solvefor x,x^2=sqrt(3)+i

解

解く

x, x^{2} = 3 + i

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

解答ステップ

x^{2} = 3 + i

代用

x = a + bi

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

拡張

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

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

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

代用を戻す

x = a + bi

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