Question

If $\alpha ,\beta$ are the roots of the $x^2-2x+4=0$ then ${\alpha }^n+{\beta }^n$ is equal to

• A. $2^n\cos \frac{n\pi }{3}$
• B. $2^n\cos \frac{(n+1)\pi }{3}$
• C. $2^{n+1}\cos \frac{n\pi }{3}$
• D. $2^{n+1}\cos \frac{(n+1)\pi }{3}$

Answer 1

The correct option is C $2^{n+1}\cos \frac{n\pi }{3}$

$\alpha$&$\beta$ are root of $x^2-2x+4=0$ we know

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4\times 4\times 1}}{2\times 1}$

$\frac{2\pm \sqrt{-12}}{2}$

$=1\pm \sqrt{-3}$ for $\sqrt{-1}=i$

So $\alpha =1+\sqrt{3}i$ $\beta =1-3L$

now convert to

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

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

as well as

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

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

$=2^n\cos \frac{n\pi }{3}+2^{n}i\sin \frac{n\pi }{3}+2^n\cos \frac{n\pi }{3}-2^{n}i\sin \frac{n\pi }{3}$

$={2.2}^n\cos \frac{n\pi }{3}$

${\alpha }^n+{\beta }^n=2^{n+1}\cos \frac{n\pi }{3}$