(1+i)z^2=-1+7i

解

(1 + i) z^{2} = - 1 + 7 i

z = 2 + i, z = - 2 - i

解答ステップ

(1 + i) z^{2} = - 1 + 7 i

代用

z = x + y i

(1 + i) (x + y i)^{2} = - 1 + 7 i

拡張

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

(x^{2} - y^{2} - 2 x y) + i (x^{2} - y^{2} + 2 x y) = - 1 + 7 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} - 2 x y = - 1 x^{2} - y^{2} + 2 x y = 7]

[x^{2} - y^{2} - 2 x y = - 1 x^{2} - y^{2} + 2 x y = 7] : (x = 2, x = - 2, y = 1 y = - 1)

代用を戻す

z = x + y i

z = 2 + i, z = - 2 - i