((a+bi)+2)^2=-2i

解答

((a + bi) + 2)^{2} = - 2 i

(a = - 3, a = - 1, b = 1 b = - 1)

求解步骤

((a + bi) + 2)^{2} = - 2 i

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((a + bi) + 2)^{2} : (a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab)

(a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab) = - 2 i

将

- 2 i

改写成标准复数形式:

0 - 2 i

(a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab) = 0 - 2 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} + 4 a - b^{2} + 4 = 0 4 b + 2 ab = - 2]

[a^{2} + 4 a - b^{2} + 4 = 0 4 b + 2 ab = - 2] : (a = - 3, a = - 1, b = 1 b = - 1)

(a = - 3, a = - 1, b = 1 b = - 1)

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

so l v e f or c, z = \frac{- 3 + 3 i}{2 - c}

z^{2} = - 15 + 8 i

\frac{( - 6 i + n - 3 n + 5 )}{4 + n ^{2}} = 0

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