与えられた複素数の計算問題を解きます。具体的には、以下の4つの問題を解きます。 (1) $\frac{2-i}{3+i} - \frac{5+10i}{1-3i}$ (2) $\frac{1}{1+i} + \frac{1}{1-i}$ (3) $\frac{1-2i}{1+2i} - \frac{1+2i}{1-2i}$ (4) $\frac{\sqrt{3}+2i}{\sqrt{3}-i} + \frac{2+\sqrt{3}i}{1-\sqrt{3}i}$

1. 問題の内容

与えられた複素数の計算問題を解きます。具体的には、以下の4つの問題を解きます。

(1) $\frac{2-i}{3+i} - \frac{5+10i}{1-3i}$

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

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

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

2. 解き方の手順

(1) $\frac{2-i}{3+i} - \frac{5+10i}{1-3i}$

まず、各分数の分母を実数化します。

$\frac{2-i}{3+i} = \frac{(2-i)(3-i)}{(3+i)(3-i)} = \frac{6-2i-3i+i^2}{9-i^2} = \frac{6-5i-1}{9+1} = \frac{5-5i}{10} = \frac{1-i}{2}$

$\frac{5+10i}{1-3i} = \frac{(5+10i)(1+3i)}{(1-3i)(1+3i)} = \frac{5+15i+10i+30i^2}{1-9i^2} = \frac{5+25i-30}{1+9} = \frac{-25+25i}{10} = \frac{-5+5i}{2}$

したがって、

$\frac{2-i}{3+i} - \frac{5+10i}{1-3i} = \frac{1-i}{2} - \frac{-5+5i}{2} = \frac{1-i+5-5i}{2} = \frac{6-6i}{2} = 3-3i$

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

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

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

$\frac{1}{1+i} + \frac{1}{1-i} = \frac{1-i}{2} + \frac{1+i}{2} = \frac{1-i+1+i}{2} = \frac{2}{2} = 1$

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

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

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

$\frac{1-2i}{1+2i} - \frac{1+2i}{1-2i} = \frac{-3-4i}{5} - \frac{-3+4i}{5} = \frac{-3-4i+3-4i}{5} = \frac{-8i}{5}$

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

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

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

$\frac{\sqrt{3}+2i}{\sqrt{3}-i} + \frac{2+\sqrt{3}i}{1-\sqrt{3}i} = \frac{1+3\sqrt{3}i}{4} + \frac{-1+3\sqrt{3}i}{4} = \frac{1+3\sqrt{3}i-1+3\sqrt{3}i}{4} = \frac{6\sqrt{3}i}{4} = \frac{3\sqrt{3}i}{2}$

3. 最終的な答え

(1) $3-3i$

(2) $1$

(3) $-\frac{8}{5}i$

(4) $\frac{3\sqrt{3}}{2}i$