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

解答

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

z = \frac{- 3 - 3}{6} + i \frac{3 + 3}{6}, z = \frac{- 3 + 3}{6} + i \frac{3 - 3}{6}

求解步骤

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

将

(z + 1)^{3}

para o lado esquerdo

(- 3 z^{2} - 6 z - 1) + i (1 - 3 z^{2}) = 0

改写成标准形式

a x^{2} + b x + c = 0

- (3 + 3 i) z^{2} - 6 z - 1 + i = 0

使用求根公式求解

z_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 3 - 3 i ) ( - 1 + i )}{2 ( - 3 - 3 i )}

(- 6)^{2} - 4 (- 3 - 3 i) (- 1 + i) = 23

z_{1, 2} = \frac{- ( - 6 ) \pm 2 3}{2 ( - 3 - 3 i )}

将解分隔开

z_{1} = \frac{- ( - 6 ) + 2 3}{2 ( - 3 - 3 i )}, z_{2} = \frac{- ( - 6 ) - 2 3}{2 ( - 3 - 3 i )}

z = \frac{- ( - 6 ) + 2 3}{2 ( - 3 - 3 i )} : \frac{- 3 - 3}{6} + i \frac{3 + 3}{6}

z = \frac{- ( - 6 ) - 2 3}{2 ( - 3 - 3 i )} : \frac{- 3 + 3}{6} + i \frac{3 - 3}{6}

二次方程组的解是:

z = \frac{- 3 - 3}{6} + i \frac{3 + 3}{6}, z = \frac{- 3 + 3}{6} + i \frac{3 - 3}{6}

2 = \frac{1}{1} - \frac{v ^{2}}{c ^{2}}

2 \cdot x^{2} - 3 x - 22 = 0

(3 x - 1)^{2} - (2 x + 3)^{2} = 0

(x - a)^{2} = b

so l v e f or y, y (y + 1) = x (x + 1) (x + 2)