Question

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$, prove that ${\alpha }^n-{\beta }^n=i{2}^{n+1}\sin \frac{n\pi }{3}$ and deduce ${\alpha }^9-{\beta }^9$.

Answer 1

Given equation is $x^2-2x+4=0$

$\Rightarrow x^2-\frac{b\pm \sqrt{b^2-4ac}}{2a}$
$\Rightarrow x=\frac{2\pm \sqrt{4-16}}{2}=\frac{2\pm i2\sqrt{3}}{2}=1\pm i\sqrt{3}$
$\Rightarrow \alpha =1+i\sqrt{3}$, $\beta =1-i\sqrt{3}$
Let $1+i\sqrt{3}=1$; $r[\cos \theta +i\sin \theta ]$
$\Rightarrow r\cos \theta =1$; $r\sin \theta =\sqrt{3}$
$\Rightarrow r=\sqrt{1+3}=2$
$\therefore \cos \theta =\frac{1}{2},\sin \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{3}$
$\therefore (1+i\sqrt{3})=2[\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}]$
So, $(1-i\sqrt{3})=2[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]$
${\alpha }^n-{\beta }^n={\left\{2[\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}]\right\}}^n-{\left\{2[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]\right\}}^n$
$=2^n[(\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3})-(\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3})]$
$=2^n[\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}-\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}]$
$=2^n\left[2i\sin \frac{n\pi }{3}\right]$
$=i{2}^{n+1}\sin \frac{n\pi }{3}$
Substitute $n=9$, we get
${\alpha }^9-{\beta }^9=i{2}^{10}\sin \frac{9\pi }{3}=i\left(2^{10}\right)\sin 3\pi$
$=i\left(2^{10}\right)\left(0\right)=0$.