$x = \frac{-1 + \sqrt{5}i}{2}$ , $y = \frac{-1 - \sqrt{5}i}{2}$のとき、次の式の値を求めよ。 (1) $xy(x+y)$ (2) $x^2 + y^2$ (3) $\frac{y}{x} + \frac{x}{y}$

1. 問題の内容

$x = \frac{-1 + \sqrt{5}i}{2}$ , $y = \frac{-1 - \sqrt{5}i}{2}$のとき、次の式の値を求めよ。

(1) $xy(x+y)$

(2) $x^2 + y^2$

(3) $\frac{y}{x} + \frac{x}{y}$

2. 解き方の手順

まず、$x$と$y$の和と積を計算します。

$x+y = \frac{-1 + \sqrt{5}i}{2} + \frac{-1 - \sqrt{5}i}{2} = \frac{-1 + \sqrt{5}i -1 - \sqrt{5}i}{2} = \frac{-2}{2} = -1$

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

(1) $xy(x+y) = \frac{3}{2} \cdot (-1) = -\frac{3}{2}$

(2) $x^2 + y^2 = (x+y)^2 - 2xy = (-1)^2 - 2 \cdot \frac{3}{2} = 1 - 3 = -2$

(3) $\frac{y}{x} + \frac{x}{y} = \frac{y^2 + x^2}{xy} = \frac{x^2 + y^2}{xy} = \frac{-2}{\frac{3}{2}} = -2 \cdot \frac{2}{3} = -\frac{4}{3}$

3. 最終的な答え

(1) $xy(x+y) = -\frac{3}{2}$

(2) $x^2 + y^2 = -2$

(3) $\frac{y}{x} + \frac{x}{y} = -\frac{4}{3}$