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

解答

z^{2} = - 1 + 3 i

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

求解步骤

z^{2} = - 1 + 3 i

替代

z = x + y i

(x + y i)^{2} = - 1 + 3 i

展开

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

(x^{2} - y^{2}) + 2 i x y = - 1 + 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} = - 1 2 x y = 3]

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

z = x + y i

代回

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

(x + 2 y) i + 2 x = 2 + 3 i

so l v e f or x, z = \frac{x + 4 i}{1 - x i}

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

(a + bi) = (1 + i) \cdot (a - bi) + 3 - 2 i

- \frac{1}{z - 2 i} + 1 - z = 0