solvefor m,(m-n)^2=-2i

解答

求解

m, (m - n)^{2} = - 2 i

m = (- 1 + n) + i, m = (n + 1) - i

求解步骤

(m - n)^{2} = - 2 i

替代

m = x + y i

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

展开

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

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

将

- 2 i

改写成标准复数形式:

0 - 2 i

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

[- 2 n x + x^{2} + n^{2} - y^{2} = 0 - 2 n y + 2 x y = - 2]

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

m = x + y i

代回

m = (- 1 + n) + i, m = (n + 1) - i

k_{1} - 3 - 27 i k_{2} = 0

so l v e f or c, i = \frac{d}{( 1 - c )}

(a + i)^{2} = 8 + bi

\frac{3 + ai}{5 + bi} = 3 + 5 i

so l v e f or x, (x + y)^{2} = 3 - 4 i