If z^2+z+1=0, where z is a complex number, then the value of (z+1z)^2+(z^2+1z^2)^2+(z^3+1z…

Question

If $z^2+z+1=0$, where $z$ is a complex number, then the value of $\left(z+\dfrac1z\right)^2+\left(z^2+\dfrac1{z^2}\right)^2+\left(z^3+\dfrac1{z^3}\right)^2+\cdots+\left(z^6+\dfrac1{z^6}\right)^2$ is

Accepted Answer

$z=\omega$ or $\omega^2$, so $\dfrac1z=\omega^2$ or $\omega$.

$z+\dfrac1z=-1$, $z^2+\dfrac1{z^2}=-1$, $z^3+\dfrac1{z^3}=2$, $z^4+\dfrac1{z^4}=-1$, $z^5+\dfrac1{z^5}=-1$, $z^6+\dfrac1{z^6}=2$.

$$\text{Sum}=1+1+4+1+1+4=12.$$

Solutions · 1