Question

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

Answer 1

Given equation is $z^2+z+1=0$
$\Rightarrow z=\omega ,{\omega }^2$

Now, $(z+\frac{1}{z})+(z^2+\frac{1}{z^2})+(z^3+\frac{1}{z^3})+\cdots +(z^6+\frac{1}{z^6})$
$=(\omega +{\omega }^2)+({\omega }^2+\omega )+({\omega }^3+{\omega }^{-3})+(\omega +{\omega }^2)+({\omega }^2+\omega )+({\omega }^6+{\omega }^{-6})$
Using $1+\omega +{\omega }^2=0$
$=-1-1+(1+1)-1-1+(1+1)=0$