If z^2 + z + 1 = 0 , where z is a complex number, then the value of ≤ft(z + 1/z)^2 + ≤ft(z^2 + 1/z^2)^2 + + ≤ft(z^6 + 1/z^6)^2 is: - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

mediumMCQAIEEE 20062026Mathematics

1 attempts0% success rate1 mark

If z^{2} + z + 1 = 0, where z is a complex number, then the value of (z + \frac{1}{z})^{2} + (z^{2} + \frac{1}{z ^{2}})^{2} + \dots + (z^{6} + \frac{1}{z ^{6}})^{2} is:

Solution & Step-by-step Explanation

The roots of z^{2} + z + 1 = 0 are ω and ω^{2} .Let z = ω . Note that ω^{3} = 1 and 1 + ω + ω^{2} = 0 . k = 1 : (ω + 1/ ω)^{2} = (ω + ω^{2})^{2} = (- 1)^{2} = 1 k = 2 : (ω^{2} + 1/ ω^{2})^{2} = (ω^{2} + ω)^{2} = (- 1)^{2} = 1 k = 3 : (ω^{3} + 1/ ω^{3})^{2} = (1 + 1)^{2} = 4 k = 4 : (ω^{4} + 1/ ω^{4})^{2} = (ω + ω^{2})^{2} = 1 k = 5 : (ω^{5} + 1/ ω^{5})^{2} = (ω^{2} + ω)^{2} = 1 k = 6 : (ω^{6} + 1/ ω^{6})^{2} = (1 + 1)^{2} = 4 Total Sum = 1 + 1 + 4 + 1 + 1 + 4 = 12 .

Try it yourself before checking the explanation above.

If z^{2} + z + 1 = 0, where z is a complex number, then the value of (z + \frac{1}{z})^{2} + (z^{2} + \frac{1}{z ^{2}})^{2} + \dots + (z^{6} + \frac{1}{z ^{6}})^{2} is:

18

54

6

12

Practice with timed mock tests and track your performance across Mathematics.