If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2px + \left(p^2 + q^2\right) = 0$ and $\tan \theta = \dfrac{q}{y + p}$, show that $\dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} = q^{n - 1} \left(\dfrac{\sin n\theta}{\sin^n \theta}\right)$
The roots of the given quadratic equation $\;$ $x^2 - 2px + \left(p^2 + q^2\right) = 0$ $\;$ are
$\begin{aligned}
x & = \dfrac{2p \pm \sqrt{4p^2 - 4p^2 - 4q^2}}{2} \\\\
& = \dfrac{2p \pm 2iq}{2} \\\\
& = p \pm iq
\end{aligned}$
$\because$ $\;$ $\alpha$ and $\beta$ are the roots of the given quadratic equation, let
$\alpha = p + i q$ $\;$ and $\;$ $\beta = p - iq$
Given: $\;$ $\tan \theta = \dfrac{q}{y + p}$
$\implies$ $y = \dfrac{q}{\tan \theta} - p$
$\begin{aligned}
\therefore \; \left(y + \alpha\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p + iq\right)^n \\\\
& = \left[q \left(\dfrac{1}{\tan \theta} + i\right)\right]^n \\\\
& = q^n \left[\dfrac{\cos \theta + i \sin \theta}{\sin \theta}\right]^n \\\\
& = \dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\left(y + \beta\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p - iq\right)^n \\\\
& = \left[q \left(\dfrac{1}{\tan \theta} - i\right)\right]^n \\\\
& = q^n \left[\dfrac{\cos \theta - i \sin \theta}{\sin \theta}\right]^n \\\\
& = \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (2)
\end{aligned}$
$\alpha - \beta = p + iq - \left(p - iq\right) = 2 \; i \; q$ $\;\;\; \cdots \; (3)$
$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,
$\begin{aligned}
\dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} & = \dfrac{\dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} - \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta}}{2 \;i \;q} \\\\
& = \dfrac{q^n \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]}{2 \;i \;q \; \sin^n \theta} \\\\
& = \dfrac{2 \; i \; q^n \; \sin \left(n \theta\right)}{2 \; i \; q \; \sin^n \theta} \\\\
& = q^{n - 1} \left[\dfrac{\sin \left(n \theta\right)}{\sin^n \theta}\right]
\end{aligned}$
Hence proved.