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1.z-2为纯虚数,设z-2=bi,则z=2+biz+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/bb=±3z=2±3i2.设z=x+yi,y≠0,z+9/(
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![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1.z-2为纯虚数,设z-2=bi,则z=2+biz+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/bb=±3z=2±3i2.设z=x+yi,y≠0,z+9/(](/uploads/image/z/6147480-48-0.jpg?t=%E6%95%B0%E5%AD%A6%E5%B7%B2%E7%9F%A5z%E4%B8%BA%E8%99%9A%E6%95%B0%2Cz%2Bz-2%E5%88%86%E4%B9%8B9%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E2%91%A0%E8%8B%A5z-2%E4%B8%BA%E7%BA%AF%E8%99%9A%E6%95%B0%2C%E6%B1%82%E8%99%9A%E6%95%B0z+%E2%91%A1%E6%B1%82lz-4l%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B41.z-2%E4%B8%BA%E7%BA%AF%E8%99%9A%E6%95%B0%2C%E8%AE%BEz-2%3Dbi%2C%E5%88%99z%3D2%2Bbiz%2B9%2F%28z-2%29%3D2%2Bbi%2B9%2F%28bi%29%3D2%2Bbi-9i%2Fb%3D2%2B%28b-9%2Fb%29i%E5%9B%A0%E4%B8%BAz%2B9%2F%28z-2%29%E4%B8%BA%E5%AE%9E%E6%95%B0%E5%88%99b%3D9%2Fbb%3D%C2%B13z%3D2%C2%B13i2.%E8%AE%BEz%3Dx%2Byi%2Cy%E2%89%A00%2Cz%2B9%2F%28)
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1.z-2为纯虚数,设z-2=bi,则z=2+biz+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/bb=±3z=2±3i2.设z=x+yi,y≠0,z+9/(
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
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].这个范围是怎么来的
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1.z-2为纯虚数,设z-2=bi,则z=2+biz+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/bb=±3z=2±3i2.设z=x+yi,y≠0,z+9/(
z=x+yi
满足(x-2)^2+y^2=9,
所以z在这个圆上
圆心C(2,0),r=3
而|z-4|就是z和A(4,0)的距离
显然A在园内
所以连接AC,和元交点就是最值点
AC=2
所以最大是r+2=5
最小r-2=1
所以是[1,5]