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) $z^2 = i$

$z = r(\cos\theta + i\sin\theta)$とおくと、

$z^2 = r^2(\cos2\theta + i\sin2\theta) = i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}$

したがって、$r^2 = 1, 2\theta = \frac{\pi}{2} + 2n\pi$ (nは整数)

$r > 0$より$r=1$

$\theta = \frac{\pi}{4} + n\pi$

$n=0$のとき、$\theta = \frac{\pi}{4}$

$n=1$のとき、$\theta = \frac{5\pi}{4}$

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

$z = \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$

(2) $z^4 = -4$

$z = r(\cos\theta + i\sin\theta)$とおくと、

$z^4 = r^4(\cos4\theta + i\sin4\theta) = -4 = 4(\cos\pi + i\sin\pi)$

したがって、$r^4 = 4, 4\theta = \pi + 2n\pi$ (nは整数)

$r > 0$より$r = \sqrt{2}$

$\theta = \frac{\pi}{4} + \frac{n\pi}{2}$

$n=0$のとき、$\theta = \frac{\pi}{4}$

$n=1$のとき、$\theta = \frac{3\pi}{4}$

$n=2$のとき、$\theta = \frac{5\pi}{4}$

$n=3$のとき、$\theta = \frac{7\pi}{4}$

よって、$z = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = \sqrt{2}(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) = 1 + i$

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

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

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

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

$z = r(\cos\theta + i\sin\theta)$とおくと、

$z^2 = r^2(\cos2\theta + i\sin2\theta) = 1 + \sqrt{3}i = 2(\frac{1}{2} + \frac{\sqrt{3}}{2}i) = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$

したがって、$r^2 = 2, 2\theta = \frac{\pi}{3} + 2n\pi$ (nは整数)

$r > 0$より$r = \sqrt{2}$

$\theta = \frac{\pi}{6} + n\pi$

$n=0$のとき、$\theta = \frac{\pi}{6}$

$n=1$のとき、$\theta = \frac{7\pi}{6}$

よって、$z = \sqrt{2}(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) = \sqrt{2}(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$

$z = \sqrt{2}(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}) = \sqrt{2}(-\frac{\sqrt{3}}{2} - \frac{1}{2}i) = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i$

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