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受欢迎的代数 >

z^2+zi=2i^2

解答

z^{2} + z i = 2 i^{2}

解答

z = - 2 i, z = i

求解步骤

z^{2} + z i = 2 i^{2}

替代

z = x + y i

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

乘开

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

将

- 2

改写成标准复数形式:

- 2 + 0 i

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

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

z = x + y i

代回

z = - 2 i, z = i

流行的例子

z^{2} + 16 = 30 i

-2iez^2=-1+i,z^1=3

- 2 i e z^{2} = - 1 + i, z^{1} = 3

(i)x^2-16=0,(ii)6x^2-24=0

(i) x^{2} - 16 = 0, (ii) 6 x^{2} - 24 = 0

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

(2i+1)^2=(m+ni)(2-i)

(2 i + 1)^{2} = (m + ni) (2 - i)