ASK - ANSWER - COMMENT - VOTE - CREATE

MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
25 views
If $z^{2}+z+1=0$, where $z$ is a complex number, then the value of

\begin{aligned}\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}+\\ \ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2} & \text { is } \end{aligned}

1) 18
2) 54
3) 6
4) 12
| 25 views

0 like 0 dislike

(4)

Explanation

$z^{2}+z+1=0 \quad \Rightarrow z=\omega$ 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 $1+1+4+1+1+4=12$

by Diamond (75,914 points)

0 like 0 dislike