First slide

Algebra of complex numbers

Question

If $z^{2} + z + 1 = 0,$ where $z$ is a complex number, then values of $S = {(z + \frac{1}{z})}^{2} + {(z^{2} + \frac{1}{z^{2}})}^{2} + {(z^{3} + \frac{1}{z^{3}})}^{2} + \dots + {(z^{6} + \frac{1}{z^{6}})}^{2}$ is

Moderate

A

12
B

18
C

54
D

6

Solution

$z^{2} + z + 1 = 0 \Rightarrow z = ω, ω^{2} .$
Let $z = ω$ then ${(ω + ω^{2})}^{2}$
$\begin{array}{l} = {(ω^{2} + ω^{4})}^{2} = {(ω^{4} + ω^{8})}^{2} \\ = {(ω^{5} + ω^{10})}^{2} = 1 \end{array}$
and ${(ω^{3} + ω^{6})}^{2} = {(ω^{6} + ω^{12})}^{2} = 4 .$
Thus, $S = 12 .$

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