zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

(2-i)z+8z^2=0

解答

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

解答

z = 0, z = - \frac{1}{4} + \frac{1}{8} i

求解步骤

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

替代

z = x + y i

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

展开

(2 - i) (x + y i) + 8 (x + y i)^{2} : (2 x + y + 8 x^{2} - 8 y^{2}) + i (- x + 2 y + 16 x y)

(2 x + y + 8 x^{2} - 8 y^{2}) + i (- x + 2 y + 16 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(2 x + y + 8 x^{2} - 8 y^{2}) + i (- x + 2 y + 16 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:

[2 x + y + 8 x^{2} - 8 y^{2} = 0 - x + 2 y + 16 x y = 0]

[2 x + y + 8 x^{2} - 8 y^{2} = 0 - x + 2 y + 16 x y = 0] : (x = 0, x = - \frac{1}{4}, y = 0 y = \frac{1}{8})

z = x + y i

代回

z = 0, z = - \frac{1}{4} + \frac{1}{8} i

流行的例子

r(t)=(2t^2-5t)i+(4t-1)

r (t) = (2 t^{2} - 5 t) i + (4 t - 1)

z^{2} = \frac{- 2}{3 i}

z^{2} + 3 i z + 4 = 0

z^2-z^1=(4+3i)-(3-2i)

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

5 x + 7 y i = - x - 7 i