**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:

$\alpha^9-\beta^9=i2^{9+1}\sin\large\frac{9\pi}{3}$

$\qquad\quad\;=i2^{10}\sin 3\pi$

$\qquad\quad\;=0$