Q50 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If $z$ is a complex number such that $z^2+z+1=0$, then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3$ $+\left(z^3+\frac{1}{z^3}\right)^3+\ldots . .+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$

  • A
    4037
  • Correct answer
    B
    -2020
  • C
    4038
  • D
    $2020+673 i$

Solution

Given, $z^2+z+1=0$ $\therefore \quad z=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \therefore z=\omega, \omega^2$ Now $z+\frac{1}{Z}=\omega+\frac{1}{\omega}=\omega+\omega^2=-1$ $z^2+\frac{1}{z^2}=\omega^2+\frac{1}{\omega^2}=\omega^2+\omega=-1$ $\therefore\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3$ $\Rightarrow(-1)^3+(-1)^3+\ldots+(-1)^3=-2020$

Hence, the correct answer is option: 'B'.

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