-i=x^2-y^2+2xyi

解

- i = x^{2} - y^{2} + 2 x y i

(x = - \frac{1}{2}, x = \frac{1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

解答ステップ

- i = x^{2} - y^{2} + 2 x y i

標準的な複素数形式で

- i

を書き換える:

0 - i

0 - i = x^{2} - y^{2} + 2 x y 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:

[0 = x^{2} - y^{2} - 1 = 2 x y]

[0 = x^{2} - y^{2} - 1 = 2 x y] : (x = - \frac{1}{2}, x = \frac{1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

(x = - \frac{1}{2}, x = \frac{1}{2}, y = \frac{1}{2} y = - \frac{1}{2})