次の複素数方程式を解き、解を複素数平面上に図示する問題です。 (1) $z^2 = i$ (2) $z^4 = -4$ (3) $z^2 = 1 + \sqrt{3}i$

1. 問題の内容

次の複素数方程式を解き、解を複素数平面上に図示する問題です。

(1) $z^2 = i$

(2) $z^4 = -4$

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

2. 解き方の手順

(1) $z^2 = i$

$z = x + yi$とおくと、

$(x+yi)^2 = i$

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

実部と虚部を比較して、

$x^2 - y^2 = 0$

$2xy = 1$

$x^2 = y^2$より、$x = y$または$x = -y$。

$2xy = 1$より、$x$と$y$は同符号なので、$x = y$。

$2x^2 = 1$

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

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

$x = y = \frac{\sqrt{2}}{2}$または$x = y = -\frac{\sqrt{2}}{2}$

よって、$z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$または$z = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$

(2) $z^4 = -4$

$z^4 = 4e^{i(\pi + 2k\pi)}$ ($k$は整数)

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

$k = 0, 1, 2, 3$を代入すると、

$k = 0: z = \sqrt{2}e^{i\frac{\pi}{4}} = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = \sqrt{2}(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) = 1 + i$

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

$k = 2: z = \sqrt{2}e^{i\frac{5\pi}{4}} = \sqrt{2}(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}) = \sqrt{2}(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = -1 - i$

$k = 3: z = \sqrt{2}e^{i\frac{7\pi}{4}} = \sqrt{2}(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}) = \sqrt{2}(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = 1 - i$

よって、$z = 1 + i, -1 + i, -1 - i, 1 - i$

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

$z = x + yi$とおくと、

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

$x^2 + 2xyi - y^2 = 1 + \sqrt{3}i$

実部と虚部を比較して、

$x^2 - y^2 = 1$

$2xy = \sqrt{3}$

$y = \frac{\sqrt{3}}{2x}$

$x^2 - (\frac{\sqrt{3}}{2x})^2 = 1$

$x^2 - \frac{3}{4x^2} = 1$

$4x^4 - 4x^2 - 3 = 0$

$(2x^2 - 3)(2x^2 + 1) = 0$

$2x^2 = 3$または$2x^2 = -1$

$x^2 = \frac{3}{2}$

$x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}$

$x = \frac{\sqrt{6}}{2}$のとき、$y = \frac{\sqrt{3}}{2 \cdot \frac{\sqrt{6}}{2}} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$x = -\frac{\sqrt{6}}{2}$のとき、$y = \frac{\sqrt{3}}{2 \cdot -\frac{\sqrt{6}}{2}} = -\frac{\sqrt{3}}{\sqrt{6}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$

よって、$z = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$または$z = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i$

3. 最終的な答え

(1) $z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$

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

(3) $z = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i, -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i$