TS EAMCET · Maths · Complex Number
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
- B -2020
- C 4038
- D \(2020+673 i\)
Answer & Solution
Correct Answer
(B) -2020
Step-by-step Solution
Detailed explanation
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 \(z+\frac{1}{Z}=\omega+\frac{1}{\omega}=\omega+\omega^2=-1\) \(z^2+\frac{1}{z^2}=\omega^2+\frac{1}{\omega^2}=\omega^2+\omega=-1\)…
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