Question

If $\cos \theta +i\sin \theta$ is a root of the equation $x^n+a_1{x}^{n-1}+a_2{x}^{n-1}+.......+a_{n-1}x+a_n=0$, then the absolute value of $\sum _{r=1}^n{a}_r\cos r\theta =$ .............

Answer 1

$x^n+a_1{x}^{n-1}+a_2{x}^{n-2}+.......+a_{n-1}x+a_n=0$
$\Rightarrow 1+a_1{x}^{-1}+a_2{x}^{-2}+.......+a_{n-1}{x}^{-n+1}+a_n{x}^{-n}=0$
Since, $cos\theta +isin\theta$ is the root
Therefore, $1+a_1{(cos\theta +isin\theta )}^{-1}+a_2{(cos\theta +isin\theta )}^{-2}+.......+a_{n-1}{(cos\theta +isin\theta )}^{-n+1}+a_n{(cos\theta +isin\theta )}^{-n}=0$
$\Rightarrow 1+a_1(cos\theta -isin\theta )+a_2(cos2\theta -isin2\theta )+.......+a_{n-1}(cos(n-1)\theta -isin(n-1)\theta )+a_n(cosn\theta -isinn\theta )=0$
On comparing real parts, we get
$a_1\left(cos\theta \right)+a_2\left(cos2\theta \right)+.......+a_{n-1}\left(cos(n-1)\theta \right)+a_n\left(cosn\theta \right)=-1$
Therefore, ${\sum }_{r=1}^n{a}_{r}cosr\theta =-1$
Ans: -1