设x,y是复数,满足（x+yi)i-2+4i=(x-yi)(1-i),求x,y设x,y是共轭复数,满足（x+y）^2-3xyi=4-6i,求x,y

Question

设x,y是复数,满足（x+yi)i-2+4i=(x-yi)(1-i),求x,y
设x,y是共轭复数,满足（x+y）^2-3xyi=4-6i,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