z^2+(2+5i)*z+(-6+6i)=0

解答

z^{2} + (2 + 5 i) \cdot z + (- 6 + 6 i) = 0

z = - 3 i, z = - 2 - 2 i

求解步骤

z^{2} + (2 + 5 i) z + (- 6 + 6 i) = 0

替代

z = x + y i

(x + y i)^{2} + (2 + 5 i) (x + y i) - 6 + 6 i = 0

展开

(x + y i)^{2} + (2 + 5 i) (x + y i) - 6 + 6 i : (x^{2} - y^{2} + 2 x - 5 y - 6) + i (6 + 2 y + 5 x + 2 x y)

(x^{2} - y^{2} + 2 x - 5 y - 6) + i (6 + 2 y + 5 x + 2 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(x^{2} - y^{2} + 2 x - 5 y - 6) + i (6 + 2 y + 5 x + 2 x y) = 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:

[x^{2} - y^{2} + 2 x - 5 y - 6 = 0 6 + 2 y + 5 x + 2 x y = 0]

[x^{2} - y^{2} + 2 x - 5 y - 6 = 0 6 + 2 y + 5 x + 2 x y = 0] : (x = 0, x = - 2, y = - 3 y = - 2)

z = x + y i

代回

z = - 3 i, z = - 2 - 2 i

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

\frac{1}{z} = \frac{- 8 + 2 i}{2 i}

x^{2} + 5 x + 2 x = 1 - 5 i

3 (x - 2) = 2 y i - (1 - i), y + 2 = x - 3 i

so l v e f or a, z = (a + bi) (a - bi)