Question

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

• A. $x^2-2x\cos n\varphi -1=0$
• B. $x^2-2x\cos n\varphi +1=0$
• C. $x^2-2x\sin n\varphi +1=0$
• D. $x^2+2x\sin n\varphi -1=0$

Answer 1

The correct option is B $x^2-2x\cos n\varphi +1=0$

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