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

(A) 18

(B) 54

(C) 6

(D) 12

Toolbox:
• $1+\omega+\omega^2=0$ where $\omega$ is cube root of 1
• $\omega^3=\omega^6=\omega^9=............=1$
• $\omega=\omega^4=\omega^7=..................$
• $\omega^2=\omega^5=\omega^8=.....................$
Since $1+z+z^2=0,$ $z=\omega$
Dividing both the sides by $z$
$\Rightarrow\:z+\large\frac{1}{z}$$=-1 z^2+\large\frac{1}{z^2}$$=(z+\large\frac{1}{z}$$)^2-2=1-2=-1 z^3+\large\frac{1}{z^3}$$=1+1=2$
$z^4+\large\frac{1}{z^4}$$=z+\large\frac{1}{z}$$=-1$
and so on