Multiple choice

If $\displaystyle \alpha $ and $\displaystyle \beta $ are the roots of the equation $\displaystyle { x }^{ 2 }-2x+4=0$, then the value of $\displaystyle { \alpha }^{ n }+{ \beta }^{ n }$ will be:

  1. $\displaystyle i{ 2 }^{ n+1 }\sin { \left( { n\pi }/{ 3 } \right) } $
  2. $\displaystyle { 2 }^{ n+1 }\cos { \left( { n\pi }/{ 3 } \right) } $
  3. $\displaystyle i{ 2 }^{ n-1 }\sin { \left( { n\pi }/{ 3 } \right) } $
  4. $\displaystyle { 2 }^{ n-1 }\cos { \left( { n\pi }/{ 3 } \right) } $
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B Correct answer
Explanation

The roots of x^2 - 2x + 4 = 0 are x = (2 +/- sqrt(4 - 16))/2 = 1 +/- i*sqrt(3). In polar form, these are 2(cos(pi/3) +/- i*sin(pi/3)). By De Moivre's Theorem, alpha^n + beta^n = 2^n * (cos(n*pi/3) + i*sin(n*pi/3)) + 2^n * (cos(n*pi/3) - i*sin(n*pi/3)) = 2 * 2^n * cos(n*pi/3) = 2^(n+1) * cos(n*pi/3).

AI explanation

Using the quadratic formula on x^2 - 2x + 4 = 0, the roots alpha and beta are found to be 1 + i*sqrt(3) and 1 - i*sqrt(3). These complex numbers can be written in polar form as 2*cos(pi/3) plus or minus i*2*sin(pi/3), or simply 2*e^(plus or minus i*pi/3). Applying De Moivre's theorem, the nth power of these roots is 2^n * e^(plus or minus i*n*pi/3). Adding the two roots together gives 2^n * 2 * cos(n*pi/3), which simplifies to 2^(n+1) * cos(n*pi/3).

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