Question

虚数Z满足Z的模=1，Z^2+2Z+1/Z<0,求Z

Answer 1

虚数z满足|z|=1,z²+2z+1/z<0,求z.
z^2+2z+1/z<0
说明z^2+2z+1/z是
实数
因此z^2+2z+1/z=z^2+2z+1/z的
共轭复数
z'代表z的共轭复数
z^2+2z+1/z=(z')^2+2z'+1/z'
[z*z'=|z|^2=1]
z^2+2z+z'=(z')^2+2z'+z
z^2-(z')2+z-z'=0
(z-z')(z+z'+1)=0
设z=a+bi
a,b是实数
当z=z'时
a+bi=a-bi
b=0
|a|=1
a=1或-1
所以z=1或-1
需满足z^2+2z+1/z<0
所以z=-1
当z+z'+1=0
2a=-1
a=-1/2
a^2+b^2=1
b=±√3/2
所以z=-1/2±√3/2i
需满足z^2+2z+1/z<0
当z=-1/2+√3/2i
z^2+2z+1/z=-2<0
当z=-1/2-√3/2i
z^2+2z+1/z=-2<0
检验得z=-1/2±√3/2i