(2-2i)x^2-(11+9i)x-16+6i=0

解答

(2 - 2 i) x^{2} - (11 + 9 i) x - 16 + 6 i = 0

x = - 1 + i, x = \frac{3}{2} + 4 i

求解步骤

(2 - 2 i) x^{2} - (11 + 9 i) x - 16 + 6 i = 0

替代

x = a + bi

(2 - 2 i) (a + bi)^{2} - (11 + 9 i) (a + bi) - 16 + 6 i = 0

展开

(2 - 2 i) (a + bi)^{2} - (11 + 9 i) (a + bi) - 16 + 6 i : (2 a^{2} - 2 b^{2} + 4 ab - 11 a + 9 b - 16) + i (6 - 11 b - 2 a^{2} + 2 b^{2} - 9 a + 4 ab)

(2 a^{2} - 2 b^{2} + 4 ab - 11 a + 9 b - 16) + i (6 - 11 b - 2 a^{2} + 2 b^{2} - 9 a + 4 ab) = 0

将

0

改写成标准复数形式:

0 + 0 i

(2 a^{2} - 2 b^{2} + 4 ab - 11 a + 9 b - 16) + i (6 - 11 b - 2 a^{2} + 2 b^{2} - 9 a + 4 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:

[2 a^{2} - 2 b^{2} + 4 ab - 11 a + 9 b - 16 = 0 6 - 11 b - 2 a^{2} + 2 b^{2} - 9 a + 4 ab = 0]

[2 a^{2} - 2 b^{2} + 4 ab - 11 a + 9 b - 16 = 0 6 - 11 b - 2 a^{2} + 2 b^{2} - 9 a + 4 ab = 0] : (a = - 1, a = \frac{3}{2}, b = 1 b = 4)

x = a + bi

代回

x = - 1 + i, x = \frac{3}{2} + 4 i

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

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

3 z - 2 w = - 5 + 10 i

x^{2} + (6 + 2 i) x + (11 + 10 i) = 0

(3 - i) z + (1 + 2 i) = 7 + i es