x^2=-1+sqrt(3)i

解

x^{2} = - 1 + 3 i

x = \frac{1}{2} + \frac{3}{2} i, x = - \frac{1}{2} - \frac{3}{2} i

解答ステップ

x^{2} = - 1 + 3 i

代用

x = a + bi

(a + bi)^{2} = - 1 + 3 i

拡張

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

(a^{2} - b^{2}) + 2 iab = - 1 + 3 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} = - 1 2 ab = 3]

[a^{2} - b^{2} = - 1 2 ab = 3] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

代用を戻す

x = a + bi

x = \frac{1}{2} + \frac{3}{2} i, x = - \frac{1}{2} - \frac{3}{2} i