Question

If $z=\cos \theta +i\sin \theta$ is a root of $x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+...+p_n=0,$ where $p_1,p_2,...,p_n\in R$, Find the value of $22+7(p_1\sin \theta +p_2\sin 2\theta +...+p_n\sin (n\theta \left)\right)$

Answer 1

$x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+.......+p_{n-1}x+p_n=0$
$\Rightarrow 1+p_1{x}^{-1}+p_2{x}^{-2}+.......+p_{n-1}{x}^{-n+1}+p_n{x}^{-n}=0$
Since, $cos\theta +isin\theta$ is the root
Therefore, $1+p_1{(cos\theta +isin\theta )}^{-1}+p_2{(cos\theta +isin\theta )}^{-2}+.......+p_{n-1}{(cos\theta +isin\theta )}^{-n+1}+p_n{(cos\theta +isin\theta )}^{-n}=0$
$\Rightarrow 1+p_1(cos\theta -isin\theta )+p_2(cos2\theta -isin2\theta )+.......+p_{n-1}(cos(n-1)\theta -isin(n-1)\theta )+p_n(cosn\theta -isinn\theta )=0$
On comparing imaginary parts, we get
$p_1\left(sin\theta \right)+p_2\left(sin2\theta \right)+.......+p_{n-1}\left(sin(n-1)\theta \right)+p_n\left(sinn\theta \right)=0$
Therefore, $22+7(p_1\sin \theta +p_2\sin 2\theta +...+p_n\sin (n\theta \left)\right)=22$
Ans: 22