次の8つの複素数の計算問題を解きます。 (1) $(2-\sqrt{3}i)(2+\sqrt{3}i)$ (2) $(\sqrt{-2}+\sqrt{5})(\sqrt{-6}-\sqrt{15})$ (3) $(1+i)^3$ (4) $(\frac{1-i}{\sqrt{2}})^2$ (5) $\frac{2}{1+\sqrt{3}i}$ (6) $1+2i+\frac{1}{4i}$ (7) $\frac{\sqrt{5}+i}{\sqrt{5}-i}$ (8) $\frac{2+\sqrt{-3}}{2-\sqrt{-3}}$

はい、承知いたしました。与えられた複素数の計算問題を解きます。

1. 問題の内容

次の8つの複素数の計算問題を解きます。

(1) $(2-\sqrt{3}i)(2+\sqrt{3}i)$

(2) $(\sqrt{-2}+\sqrt{5})(\sqrt{-6}-\sqrt{15})$

(3) $(1+i)^3$

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

(5) $\frac{2}{1+\sqrt{3}i}$

(6) $1+2i+\frac{1}{4i}$

(7) $\frac{\sqrt{5}+i}{\sqrt{5}-i}$

(8) $\frac{2+\sqrt{-3}}{2-\sqrt{-3}}$

2. 解き方の手順

(1) $(2-\sqrt{3}i)(2+\sqrt{3}i)$

これは和と差の積の形なので、$a^2 - b^2$を利用します。

$(2-\sqrt{3}i)(2+\sqrt{3}i) = 2^2 - (\sqrt{3}i)^2 = 4 - 3i^2 = 4 - 3(-1) = 4 + 3 = 7$

(2) $(\sqrt{-2}+\sqrt{5})(\sqrt{-6}-\sqrt{15})$

$\sqrt{-2} = \sqrt{2}i$, $\sqrt{-6} = \sqrt{6}i$であるから、

$(\sqrt{2}i + \sqrt{5})(\sqrt{6}i - \sqrt{15}) = \sqrt{12}i^2 - \sqrt{30}i + \sqrt{30}i - \sqrt{75} = -2\sqrt{3} - 5\sqrt{3} = -7\sqrt{3}$

(3) $(1+i)^3$

$(1+i)^3 = (1+i)^2(1+i) = (1 + 2i + i^2)(1+i) = (1+2i-1)(1+i) = 2i(1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$

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

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

(5) $\frac{2}{1+\sqrt{3}i}$

分母の共役複素数をかけて実数化します。

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

(6) $1+2i+\frac{1}{4i}$

$\frac{1}{4i} = \frac{1}{4i} \times \frac{-4i}{-4i} = \frac{-4i}{-16i^2} = \frac{-4i}{16} = -\frac{1}{4}i$

$1+2i-\frac{1}{4}i = 1 + \frac{8}{4}i - \frac{1}{4}i = 1 + \frac{7}{4}i$

(7) $\frac{\sqrt{5}+i}{\sqrt{5}-i}$

分母の共役複素数をかけて実数化します。

$\frac{\sqrt{5}+i}{\sqrt{5}-i} = \frac{(\sqrt{5}+i)(\sqrt{5}+i)}{(\sqrt{5}-i)(\sqrt{5}+i)} = \frac{5 + 2\sqrt{5}i + i^2}{5 - i^2} = \frac{5 + 2\sqrt{5}i - 1}{5 + 1} = \frac{4 + 2\sqrt{5}i}{6} = \frac{2 + \sqrt{5}i}{3} = \frac{2}{3} + \frac{\sqrt{5}}{3}i$

(8) $\frac{2+\sqrt{-3}}{2-\sqrt{-3}}$

$\sqrt{-3} = \sqrt{3}i$であるから、

$\frac{2+\sqrt{3}i}{2-\sqrt{3}i} = \frac{(2+\sqrt{3}i)(2+\sqrt{3}i)}{(2-\sqrt{3}i)(2+\sqrt{3}i)} = \frac{4 + 4\sqrt{3}i + 3i^2}{4 - 3i^2} = \frac{4 + 4\sqrt{3}i - 3}{4 + 3} = \frac{1 + 4\sqrt{3}i}{7} = \frac{1}{7} + \frac{4\sqrt{3}}{7}i$

3. 最終的な答え

(1) 7

(2) $-7\sqrt{3}$

(3) $-2+2i$

(4) $-i$

(5) $\frac{1}{2}-\frac{\sqrt{3}}{2}i$

(6) $1+\frac{7}{4}i$

(7) $\frac{2}{3}+\frac{\sqrt{5}}{3}i$

(8) $\frac{1}{7}+\frac{4\sqrt{3}}{7}i$