z^2=sqrt(3)+i

解答

z^{2} = 3 + i

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

求解步骤

z^{2} = 3 + i

替代

z = x + y i

(x + y i)^{2} = 3 + i

展开

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

(x^{2} - y^{2}) + 2 i x y = 3 + i

复数仅在实部和虚部均相等时才相等改写为方程组:

[x^{2} - y^{2} = 3 2 x y = 1]

[x^{2} - y^{2} = 3 2 x y = 1] : x = \frac{3 + 2}{2}, x = - \frac{3 + 2}{2}, y = \frac{1}{2 2 + 3} y = - \frac{1}{2 2 + 3}

z = x + y i

代回

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

z^{2} + (2 i - 3) + 5 - i = 0

7 + i^{3} - x i = 9 - i^{3} - x i

x + 3 y + (3 x - y) i = 6 + 6 i

z^{2} - 2 z i - 2 = 0

(x + i y) - 2 (x - i y) + 7 - 6 i = 0