Question Number 100730 by john santu last updated on 28/Jun/20

((2z^2 )/(z^2 +∣z+1∣)) < 1

Commented bybemath last updated on 28/Jun/20

since z^2 +∣z+1∣ > 0 then 2z^2  < z^2 +∣z+1∣  z^2  > ∣z+1∣ ⇒(z^2 +z+1)(z^2 −z−1) <0  first term z^2 +z+1 > 0 for ∀z∈R  so z^2 −z−1 <0  (z−(1/2))^2 −(5/4) < 0  (z−(1/2)−((√5)/2))(z−(1/2)+((√5)/2)) <0  ⇔∴ (1/2)−((√5)/2) < z < (1/2)+((√5)/2)