(-3+2i)^2=(x+yi)^2

解

(- 3 + 2 i)^{2} = (x + y i)^{2}

(x = - 3, x = 3, y = 2 y = - 2)

解答ステップ

(- 3 + 2 i)^{2} = (x + y i)^{2}

拡張

5 - 12 i = (x^{2} - y^{2}) + 2 i x y

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:

[5 = x^{2} - y^{2} - 12 = 2 x y]

[5 = x^{2} - y^{2} - 12 = 2 x y] : (x = - 3, x = 3, y = 2 y = - 2)

(x = - 3, x = 3, y = 2 y = - 2)