Let $\mathrm{\alpha },\mathrm{\beta }\text{ be the roots of }{\mathrm{x}}^2-2\mathrm{xcos}\mathrm{\varphi }+1=0$  then the equation whose roots are ${\mathrm{\alpha }}^{\mathrm{n}}\text{ and }{\mathrm{\beta }}^{\mathrm{n}}$ is

Solution:

The given equation is  ${\mathrm{x}}^2-2\mathrm{xcos}\mathrm{\varphi }+1=0$
$\therefore \mathrm{x}=\frac{2\cos \mathrm{\varphi }\pm \sqrt{4{\cos }^2\mathrm{\varphi }-4}}{2}=\cos \mathrm{\varphi }\pm \mathrm{isin}\mathrm{\varphi }$
$$\begin{gathered} \text{ Let }\alpha =\cos \varphi +i\sin \varphi \text{, then }\beta =\cos \varphi -i\sin \varphi \\ \therefore {\alpha }^n+{\beta }^n=(\cos \varphi +i\sin \varphi {)}^n+(\cos \varphi -i\sin \varphi {)}^n \\ =2\cos n\varphi \\ \text{ and }\begin{array}{r}{\alpha }^n{\beta }^n=(\cos n\varphi +i\sin n\varphi )(\cos n\varphi -i\sin n\varphi )\\ ={\cos }^{2}n\varphi +{\sin }^{2}n\varphi =1\end{array} \end{gathered}$$
$\therefore$ Required equation is
${\mathrm{x}}^2-2\mathrm{xcos}\mathrm{n\varphi }+1=0$