(x-yi)^2=9(x+yi)

解答

(x - y i)^{2} = 9 (x + y i)

x = 9, x = 0, x = - \frac{9}{2}, x = - \frac{9}{2}, y = 0 y = 0 y = \frac{9 3}{2} y = - \frac{9 3}{2}

求解步骤

(x - y i)^{2} = 9 (x + y i)

乘开

(x^{2} - y^{2}) - 2 i x y = 9 x + 9 i y

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} = 9 x - 2 x y = 9 y]

[x^{2} - y^{2} = 9 x - 2 x y = 9 y] : x = 9, x = 0, x = - \frac{9}{2}, x = - \frac{9}{2}, y = 0 y = 0 y = \frac{9 3}{2} y = - \frac{9 3}{2}

x = 9, x = 0, x = - \frac{9}{2}, x = - \frac{9}{2}, y = 0 y = 0 y = \frac{9 3}{2} y = - \frac{9 3}{2}

\frac{2 + i}{1 + i} = a + bi

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

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

3 x + y + (2 x + 2 y) i = 13

\frac{x ^{2} - 5}{2 x + i} = 0