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受欢迎的代数 >

z^2-4iz-1=0

解答

z^{2} - 4 i z - 1 = 0

解答

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

求解步骤

z^{2} - 4 i z - 1 = 0

替代

z = x + y i

(x + y i)^{2} - 4 i (x + y i) - 1 = 0

展开

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

(x^{2} - y^{2} + 4 y - 1) + i (- 4 x + 2 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(x^{2} - y^{2} + 4 y - 1) + i (- 4 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} + 4 y - 1 = 0 - 4 x + 2 x y = 0]

[x^{2} - y^{2} + 4 y - 1 = 0 - 4 x + 2 x y = 0] : (x = 0, x = 0, y = 2 - 3 y = 2 + 3)

z = x + y i

代回

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

流行的例子

(2x+y)+6i=5+(x+4y)i

(2 x + y) + 6 i = 5 + (x + 4 y) i

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

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

i (x^{2} + 2) = x

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

3+2i+3iy=8i+x-2y

3 + 2 i + 3 i y = 8 i + x - 2 y