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

解答

(x - i) \cdot (x - 1 - 2 i) = 0

x = i, x = 1 + 2 i

求解步骤

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

替代

x = a + bi

(a + bi - i) (a + bi - 1 - 2 i) = 0

展开

(a + bi - i) (a + bi - 1 - 2 i) : (a^{2} - a - b^{2} + 3 b - 2) + i (1 - b - 3 a + 2 ab)

(a^{2} - a - b^{2} + 3 b - 2) + i (1 - b - 3 a + 2 ab) = 0

将

0

改写成标准复数形式:

0 + 0 i

(a^{2} - a - b^{2} + 3 b - 2) + i (1 - b - 3 a + 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} - a - b^{2} + 3 b - 2 = 0 1 - b - 3 a + 2 ab = 0]

[a^{2} - a - b^{2} + 3 b - 2 = 0 1 - b - 3 a + 2 ab = 0] : (a = 0, a = 1, b = 1 b = 2)

x = a + bi

代回

x = i, x = 1 + 2 i

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

i (x) = x^{2} + 1

z^{2} = 4 (- 1 - i 3)

i (t) = 4 t^{2} - 48 t + 154

i (t) = 8 t^{2} - 4 t A