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# If $\alpha$ and $\beta$ are the roots of $x^{2}-2x+4=0$. Prove that $\alpha ^{n}-\beta ^{n}=i2^{n+1}sin\large\frac{n\pi }{3}$ and calculate $\alpha ^{9}-\beta ^{9}$, where n $\in$ N?

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A)
Toolbox:
• From De moivre's theorem we have
• (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
• (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
• (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
• (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
• $e^{i\theta}=\cos\theta+i\sin\theta$
• $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
$\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$
Solving the equation $x=\large\frac{2\pm\sqrt{4-16}}{2}$
$\Rightarrow \large\frac{2\pm 2i\sqrt 3}{2}$
$\Rightarrow 1\pm i\sqrt 3$
Step 2:
Let $\alpha=1+i\sqrt 3$ and $\beta=1-i\sqrt 3$
Let $1+i\sqrt 3=r(\cos\theta+i\sin\theta).$
Equating the real and imaginary parts separately,
$r\cos\theta=1$
$r\sin\theta=\sqrt 3$
Step 3:
Squaring and adding $r^2=1+3=4$
$\Rightarrow r=2$
Also $\tan^{-1}\sqrt 3=\large\frac{\pi}{3}$
Therefore $\alpha=2(\cos\large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3}) \beta=2(\cos\large\frac{\pi}{3}$$-i\sin\large\frac{\pi}{3})$
Step 4:
$\alpha^n-\beta^n=2^n(\cos\large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})^n$$-2^n(\cos\large\frac{\pi}{3}-$$i\sin\large\frac{\pi}{3})^n \qquad\quad\;=2^n[(\cos\large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})^n$$-(\cos\large\frac{\pi}{3}-$$i\sin\large\frac{\pi}{3})^n$
$\qquad\quad\;=2^n.2i\sin\large\frac{n\pi}{3}$
$\qquad\quad\;=i2^{n+1}\sin\large\frac{n\pi}{3}$
Hence proved.
Step 5:
$\alpha^9-\beta^9=i2^{9+1}\sin\large\frac{9\pi}{3}$
$\qquad\quad\;=i2^{10}\sin 3\pi$
$\qquad\quad\;=0$