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Question

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


A
18
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B
54
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C
6
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D
12
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Solution

The correct option is D $$12$$
$$\displaystyle z^2+z+1=0 \Rightarrow z = \omega, \omega^2$$
$$\displaystyle \therefore (z+\frac {1}{z})^2 + (z^2 + \frac {1}{z^2})^2 + ...... (z^6 + \frac {1}{z^6})^2$$
$$\displaystyle = (\omega + \omega^2)^2  + (\omega + \omega^2)^2 + (\omega^3 + \frac{1}{\omega^3})^2 + (\omega + \omega^2)^2 + (\omega + \omega^2)^2 + (\omega^6 + \frac {1}{\omega^6})^2$$
$$\displaystyle 4 (\omega + \omega^2)^2 + 2(\omega^3 + \frac{1}{\omega^3})^2 = 4(1) + 2(2^2) = 4 + 8 = 12$$.

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