与えられた複素数方程式を解きます。 (1) $z^2 = -1 + i$ (2) $z^3 = -2$ (3) $z^3 = -2 + 2i$

1. 問題の内容

与えられた複素数方程式を解きます。

(1) $z^2 = -1 + i$

(2) $z^3 = -2$

(3) $z^3 = -2 + 2i$

2. 解き方の手順

(1) $z^2 = -1 + i$

$z = x + yi$とおくと、

$(x + yi)^2 = -1 + i$

$x^2 - y^2 + 2xyi = -1 + i$

実部と虚部を比較して、

$x^2 - y^2 = -1$

$2xy = 1$

$y = \frac{1}{2x}$を代入して、

$x^2 - \frac{1}{4x^2} = -1$

$4x^4 + 4x^2 - 1 = 0$

$x^2 = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2}$

$x^2 > 0$なので、$x^2 = \frac{-1 + \sqrt{2}}{2}$

$x = \pm \sqrt{\frac{-1 + \sqrt{2}}{2}}$

$x = \sqrt{\frac{-1 + \sqrt{2}}{2}}$のとき、$y = \frac{1}{2\sqrt{\frac{-1 + \sqrt{2}}{2}}} = \sqrt{\frac{1}{4 \cdot \frac{-1 + \sqrt{2}}{2}}} = \sqrt{\frac{1}{2(-1+\sqrt{2})}} = \sqrt{\frac{1}{-2 + 2\sqrt{2}}} = \sqrt{\frac{2 + 2\sqrt{2}}{4(2-1)}} = \sqrt{\frac{1 + \sqrt{2}}{2}}$

$x = -\sqrt{\frac{-1 + \sqrt{2}}{2}}$のとき、$y = -\sqrt{\frac{1 + \sqrt{2}}{2}}$

よって、

$z = \pm \left( \sqrt{\frac{-1 + \sqrt{2}}{2}} + i\sqrt{\frac{1 + \sqrt{2}}{2}} \right)$

(2) $z^3 = -2$

$z^3 = 2e^{i\pi}$

$z = \sqrt[3]{2} e^{i\frac{\pi + 2k\pi}{3}}$, $k=0,1,2$

$z_0 = \sqrt[3]{2} e^{i\frac{\pi}{3}} = \sqrt[3]{2} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \sqrt[3]{2} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$

$z_1 = \sqrt[3]{2} e^{i\frac{\pi + 2\pi}{3}} = \sqrt[3]{2} e^{i\pi} = \sqrt[3]{2}(\cos \pi + i \sin \pi) = -\sqrt[3]{2}$

$z_2 = \sqrt[3]{2} e^{i\frac{\pi + 4\pi}{3}} = \sqrt[3]{2} e^{i\frac{5\pi}{3}} = \sqrt[3]{2} \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = \sqrt[3]{2} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)$

(3) $z^3 = -2 + 2i$

$z^3 = 2\sqrt{2} e^{i\frac{3\pi}{4}}$

$z = \sqrt[3]{2\sqrt{2}} e^{i \frac{\frac{3\pi}{4} + 2k\pi}{3}}$, $k=0,1,2$

$z = \sqrt{2} e^{i (\frac{\pi}{4} + \frac{2k\pi}{3})}$

$z_0 = \sqrt{2} e^{i \frac{\pi}{4}} = \sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 1 + i$

$z_1 = \sqrt{2} e^{i (\frac{\pi}{4} + \frac{2\pi}{3})} = \sqrt{2} e^{i \frac{11\pi}{12}} = \sqrt{2} \left( \cos \frac{11\pi}{12} + i \sin \frac{11\pi}{12} \right)$

$z_2 = \sqrt{2} e^{i (\frac{\pi}{4} + \frac{4\pi}{3})} = \sqrt{2} e^{i \frac{19\pi}{12}} = \sqrt{2} \left( \cos \frac{19\pi}{12} + i \sin \frac{19\pi}{12} \right)$

3. 最終的な答え

(1) $z = \pm \left( \sqrt{\frac{-1 + \sqrt{2}}{2}} + i\sqrt{\frac{1 + \sqrt{2}}{2}} \right)$

(2) $z = \sqrt[3]{2} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right), -\sqrt[3]{2}, \sqrt[3]{2} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)$

(3) $z = 1 + i, \sqrt{2} e^{i \frac{11\pi}{12}}, \sqrt{2} e^{i \frac{19\pi}{12}}$