数学已知z为虚数,z+z-2分之9为实数,①若z-2为纯虚数,求虚数z ②求lz-4l的取值范围
问题描述:
数学已知z为虚数,z+z-2分之9为实数,①若z-2为纯虚数,求虚数z ②求lz-4l的取值范围
答
1.
z-2为纯虚数,设z-2=bi,则z=2+bi
z+9/(z-2)
=2+bi+9/(bi)
=2+bi-9i/b
=2+(b-9/b)i
因为z+9/(z-2)为实数
则b=9/b
b=±3
z=2±3i
2.
设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
因为z+9/(z-2)为实数
y-9y/[(x-2)^2+y^2]=0,
(x-2)^2+y^2=9,
|z-4|的取值范围是[1,5].
答
z-2为纯虚数则设z-2=bi,z=2+biz+z-2分之9=2+bi+9/(bi)=2+bi-9/bi=2+(b-9/b)i为实数则b=9/bb=±3z=2±3i设z=x+yi,y≠0,z+9/(z-2)=x+yi+9/(x-2+yi),y-9y/[(x-2)^2+y^2]=0,(x-2)^2+y^2=9,|z-4|的取值范围是[1,5]....