x^2-9i=0

解答

x^{2} - 9 i = 0

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

求解步骤

x^{2} - 9 i = 0

替代

x = a + bi

(a + bi)^{2} - 9 i = 0

展开

(a + bi)^{2} - 9 i : (a^{2} - b^{2}) + i (- 9 + 2 ab)

(a^{2} - b^{2}) + i (- 9 + 2 ab) = 0

将

0

改写成标准复数形式:

0 + 0 i

(a^{2} - b^{2}) + i (- 9 + 2 ab) = 0 + 0 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = 0 - 9 + 2 ab = 0]

[a^{2} - b^{2} = 0 - 9 + 2 ab = 0] : (a = \frac{3}{2}, a = - \frac{3}{2}, b = \frac{3}{2} b = - \frac{3}{2})

x = a + bi

代回

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

(x + 2 - 1) (x - 2 i) = 0

(x - i y)^{2} = 5 - 12 i

(x + 4 i) (7 + y i) = 23 + 43 i

\frac{1}{2} - (y - 4) i = (x + 1) + 2 i

z = 5 i^{131} + 3 i^{127} - 8^{343}