If z=cosθ+isinθ is a root of equation a_0 z^n +a_1 z^(n−1) +a_2 z^(n−2) +.....+a_(n−1) z+a_n =0 then prove that: i) a_0 +a_1 cos θ+a_2 cos 2θ+.....+a_n cos nθ=0 ii) a_1 sin θ + a_2 sin 2θ+....+a


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Question Number 32160 by rahul 19 last updated on 20/Mar/18

$I f z = c o s θ + i s i n θ i s a r o o t o f e q u a t i o n$ $a_{0} z^{n} + a_{1} z^{n - 1} + a_{2} z^{n - 2} + . . . . . + a_{n - 1} z + a_{n} = 0$ $t h e n p r o v e t h a t :$ $i) a_{0} + a_{1} \cos θ + a_{2} \cos 2 θ + . . . . . + a_{n} \cos n θ = 0$ $i i) a_{1} \sin θ + a_{2} \sin 2 θ + . . . . + a_{n} \sin n θ = 0 .$
	Answered by rahul 19 last updated on 20/Mar/18

	$M y t r y :$ $D i v i d i n g t h e w h o l e {e q}^{n} b y z^{n},$ $\Rightarrow a_{0} + a_{1} z^{- 1} + . . . . . . . + a_{n - 1} z^{1 - n} + a_{n} z^{- n} = 0$ $N o w z s a t i s f i e s t h e a b o v e {e q}^{n},$ $\Rightarrow a_{0} + a_{1} (\cos θ - i \sin θ) + . . . . . . . . . . .$ $\Rightarrow . . . + a_{n - 1} (c o s (n - 1) θ - i s i n (n - 1) θ)$ $\Rightarrow + a_{n} (c o s n θ - i s i n n θ) = 0 .$ $(b y d e {m o i v r e}^{'} s t h e o r e m) .$ $C o l l e c t i n g t h e r e a l a n d i m a g i n a r y$ $p a r t s e p e r a t e l y,$ $\Rightarrow (a_{0} + a_{1} \cos θ + a_{2} \cos 2 θ + . . . . . + a_{n} \cos n θ)$ $- i (a_{1} \sin θ + a_{2} \sin 2 θ + . . . . . + a_{n} \sin n θ) = 0$ $H o w t o p r o c e e d f u r t h e r ?$
	Commented by rahul 19 last updated on 21/Mar/18

	$h e y, @ a b o v e b u t h e r e m y n o . i s$ $a - i b = 0 \Rightarrow a = i b s o i s i t c o m p u l s o r y t h a t$ $a = b = 0 .$
	Commented by Tinkutara last updated on 20/Mar/18

	$A c o m p l e x n u m b e r a + i b = 0 o n l y w h e n$ $a = b = 0 . S o y o u h a v e d o n e c o m p l e t e$ $s o l u t i o n!$
	Commented by rahul 19 last updated on 20/Mar/18

	$H a h a h a . . . . .$ $t h a n k s f o r c h e c k i n g!$
	Commented by mrW2 last updated on 21/Mar/18

	$i n t h e n o t a t i o n f o r c o m p l e x n u m b e r s$ $z = a + i b$ $t h e s i g n + {d o e s n}^{'} t m e a n t h e s u m o f$ $a a n d i b .$
	Commented by mrW2 last updated on 21/Mar/18

	Commented by rahul 19 last updated on 21/Mar/18

	$o k, t h a n k y o u s i r .$
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