iz^2+2z+8-i=0

解答

i z^{2} + 2 z + 8 - i = 0

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

求解步骤

i z^{2} + 2 z + 8 - i = 0

替代

z = x + y i

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

展开

i (x + y i)^{2} + 2 (x + y i) + 8 - i : (- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 y)

(- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 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:

[- 2 x y + 2 x + 8 = 0 - 1 + x^{2} - y^{2} + 2 y = 0]

[- 2 x y + 2 x + 8 = 0 - 1 + x^{2} - y^{2} + 2 y = 0] : (x = 2, x = - 2, y = 3 y = - 1)

z = x + y i

代回

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

so l v e f or i = \frac{x}{a x + b}, x

(1 + i) (a + bi) + 5 = 2 \cdot 3 i

r = 2 t i + (t^{2} - 1) j

- 6 i = \frac{- i}{120 π \cdot C}

(x + y i) (2 + i) = 6 - 2 i