Multiple choice

let $\displaystyle a= e^{i\frac{2\pi }{13}}$ then the quadratic equation whose roots are $\displaystyle \alpha = a+a^{3}+a^{4}+a^{-4}+a^{-3}+a^{-1},\beta = a^{2}+a^{5}+a^{6}+a^{-6}+b^{-5}+a^{-2}$ is given by

  1. $\displaystyle x^{2}-x-3= 0$
  2. $\displaystyle x^{2}-x+2= 0$
  3. $\displaystyle x^{2}+x+3= 0$
  4. $\displaystyle x^{2}+x-3= 0$
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AI explanation

Let a = e^(i*2*pi/13), making it a primitive 13th root of unity, so the sum 1 + a + a^2 + ... + a^12 = 0. We group the sum into alpha + beta + 1 = 0, which means the sum of the roots is alpha + beta = -1. Using the properties of the minimal polynomial over the quadratic subfield, the product of these periods is alpha*beta = -3. A quadratic equation with these roots is x^2 - (sum)x + product = 0, resulting in x^2 + x - 3 = 0.

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