Question

Show that for an equation whose roots are nth power of the roots of the equation $x^2-2xcos\theta +1=0$ is $x^2-2xcosn\theta +1=0$.

Answer 1

For the equation
$x^2-2xcos\theta +1=0$
$x=\frac{2cos\theta \pm \sqrt{4co{s}^2\theta -4}}{2}$
$=\frac{2cos\theta \pm i2sin\theta }{2}$
$=cos\theta \pm isin\theta$
Hence
$z_1=e^{i\theta }$ and $z_2=e^{-i\theta }$ are the roots.
Hence
$a=z_1^n=e^{in\theta }$ and $b=z_2^n=e^{-ni\theta }$
Now
Hence the required equation will be
$x^2-(a+b)x+ab=$
$x^2-(e^{in\theta }+e^{-in\theta })x+e^{in\theta }.e^{-in\theta }=0$
$x^2-2cosn\theta .x+1=0$