y^2=1-i

解

y^{2} = 1 - i

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

解答ステップ

y^{2} = 1 - i

代用

y = x + v i

(x + v i)^{2} = 1 - i

拡張

(x + v i)^{2} : (x^{2} - v^{2}) + 2 i xv

(x^{2} - v^{2}) + 2 i xv = 1 - i

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:

[x^{2} - v^{2} = 1 2 xv = - 1]

[x^{2} - v^{2} = 1 2 xv = - 1] : x = \frac{1 + 2}{2}, x = - \frac{1 + 2}{2}, v = - \frac{1}{2 1 + 2} v = \frac{1}{2 1 + 2}

代用を戻す

y = x + v i

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