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]....