(x+iy)^2=3-4i

解答

(x + i y)^{2} = 3 - 4 i

(x = - 2, x = 2, y = 1 y = - 1)

求解步骤

(x + i y)^{2} = 3 - 4 i

展开

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

(x^{2} - y^{2}) + 2 i x y = 3 - 4 i

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} = 3 2 x y = - 4]

[x^{2} - y^{2} = 3 2 x y = - 4] : (x = - 2, x = 2, y = 1 y = - 1)

(x = - 2, x = 2, y = 1 y = - 1)

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

\frac{9}{2} x + (3 i) y = 0

so l v e f or y, (- 2 + 5 i) (1 - i) + (4 + 3 i) = - 6 x + 2 y - 3

x^{2} - 4 x + 4 i = - 1

(x - y) + 3 i = - 6 + y i