(x+iy)^2=i

解

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

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

解答ステップ

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

拡張

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

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

標準的な複素数形式で

i

を書き換える:

0 + i

(x^{2} - y^{2}) + 2 i x y = 0 + 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} = 0 2 x y = 1]

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