iy(2-i)+1=4i-x(1-i)

解答

i y (2 - i) + 1 = 4 i - x (1 - i)

x = - 2, y = 1

求解步骤

i y (2 - i) + 1 = 4 i - x (1 - i)

乘开

(y + 1) + 2 i y = - x + i (4 + x)

复数仅在实部和虚部均相等时才相等改写为方程组:

[y + 1 = - x 2 y = 4 + x]

[y + 1 = - x 2 y = 4 + x] : x = - 2, y = 1

x = - 2, y = 1

x^{2} + (i - 6) x + 10 = 0

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

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

(ii) 6 x^{2} - 24 = 0

x^{2} + (- 2 + 3 i) x - 2 - 4 i = 0