(x+iy)^2=3-4i

解

(x + i y)^{2} = 3 - 4 i

(x = - 2, x = 2, y = 1 y = - 1)

解答ステップ

(x + i y)^{2} = 3 - 4 i

拡張

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

(x^{2} - y^{2}) + 2 i x y = 3 - 4 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} = 3 2 x y = - 4]

[x^{2} - y^{2} = 3 2 x y = - 4] : (x = - 2, x = 2, y = 1 y = - 1)

(x = - 2, x = 2, y = 1 y = - 1)