z^2=-8i

解答

z^{2} = - 8 i

z = 2 - 2 i, z = - 2 + 2 i

求解步骤

z^{2} = - 8 i

替代

z = x + y i

(x + y i)^{2} = - 8 i

展开

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

(x^{2} - y^{2}) + 2 i x y = - 8 i

将

- 8 i

改写成标准复数形式:

0 - 8 i

(x^{2} - y^{2}) + 2 i x y = 0 - 8 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} = 0 2 x y = - 8]

[x^{2} - y^{2} = 0 2 x y = - 8] : (x = 2, x = - 2, y = - 2 y = 2)

z = x + y i

代回

z = 2 - 2 i, z = - 2 + 2 i

x - 2 i = 3 + 3 i - y i

A = P (1 + \frac{i}{100})^{2}

2 x - y + (2 x - 3 y - 9) i = 0

2 x + 5 i = 4 (2 + y i) - 12

x^{2} - y^{2} + 2 x y i = 3 + 4 i