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

1. 問題の内容

次の複素数の計算問題を解きます。

(1) $(1+\sqrt{-2})(3-\sqrt{-8})$

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

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

2. 解き方の手順

(1) $(1+\sqrt{-2})(3-\sqrt{-8})$

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

したがって、

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

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

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

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

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

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

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

したがって、

$\frac{1}{1+i} + \frac{1}{1-2i} = \frac{1-i}{2} + \frac{1+2i}{5} = \frac{5(1-i) + 2(1+2i)}{10} = \frac{5 - 5i + 2 + 4i}{10} = \frac{7 - i}{10} = \frac{7}{10} - \frac{1}{10}i$

3. 最終的な答え

(1) $7 + \sqrt{2}i$

(2) $-2 - 2i$

(3) $\frac{7}{10} - \frac{1}{10}i$