If $$z^2 + z +1 = 0,$$    where $$z$$ is complex number, then the value of $${\left( {z + \frac{1}{z}} \right)^2} + {\left( {{z^2} + \frac{1}{{{z^2}}}} \right)^2} + {\left( {{z^3} + \frac{1}{{{z^3}}}} \right)^2} + ..... + {\left( {{z^6} + \frac{1}{{{z^6}}}} \right)^2}\,{\text{is}}$$

Solution :
$$\eqalign{ & {z^2} + z + 1 = 0 \cr & \Rightarrow z = \omega \,\,{\text{or }}{\omega ^2} \cr & {\text{So, }}z + \frac{1}{z} = \omega + {\omega ^2} = - 1 \cr & {z^2} + \frac{1}{{{z^2}}} = {\omega ^2} + \omega = - 1, \cr & {z^3} + \frac{1}{{{z^3}}} = {\omega ^3} + {\omega ^3} = 2 \cr & {z^4} + \frac{1}{{{z^4}}} = - 1,{z^5} + \frac{1}{{{z^5}}} = - 1{\text{ and }}{z^6} + \frac{1}{{{z^6}}} = 2 \cr} $$
∴ The given sum $$ = 1 + 1 + 4 +1 +1 + 4 = 12$$