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受欢迎的代数 >

(2+i)z^2+(1-7i)z-5=0

解答

(2 + i) z^{2} + (1 - 7 i) z - 5 = 0

解答

z = i, z = 1 + 2 i

求解步骤

(2 + i) z^{2} + (1 - 7 i) z - 5 = 0

替代

z = x + y i

(2 + i) (x + y i)^{2} + (1 - 7 i) (x + y i) - 5 = 0

展开

(2 + i) (x + y i)^{2} + (1 - 7 i) (x + y i) - 5 : (2 x^{2} - 2 y^{2} - 2 x y + x + 7 y - 5) + i (x^{2} + y - y^{2} - 7 x + 4 x y)

(2 x^{2} - 2 y^{2} - 2 x y + x + 7 y - 5) + i (x^{2} + y - y^{2} - 7 x + 4 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(2 x^{2} - 2 y^{2} - 2 x y + x + 7 y - 5) + i (x^{2} + y - y^{2} - 7 x + 4 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^{2} - 2 y^{2} - 2 x y + x + 7 y - 5 = 0 x^{2} + y - y^{2} - 7 x + 4 x y = 0]

[2 x^{2} - 2 y^{2} - 2 x y + x + 7 y - 5 = 0 x^{2} + y - y^{2} - 7 x + 4 x y = 0] : (x = 0, x = 1, y = 1 y = 2)

z = x + y i

代回

z = i, z = 1 + 2 i

流行的例子

6 i - x + y i = 4 + 2 i

∣ z ∣ - z = 2 + i

solvefor x,x^2y^2+xsin(y)=4

so l v e f or x, x^{2} y^{2} + x sin (y) = 4

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

z^2-(1+4i)z-(5+i)=0

z^{2} - (1 + 4 i) z - (5 + i) = 0