Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$, such that ${\alpha }^n+{\beta }^n=2^k\cos \frac{n\pi }{3}$, then value of $k$ is

• A. $n-1$
• B. $n$
• C. $n+1$
• D. $2n$

Answer 1

The correct option is C $n+1$

$x^2-2x+4=0$
$\Rightarrow x=\frac{2\pm \sqrt{4-16}}{2}=1\pm i\sqrt{3}$
Let $\alpha =1+i\sqrt{3}=2[\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}]$
and $\beta =1-i\sqrt{3}=2[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]$
So, ${\alpha }^n+{\beta }^n=2^n{[\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}]}^n+2^n{[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]}^n$
$=2^n[\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3}]$
$=2^n\cdot 2\cos \frac{n\pi }{3}=2^{n+1}\cdot \cos \frac{n\pi }{3}$
$\therefore k=n+1$