(a+bi)^2=i

解

(a + bi)^{2} = i

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

解答ステップ

(a + bi)^{2} = i

拡張

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a^{2} - b^{2}) + 2 iab = i

標準的な複素数形式で

i

を書き換える:

0 + i

(a^{2} - b^{2}) + 2 iab = 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:

[a^{2} - b^{2} = 0 2 ab = 1]

[a^{2} - b^{2} = 0 2 ab = 1] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})