Question

If $z^2+z+1=0$ then the value of $(z+\frac{1}{z}{)}^2+(z^2+\frac{1}{z^2}{)}^2+(z^3+\frac{1}{z^3}{)}^2+........+(z^{21}+\frac{1}{z^{21}}{)}^2$ is equal to

• A. $21$
• B. $42$
• C. $0$
• D. $11$

Answer 1

The correct option is A $21$

$\begin{array}{c}\text{Given,}{z}^2+z+1=0\\ \text{Then}\\ \begin{array}{c}{(2+\frac{1}{2})}^2+{(z^2+\frac{1}{z^2})}^2+{(z^3+\frac{1}{z^3})}^2+\cdots +{(z^{21}+\frac{1}{z^2})}^2\\ \text{let}z=w\\ \Rightarrow w^2+w+1=0\end{array}\end{array}$

$\begin{array}{c}\Rightarrow {(w+\frac{1}{\omega })}^2+{(w^2+\frac{1}{{\omega }^2})}^2+{({\omega }^3+\frac{1}{{\omega }^3})}^2+\dots +{({\omega }^{21}+\frac{1}{{\omega }^{21}})}^2\\ \Rightarrow {(\omega +{\omega }^2)}^2+{({\omega }^2+\omega )}^2+(1+1{)}^2+\dots +{({\left({\omega }^3\right)}^7+\frac{1}{{\left({\omega }^3\right)}^7})}^2\\ \Rightarrow (-1{)}^2+(-1{)}^2+2^2+\dots +2^2\\ \Rightarrow 14(-1{)}^2+7(2{)}^2\\ \Rightarrow 14+28=42\\ \text{Hence (B) is the correct option}\end{array}$