If a = cos (2pi /7) + isin (2pi /7),then the | vedclass

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(D) Given $a = \cos (2\pi /7) + i\sin (2\pi /7)$.Since $a^7 = \cos (2\pi) + i\sin (2\pi) = 1$,we have $a^7 = 1$.Sum of roots $S = \alpha + \beta = (a

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$x^2 + x + 2 = 0$

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(D) Given $a = \cos (2\pi /7) + i\sin (2\pi /7)$.Since $a^7 = \cos (2\pi) + i\sin (2\pi) = 1$,we have $a^7 = 1$.Sum of roots $S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6) = a + a^2 + a^3 + a^4 + a^5 + a^6$.Using the sum of a geometric series,$S = \frac{a(1 - a^6)}{1 - a} = \frac{a - a^7}{1 - a} = \frac{a - 1}{1 - a} = -1$.Product of roots $P = \alpha \beta = (a + a^2 + a^4)(a^3 + a^5 + a^6) = a^4 + a^6 + a^7 + a^5 + a^7 + a^8 + a^7 + a^9 + a^{10}$.Using $a^7 = 1$,we get $P = a^4 + a^6 + 1 + a^5 + 1 + a + 1 + a^2 + a^3 = 3 + (a + a^2 + a^3 + a^4 + a^5 + a^6) = 3 + S$.Substituting $S = -1$,we get $P = 3 - 1 = 2$.The quadratic equation is $x^2 - Sx + P = 0$,which becomes $x^2 - (-1)x + 2 = 0$,or $x^2 + x + 2 = 0$.

If $a = \cos (2\pi /7) + i\sin (2\pi /7)$,then the quadratic equation whose roots are $\alpha = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$ is

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