z^2=(cosθ)^2-(sinθ)^2+2icosθsinθ

= cos 2θ+is N2θ，

Z^2+Z=cos2θ+cosθ+i(sin2θ+sinθ)

= 2 cos(3θ/2)cos(θ/2)+2 isin(3θ/2)cos(θ/2)

=2cos(θ/2)[cos(3θ/2)+isin(3θ/2)，

So | z 2+z | =| 2cos (θ/2) |,

Because1>; =|cos(θ/2)| >=0, so 2 >; =|z^2+z|>； =0