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 $12$

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

$\therefore (z+\frac{1}{z}{)}^2+(z^2+\frac{1}{z^2}{)}^2+......(z^6+\frac{1}{z^6}{)}^2$

$=(\omega +{\omega }^2{)}^2+(\omega +{\omega }^2{)}^2+({\omega }^3+\frac{1}{{\omega }^3}{)}^2+(\omega +{\omega }^2{)}^2+(\omega +{\omega }^2{)}^2+({\omega }^6+\frac{1}{{\omega }^6}{)}^2$

$4(\omega +{\omega }^2{)}^2+2({\omega }^3+\frac{1}{{\omega }^3}{)}^2=4\left(1\right)+2\left(2^2\right)=4+8=12$.