Complex no. inequality

[man111]

If [tex]Z[/tex] is a Complex no. such that [tex]|Z|=1[/tex]. Then show that [tex]\sqrt{2}\leq |1-z|+|1+z^2|\leq 4[/tex]

[drago]

From above:
[tex]sup_z(|z-1|+|z^2-(-1)|) \le sup_u(|u-1|)+sup_v(v-(-1))=2+2=4; (|z|=|u|=|v|=1);[/tex]
Becomes equality when z=-1:
Below:
we use:[tex]|z^2-z|=|z||z-1|=|z-1|, when |z|=1[/tex]
[tex]|z-1|+|z^2-(-1)|)\ge min \{ |z-1| , |z-z^2|+|z^2-(-1)|\ } \ge min \{|z-1|, |z-(-1)|\} \ge \sqrt{2}.[/tex]
Becomes equality when z=i, z=-i.