4x+4iy-23=5ix+5i^2y+2i

解答

4 x + 4 i y - 23 = 5 i x + 5 i^{2} y + 2 i

x = 2, y = 3

求解步骤

4 x + 4 i y - 23 = 5 i x + 5 i^{2} y + 2 i

化简

(4 x - 23) + 4 i y = - 5 y + i (2 + 5 x)

复数仅在实部和虚部均相等时才相等改写为方程组:

[4 x - 23 = - 5 y 4 y = 2 + 5 x]

[4 x - 23 = - 5 y 4 y = 2 + 5 x] : x = 2, y = 3

x = 2, y = 3

z^{2} - (3 + 2 i) z + (1 + 3 i) = 0

(x - (3 - 4 i)) (x - (3 + 4 i)) = 0

(x + i) (3 - i y) = 1 + 13 i

(a - bi) + (b + ai) = 5 + i

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