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受欢迎的代数 >

z^2+2iz+i-1=0

解答

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

解答

z = \frac{1}{2} + \frac{- 1 - 2}{2} i, z = - \frac{1}{2} - \frac{- 1 + 2}{2} i

求解步骤

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

替代

z = x + y i

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

展开

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

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

z = x + y i

代回

z = \frac{1}{2} + \frac{- 1 - 2}{2} i, z = - \frac{1}{2} - \frac{- 1 + 2}{2} i

流行的例子

(a-bi)^2=13-12i

(a - bi)^{2} = 13 - 12 i

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

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

(5i)/(4+9i)= a/b

\frac{5 i}{4 + 9 i} = \frac{a}{b}

(z-(1+2i))(z+(1+2i))=0

(z - (1 + 2 i)) (z + (1 + 2 i)) = 0

5 x - 2 y i = - 8 x + 9