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# If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2px+\left ( p^{2} + q^{2}\right )=0$ and $tan \; \theta =\large \frac{q}{y+p}$ show that $\large\frac{\left (y+\alpha \right )^{n}-\left ( y+\beta \right )^{n}}{\alpha -\beta }$ = $q^{n-1}\large\frac{sin \;n\theta }{sin\;^{n}\theta }$

n N

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 the equation $x^2-2px+(p^2+q^2)=0$
Solving $x=\large\frac{2p\pm\sqrt{4p^2-4(p^2+q^2)}}{2}$
$\Rightarrow \large\frac{2p\pm i2q}{2}$
$\Rightarrow p\pm iq$
Step 2:
Let $\alpha=p+iq,$ then $\beta=p-iq\Rightarrow \alpha-\beta=2qi$
We also have $\tan\theta=\large\frac{q}{y+p}$
$\Rightarrow y+p=\large\frac{q}{\tan\theta}=\large\frac{q\cos\theta}{\sin\theta}$
Step 3:
Now $\large\frac{(y+\alpha)^n-(y+\beta)^n}{\alpha-\beta}=\large\frac{(y+p+iq)^n-(y+p-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{(\Large\frac{q\cos\theta}{\sin\theta}\normalsize+iq)^n-(\Large\frac{q\cos\theta}{\sin\theta}\normalsize-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{(\cos \theta+i\sin\theta)^n-(\cos\theta-i\sin\theta)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{\cos n\theta+i\sin n\theta-\cos n\theta+i\sin n\theta}{2qi}$
$\qquad\qquad=\large\frac{q^n}{2qi\sin^n\theta}$$2i\sin n\theta \qquad\qquad=\large\frac{q^n}{q\sin^n\theta}$$\sin n\theta$