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Question: If a = cos \(\frac{2\pi}{7}\) + i sin \(\frac{2\pi}{7}\) and p = a + a<sup>2</sup> + a<sup>4</sup>, ...

If a = cos 2π7\frac{2\pi}{7} + i sin 2π7\frac{2\pi}{7} and p = a + a2 + a4, q = a3 + a5 + a6 then the equation whose roots are p and q –

A

x2 + x + 4 = 0

B

x2 + x + 2 = 0

C

x2 + x – 2 = 0

D

x2 + x – 4 = 0

Answer

x2 + x + 2 = 0

Explanation

Solution

Sol. a7 = cos 2p + i sin 2p = 1

Ž sum = a + a 2 + a 3 + a 4 + a 5 + a6

= α(1−α6)1−α\frac{\alpha(1 - \alpha^{6})}{1 - \alpha}

s = α−α71−α\frac{\alpha - \alpha^{7}}{1 - \alpha}= α−11−α\frac{\alpha - 1}{1 - \alpha} = –1

Product of roots

pq = (a 4 + a 6 + a 7) + (a 5 + a 7 + a 8) + (a 7 + a 9 + a10)

= (a 4 + a 6 + 1) + (a 5 + 1 + a) + (1 + a 2 + a 3)

= 3 + (a + a2 + …. + a6) = 3 + (–1)

= 3 – 1 = 2

Equation x2 + x + 2 = 0.