设|z|=1,且z²+2z+(1/z)

Question

设|z|=1,且z²+2z+(1/z)<0,则z=

Answer 1

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