已知z=x+yi(x,y∈R),且 2x+y+ilog2x-8=(1-log2y)i,则z=(  )A.2+iB.1+2iC.2+i或1+2iD.无

2025-06-27 07:08:15
推荐回答(1个)
回答1:

由 2x+y+ilog2x-8=(1-log2y)i,得:2x+y-8+ilog2x=(1-log2y)i
所以

2x+y?8=0
log2x=1?log2y
,即
2x+y23
log2xy=1
,也就是
x+y=3
xy=2

解得:
x=2
y=1
,或
x=1
y=2

由z=x+yi,所以,z=2+i或1+2i.
故选C.