Question

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

• A. $i{2}^{n+1}\sin \left(n\pi /3\right)$
• B. $2^{n+1}\cos \left(n\pi /3\right)$
• C. $i{2}^{n-1}\sin \left(n\pi /3\right)$
• D. $2^{n-1}\cos \left(n\pi /3\right)$

Answer 1

The correct option is B $2^{n+1}\cos \left(n\pi /3\right)$

Since, $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$

$\therefore \alpha +\beta =2$ and $\alpha \beta =4$

Now, $(\alpha -\beta )=\sqrt{{(\alpha +\beta )}^2-4\alpha \beta }$

$=\sqrt{4-16}=2\sqrt{3}i$

On solving, we get

$2\alpha =2+2\sqrt{3}i$

$\Rightarrow \alpha =2(\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

$=2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$

and $\beta =\frac{2-2\sqrt{3}i}{2}=2(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})$

$\therefore {\alpha }^n+{\beta }^n={\left[2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\right]}^n+{\left[2(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})\right]}^n$

$=2^n\left[2\cos \frac{n\pi }{3}\right]=2^{n+1}\cos \frac{n\pi }{3}$