Explanation

(b)
The given equation is $x^{2} - 2x\cos{\phi + 1 = 0.}$
$\therefore x = \frac{2\cos{\phi \pm \sqrt{4\cos^{2}{\phi - 4}}}}{2} = \cos{\phi \pm i\sin\phi}$
Let $\alpha = \cos{\phi +}i\sin{\phi,\ then\ }\beta = \cos{\phi -}i\sin\phi$
$\therefore\ \alpha^{n} + \beta^{n} = (\cos{n\phi +}i{\sin{\phi)}}^{n} + \left( \cos{\phi -}i\sin\phi \right)^{n}$
$= 2\cos{n\phi}$
and $\alpha^{n}\beta^{n} = (\cos{n\phi +}i\sin{n\phi)(cos}\ n\phi - i\sin{n\phi)}$
$= \cos^{2}{n\phi +}\sin^{2}{n\phi = 1}$
$\therefore$ Required equation is $x^{2} - 2x\cos{n\ \phi + 1 = 0}$