = \arctan(1) = \frac{\pi}{4}$

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

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

偏角は $\theta = \arctan(\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}) = \arctan(-1) = -\frac{\pi}{4}$

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

(2)

$z = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})$ のとき、

$z^{18} = (\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))^{18} = \cos(\frac{18\pi}{4}) + i\sin(\frac{18\pi}{4}) = \cos(\frac{9\pi}{2}) + i\sin(\frac{9\pi}{2}) = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = i$

$\frac{1}{z^{12}} = \frac{1}{(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))^{12}} = \frac{1}{\cos(\frac{12\pi}{4}) + i\sin(\frac{12\pi}{4})} = \frac{1}{\cos(3\pi) + i\sin(3\pi)} = \frac{1}{-1} = -1$

$z^{18} + \frac{1}{z^{12}} = i + (-1) = -1 + i$

$z = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})$ のとき、

$z^{18} = (\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))^{18} = \cos(-\frac{18\pi}{4}) + i\sin(-\frac{18\pi}{4}) = \cos(-\frac{9\pi}{2}) + i\sin(-\frac{9\pi}{2}) = \cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}) = -i$

$\frac{1}{z^{12}} = \frac{1}{(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))^{12}} = \frac{1}{\cos(-\frac{12\pi}{4}) + i\sin(-\frac{12\pi}{4})} = \frac{1}{\cos(-3\pi) + i\sin(-3\pi)} = \frac{1}{-1} = -1$

$z^{18} + \frac{1}{z^{12}} = -i + (-1) = -1 - i$

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)

$-1 + i$ または $-1 - i$