Question

设x,y是复数,满足（x+yi)i-2+4i=(x-yi)(1-i),求x,y

Answer 1

x,y是复数,满足（x+yi)i-2+4i=(x-yi)(1-i),求x,y

xi-y-2+4i=x-xi-yi+y

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

x+y=-y-2

x+4=-x-y

x+2y=-2

2x+y=-4

解得:x=-2,y=0

设x,y是共轭复数,满足（x+y）^2-3xyi=4-6i,求x,y

若设x=a+bi ,则y=a-bi 代入方程中,得出a^2=1,b^=1所以可知

若x=1+i,y=1-i

x=-1+i,y=-1-i

x=1-i y=1+i

x=-1-i,y=-1+i