Multiple choice

Let $\displaystyle \alpha ,\beta $ be the roots of $\displaystyle { x }^{ 2 }-2x\cos { \phi } +1=0$, then the equation whose roots are $\displaystyle { \alpha }^{ n },{ \beta }^{ n }$ is

  1. $\displaystyle { x }^{ 2 }-2x\cos { n\phi } -1=0$
  2. $\displaystyle { x }^{ 2 }-2x\cos { n\phi } +1=0$
  3. $\displaystyle { x }^{ 2 }-2x\sin { n\phi } +1=0$
  4. $\displaystyle { x }^{ 2 }+2x\sin { n\phi } -1=0$
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B Correct answer
Explanation

Roots are cos(phi) +/- i*sin(phi) = e^(i*phi) and e^(-i*phi). The roots of the new equation are e^(in*phi) and e^(-in*phi). The equation is x^2 - (sum)x + (product) = 0. Sum = 2*cos(n*phi), Product = 1. Equation: x^2 - 2*x*cos(n*phi) + 1 = 0.

AI explanation

Let the roots alpha and beta be cos(phi) + i sin(phi) and cos(phi) - i sin(phi). Raising these roots to the nth power using De Moivre's Theorem gives alpha^n = cos(n phi) + i sin(n phi) and beta^n = cos(n phi) - i sin(n phi). The sum of the new roots is alpha^n + beta^n = 2 cos(n phi) and the product is alpha^n * beta^n = cos^2(n phi) + sin^2(n phi) = 1. Forming the quadratic equation with these roots gives x^2 - (sum)x + product = 0, which simplifies to x^2 - 2x cos(n phi) + 1 = 0.

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