x^2-9i+11=3i

解答

x^{2} - 9 i + 11 = 3 i

x = \frac{- 11 + 265}{2} + \frac{6 2}{265 - 11} i, x = - \frac{- 11 + 265}{2} - \frac{6 2}{265 - 11} i

求解步骤

x^{2} - 9 i + 11 = 3 i

替代

x = a + bi

(a + bi)^{2} - 9 i + 11 = 3 i

展开

(a + bi)^{2} - 9 i + 11 : (a^{2} - b^{2} + 11) + i (- 9 + 2 ab)

(a^{2} - b^{2} + 11) + i (- 9 + 2 ab) = 3 i

将

3 i

改写成标准复数形式:

0 + 3 i

(a^{2} - b^{2} + 11) + i (- 9 + 2 ab) = 0 + 3 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} - b^{2} + 11 = 0 - 9 + 2 ab = 3]

[a^{2} - b^{2} + 11 = 0 - 9 + 2 ab = 3] : a = \frac{- 11 + 265}{2}, a = - \frac{- 11 + 265}{2}, b = \frac{6 2}{265 - 11} b = - \frac{6 2}{265 - 11}

x = a + bi

代回

x = \frac{- 11 + 265}{2} + \frac{6 2}{265 - 11} i, x = - \frac{- 11 + 265}{2} - \frac{6 2}{265 - 11} i

(7 + 3 i) x + (3 - 2 i) y = 1 + 7 i

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

so l v e f or x, (i x - i y)^{2} = 2

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

z^{2} = - 2 - 23 i