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If $z^2+z+1=0$, where $z$ is a complex number, then the value of
$$
\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\cdots+\left(z^6+\frac{1}{z^6}\right)^2
$$
Options:
$$
\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\cdots+\left(z^6+\frac{1}{z^6}\right)^2
$$
Solution:
1646 Upvotes
Verified Answer
The correct answer is:
12
12
$z^2+z+1=0 \quad \Rightarrow z=\omega$ or $\omega^2$ so, $z+\frac{1}{z}=\omega+\omega^2=-1, z^2+\frac{1}{z^2}=\omega^2+\omega=-1, z^3+\frac{1}{z^3}=\omega^3+\omega^3=2$ $z^4+\frac{1}{z^4}=-1, z^5+\frac{1}{z^5}=-1$ and $z^6+\frac{1}{z^6}=2$ $\therefore$ The given sum $=1+1+4+1+1+4=12$
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