(x+iy)^2+(1+4i)(x+iy)=5+i

解答

(x + i y)^{2} + (1 + 4 i) (x + i y) = 5 + i

(x = 1, x = - 2, y = - 1 y = - 3)

求解步骤

(x + i y)^{2} + (1 + 4 i) (x + i y) = 5 + i

展开

(x + i y)^{2} + (1 + 4 i) (x + i y) : (x^{2} - y^{2} + x - 4 y) + i (y + 4 x + 2 x y)

(x^{2} - y^{2} + x - 4 y) + i (y + 4 x + 2 x y) = 5 + 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} + x - 4 y = 5 y + 4 x + 2 x y = 1]

[x^{2} - y^{2} + x - 4 y = 5 y + 4 x + 2 x y = 1] : (x = 1, x = - 2, y = - 1 y = - 3)

(x = 1, x = - 2, y = - 1 y = - 3)

z^{2} + (1 - i) z + 1 - i = 0

\frac{1}{i C} = - 0.035 i

x + i y + i x + y = 2 + i

(3 x - 2 y i)^{2} = \frac{200}{4 + 3 i}

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