4z^2+(4-4i)z+5-2i=0

解答

4 z^{2} + (4 - 4 i) z + 5 - 2 i = 0

z = - \frac{1}{2} - \frac{- 1 + 5}{2} i, z = - \frac{1}{2} + \frac{1 + 5}{2} i

求解步骤

4 z^{2} + (4 - 4 i) z + 5 - 2 i = 0

替代

z = x + y i

4 (x + y i)^{2} + (4 - 4 i) (x + y i) + 5 - 2 i = 0

展开

4 (x + y i)^{2} + (4 - 4 i) (x + y i) + 5 - 2 i : (4 x^{2} - 4 y^{2} + 4 x + 4 y + 5) + i (- 2 - 4 x + 4 y + 8 x y)

(4 x^{2} - 4 y^{2} + 4 x + 4 y + 5) + i (- 2 - 4 x + 4 y + 8 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(4 x^{2} - 4 y^{2} + 4 x + 4 y + 5) + i (- 2 - 4 x + 4 y + 8 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:

[4 x^{2} - 4 y^{2} + 4 x + 4 y + 5 = 0 - 2 - 4 x + 4 y + 8 x y = 0]

[4 x^{2} - 4 y^{2} + 4 x + 4 y + 5 = 0 - 2 - 4 x + 4 y + 8 x y = 0] : (x = - \frac{1}{2}, x = - \frac{1}{2}, y = - \frac{- 1 + 5}{2} y = \frac{1 + 5}{2})

z = x + y i

代回

z = - \frac{1}{2} - \frac{- 1 + 5}{2} i, z = - \frac{1}{2} + \frac{1 + 5}{2} i

- 1 + 2 i = \frac{2 i}{( z - 1 )}

0 = z^{2} + i z - 6

(7 + m) + (4 l - 10) i = 3 - 6 i

(- 2 - 6 i) - p (4 - 2 i) = q i

(x - y i) + (y + x i) = (5 + i)