x^2=i+1

解

x^{2} = i + 1

x = \frac{1 + 2}{2} + \frac{1}{2 1 + 2} i, x = - \frac{1 + 2}{2} - \frac{1}{2 1 + 2} i

解答ステップ

x^{2} = i + 1

代用

x = a + bi

(a + bi)^{2} = i + 1

拡張

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a^{2} - b^{2}) + 2 iab = i + 1

複素数は, その実数部と虚数部が等しい場合にのみ等しくなるequation系として書き換える:

[a^{2} - b^{2} = 1 2 ab = 1]

[a^{2} - b^{2} = 1 2 ab = 1] : a = \frac{1 + 2}{2}, a = - \frac{1 + 2}{2}, b = \frac{1}{2 1 + 2} b = - \frac{1}{2 1 + 2}

代用を戻す

x = a + bi

x = \frac{1 + 2}{2} + \frac{1}{2 1 + 2} i, x = - \frac{1 + 2}{2} - \frac{1}{2 1 + 2} i