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

$\begin{array}{l}\because z^2+z+1=0\\ \therefore z=\omega ,{\omega }^2\\ \because z+\frac{1}{z}=\omega +\frac{1}{\omega }=\omega +{\omega }^2=-1\\ \begin{array}{l}\Rightarrow z^2+\frac{1}{z^2}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1\\ \therefore z^3+\frac{1}{z^3}={\omega }^3+\frac{1}{{\omega }^3}=1+1=2\\ z^4+\frac{1}{z^4}={\omega }^4+\frac{1}{{\omega }^4}=\omega +\frac{1}{\omega }=-1\\ z^5+\frac{1}{z^5}={\omega }^5+\frac{1}{{\omega }^5}={\omega }^2+\omega =-1\\ \end{array}\end{array}$

and $z^6+\frac{1}{z^6}={\omega }^6+\frac{1}{{\omega }^6}=2$

$\therefore$ Required sum $=(-1{)}^2+(-1{)}^2+(2{)}^2+(-1{)}^2+(-1{)}^2+(2{)}^2=12$