Question

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

• A. 18
• B. 54
• C. 6
• D. 12

Answer 1

The correct option is D 12

$z^2+z+1=0\Rightarrow z=\omega or{\omega }^2,$ where $\omega$ is a complex cube root of unity.
$\therefore z+\frac{1}{z}=\omega +{\omega }^2=-1z^2+\frac{1}{z^2}={\omega }^2+\omega =-1,z^3+\frac{1}{z^3}={\omega }^3+{\omega }^3=2,z^4+\frac{1}{z^4}=-1,z^5+\frac{1}{z^5}=-1,and{z}^6+\frac{1}{z^6}=2$
Hence, the given sum is
$1+1+4+1+1+4=12$