If $z^2+z+1=0$ then $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2=$ where $\mathrm{z}=\mathrm{w}=$ complex cube root of unity

Question

Aicharya · Accepted Answer

Z is a complex cube root of unity $	herefore \quad z^3=1...(i)$ and $1+z+z^2=0$ $	herefore \quad z^2+1=-z$...(ii) Consider, $\begin{aligned} & \left(z^3+\frac{1}{z^3}ight)^2+\left(z^4+\frac{1}{z^4}ight)^2 \ & =\left(1+\frac{1}{1}ight)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}ight)^2 \ & =4+\left[z+\frac{1}{z}ight]^2 \end{aligned}$...[From (i)] $\begin{aligned} & =4+\left[\frac{z^2+1}{z}ight]^2 \ & =4+\left[\frac{-z}{z}ight]^2 ...[From(ii)]\ & =4+(-1)^2 =5 \end{aligned}$

If $z^2+z+1=0$ then $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2=$ where…

Solution