次の複素数を含む式を計算する問題です。 (1) $\sqrt{-2} \sqrt{-8}$ (2) $-\sqrt{-3} \sqrt{-4}$ (3) $(2+\sqrt{-3})^2$ (4) $(-1+\sqrt{-2})^2$ (5) $\frac{\sqrt{-25}}{\sqrt{-5}}$ (6) $\frac{\sqrt{6}}{\sqrt{-2}}$

1. 問題の内容

次の複素数を含む式を計算する問題です。

(1) $\sqrt{-2} \sqrt{-8}$

(2) $-\sqrt{-3} \sqrt{-4}$

(3) $(2+\sqrt{-3})^2$

(4) $(-1+\sqrt{-2})^2$

(5) $\frac{\sqrt{-25}}{\sqrt{-5}}$

(6) $\frac{\sqrt{6}}{\sqrt{-2}}$

2. 解き方の手順

(1) $\sqrt{-2} \sqrt{-8}$

$\sqrt{-2} = \sqrt{2}i$

$\sqrt{-8} = \sqrt{8}i = 2\sqrt{2}i$

$\sqrt{-2} \sqrt{-8} = \sqrt{2}i \cdot 2\sqrt{2}i = 2 \cdot 2 \cdot i^2 = 4(-1) = -4$

(2) $-\sqrt{-3} \sqrt{-4}$

$\sqrt{-3} = \sqrt{3}i$

$\sqrt{-4} = 2i$

$-\sqrt{-3} \sqrt{-4} = -\sqrt{3}i \cdot 2i = -2\sqrt{3}i^2 = -2\sqrt{3}(-1) = 2\sqrt{3}$

(3) $(2+\sqrt{-3})^2$

$\sqrt{-3} = \sqrt{3}i$

$(2+\sqrt{-3})^2 = (2+\sqrt{3}i)^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3}i + (\sqrt{3}i)^2 = 4 + 4\sqrt{3}i + 3i^2 = 4 + 4\sqrt{3}i - 3 = 1 + 4\sqrt{3}i$

(4) $(-1+\sqrt{-2})^2$

$\sqrt{-2} = \sqrt{2}i$

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

(5) $\frac{\sqrt{-25}}{\sqrt{-5}}$

$\sqrt{-25} = 5i$

$\sqrt{-5} = \sqrt{5}i$

$\frac{\sqrt{-25}}{\sqrt{-5}} = \frac{5i}{\sqrt{5}i} = \frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}$

(6) $\frac{\sqrt{6}}{\sqrt{-2}}$

$\sqrt{-2} = \sqrt{2}i$

$\frac{\sqrt{6}}{\sqrt{-2}} = \frac{\sqrt{6}}{\sqrt{2}i} = \frac{\sqrt{3}}{i} = \frac{\sqrt{3}i}{i^2} = \frac{\sqrt{3}i}{-1} = -\sqrt{3}i$

3. 最終的な答え

(1) -4

(2) $2\sqrt{3}$

(3) $1+4\sqrt{3}i$

(4) $-1-2\sqrt{2}i$

(5) $\sqrt{5}$

(6) $-\sqrt{3}i$