<s>[INST] 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?
 [/INST] Since $z \neq 0$, you have
$$\begin{equation}\begin{aligned}
& z^2 + z + 2 = 0 \\
& z + 1 + \frac{2}{z} = 0 \\
& z + \frac{2}{z} = -1 \\
& \left(z + \frac{2}{z}\right)^2 = (-1)^2 \\
& z^2 + 4 + \frac{4}{z^2} = 1 \\
& z^2 + \frac{4}{z^2} = -3
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
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