Question

If $z=\cos \theta +i\sin \theta$ be a root of the equation $a_0{z}^n+a_1{z}^{n-1}+a_2{z}^{n-2}+\dots \dots +a_{n-1}z+a_n=0$, then

• A. $a_1\sin \theta +a_2\sin 2\theta +\dots \dots +a_n\sin n\theta =0$
• B. $a_0+a_1\cos \theta +a_2\cos 2\theta +...+a_n\cos n\theta =-1$
• C. $a_0+a_1\cos \theta +a_2\cos 2\theta +...+a_n\cos n\theta =0$
• D. $a_0+a_1\cos \theta +a_2\cos 2\theta +...+a_n\cos n\theta =1$

Answer 1

Answer: (A), (C)

The correct options are
A $a_1\sin \theta +a_2\sin 2\theta +\dots \dots +a_n\sin n\theta =0$
C $a_0+a_1\cos \theta +a_2\cos 2\theta +...+a_n\cos n\theta =0$

Dividing the given equation by $z^n$, we get
$a_0+a_1{z}^{-1}+a_2{z}^{-2}+\dots \dots +a_{n-1}{z}^{1-n}+a_n{z}^{-n}=0$
Now, $z=\cos \theta +i\sin \theta =e^{i\theta }$ satisfies the above equation.
Hence,
$a_0+a_1{e}^{-i\theta }+a_2{e}^{-2i\theta }+\dots \dots +a_{n-1}{e}^{-i(n-1)\theta }+a_n{e}^{-in\theta }=0$
$\Rightarrow (a_0+a_1\cos \theta +a_2\cos 2\theta +\dots \dots +a_n\cos n\theta )-i(a_1\sin \theta +a_2\sin 2\theta +\dots \dots +a_n\sin n\theta )=0$

$\therefore a_0+a_1\cos \theta +a_2\cos 2\theta +...+a_n\cos n\theta =0$
and $a_1\sin \theta +a_2\sin 2\theta +\dots \dots +a_n\sin n\theta =0$