z^2=1-i

解

z^{2} = 1 - i

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

解答ステップ

z^{2} = 1 - i

代用

z = x + y i

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

拡張

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

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

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

代用を戻す

z = x + y i

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