If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$ 
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate: 
  $$z^2 + \frac{4}{z^2} = -3$$

$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?

$z^2+z+2=0$ means $z^2=-z-2$  and  $1+z+\dfrac 2z=0$ so $\dfrac2z=-1-z$ so $\dfrac4{z^2}=z^2+2z+1$.
Therefore, $z^2+\dfrac4{z^2}=z^2+z-1=z^2+z+2-3=-3$.
