數(shù)學(xué)已知z為虛數(shù),z+z-2分之9為實(shí)數(shù),①若z-2為純虛數(shù),求虛數(shù)z ②求lz-4l的取值范圍
1.
z-2為純虛數(shù),設(shè)z-2=bi,則z=2+bi
z+9/(z-2)
=2+bi+9/(bi)
=2+bi-9i/b
=2+(b-9/b)i
因?yàn)閦+9/(z-2)為實(shí)數(shù)
則b=9/b
b=±3
z=2±3i
2.
設(shè)z=x+yi,y≠0,
z+9/(z-2)
=x+yi+9/(x-2+yi)
=x+yi+9(x-2-yi)/[(x-2)^2+y^2]
=x+9(x-2)/[(x-2)^2+y^2]+{y-9y/[(x-2)^2+y^2]}i
因?yàn)閦+9/(z-2)為實(shí)數(shù)
y-9y/[(x-2)^2+y^2]=0,
（x-2)^2+y^2=9,
|z-4|的取值范圍是[1,5].這個范圍是怎么來的