Question

If $z=cos\theta +isin\theta +isin\theta$ is a root of the equation $a_0{z}^n+a_1{z}^{n-1}+a_2{z}^{n-2}+...+a_{n-1}z+a_n=0$, then prove the following
(i) $a_0+a_{1}cos\theta +a_{2}cos2\theta +...+a_{n}cosn\theta =0$
(ii) $a_{1}sin\theta +a_{2}sin2\theta +...+a_{n}sinn\theta =0$.

Answer 1

$z=\cos \theta +i\sin \theta +i\sin \theta$

$\Rightarrow a_0{z}^n+a_1{z}^{n-1}+a_2{z}^{n-2}+.....+a_{n-1}z+a_n=0$

When $z^n=\cos \theta$

$\Rightarrow a_0+a_1\cos \theta +a_2\cos 2\theta +......+a_n\cos n\theta$

$a_0{z}^n+a_1{z}^{n-1}+a_2{z}^{n-2}+.....+a_{n-1}+z+a_n=0$

$a_0{z}^n=a_0$

$a_0{z}^{n-1}=a_1\cos \theta .....$

$a_0+a_1\cos \theta +.....+a_n\cos n\theta =0$

Similarly when $z^n=\sin \theta$

$\Rightarrow a_0\sin \theta +a_1\sin \theta +a_2\sin 2\theta +.......+a_n\sin n\theta =0$

$a_0+a_1\sin \theta +a_2\sin 2\theta +.......+a_n\sin n\theta =0.$

Hence, the answer is 1 statement.