If 1,a_(1,) a_2 ,...,a_(n−1) are n^(th) roots of unity, then prove that. (1+a_1 )(1+a_2 )..(1+a


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Question Number 52086 by prakash jain last updated on 03/Jan/19

$If 1, a_{1,} a_{2}, . . ., a_{n - 1} are n^{t h} roots of$ $unity, then prove that .$ $(1 + a_{1}) (1 + a_{2}) . . (1 + a_{n - 1}) =$ $n if n is even$ $0 if n is odd$
Commented by mr W last updated on 03/Jan/19

$s h o u l d i t b e$ $= n i f n i s o d d$ $= 0 i f n i s e v e n ?$
Commented by mr W last updated on 03/Jan/19

$p l e a s e c h e c k t h e q u e s t i o n s i r :$ $n = 3 :$ $a_{1} = ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $a_{2} = ω^{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2}$ $(1 + a_{1}) (1 + a_{2}) = (\frac{1}{2} - \frac{\sqrt{3}}{2} i) (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $= \frac{1}{4} + \frac{3}{4}$ $= 1 \neq n = 3$
Commented by prakash jain last updated on 03/Jan/19
You are correct. The question should 1 for n odd. And 0 for n even
	Answered by mr W last updated on 04/Jan/19

	$s e e q u e s t i o n a b o v e$ $a_{k} = \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n}, k = 0, 1, 2, . . ., n - 1$ $1 + a_{k} = 1 + \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n}$ $1 + a_{k} = {2 \cos}^{2} \frac{k π}{n} + i 2 \sin \frac{k π}{n} \cos \frac{k π}{n}$ $1 + a_{k} = 2 \cos \frac{k π}{n} (\cos \frac{k π}{n} + i \sin \frac{k π}{n})$ $P = \underset{k = 1}{\prod^{n - 1}} (1 + a_{k}) = 2^{n - 1} \underset{k = 1}{\prod^{n - 1}} \cos \frac{k π}{n} \underset{k = 1}{\prod^{n - 1}} (\cos \frac{k π}{n} + i \sin \frac{k π}{n})$ $i f n = e v e n :$ $a t k = \frac{n}{2} :$ $\cos \frac{k π}{n} = \cos \frac{π}{2} = 0$ $\Rightarrow P = 0$ $i f n = o d d :$ $\cos \frac{(n - 1) π}{n} = \cos (π - \frac{π}{n}) = - \cos \frac{π}{n}$ $\sin \frac{(n - 1) π}{n} = \sin (π - \frac{π}{n}) = \sin \frac{π}{n}$ $[\cos \frac{(n - 1) π}{n} + i \sin \frac{(n - 1) π}{n}] [\cos \frac{π}{n} + i \sin \frac{π}{n}]$ $= [- \cos \frac{π}{n} + i \sin \frac{π}{n}] [\cos \frac{π}{n} + i \sin \frac{π}{n}]$ $= - [\cos \frac{π}{n} - i \sin \frac{π}{n}] [\cos \frac{π}{n} + i \sin \frac{π}{n}]$ $= - [\cos^{2} \frac{π}{n} + \sin^{2} \frac{π}{n}]$ $= - 1$ $\Rightarrow \underset{k = 1}{\prod^{n - 1}} (\cos \frac{k π}{n} + i \sin \frac{k π}{n}) = {(- 1)}^{\frac{n - 1}{2}} = 1$ $\cos \frac{π}{n} \cos \frac{(n - 1) π}{n} = - \cos^{2} \frac{π}{n}$ $\cos \frac{2 π}{n} \cos \frac{(n - 2) π}{n} = - \cos^{2} \frac{2 π}{n}$ $P = 2^{n - 1} \cos^{2} \frac{π}{n} \cos^{2} \frac{2 π}{n} \cos^{2} \frac{3 π}{n} . . . \cos^{2} \frac{(n - 1) π}{2 n}$ $P = 2^{n - 1} {[\cos \frac{π}{n} \cos \frac{2 π}{n} \cos \frac{3 π}{n} . . . \cos \frac{(n - 1) π}{2 n}]}^{2}$ $. . . . . . ? ? ? ?$
	Commented by prakash jain last updated on 03/Jan/19

	$I was able to prove for odd .$ $(1 + x) (1 + x^{2}) . . . (1 + x^{n - 1})$ $= \underset{k = 0}{\sum^{\frac{n (n + 1)}{2}}} x^{k} = \frac{x^{1 + \frac{n (n + 1)}{2}} - 1}{x - 1}$ $x is first root of unity .$ $\frac{x . {(x^{n})}^{(n + 1) / 2} - 1}{x - 1} = \frac{x - 1}{x - 1} = 1$
	Commented by mr W last updated on 03/Jan/19

	$t h a n k s s i r!$ $i h a v e t o s t u d y i n d e t a i l . i {c a n}^{'} t g e t t h i s r e s u l t y e t .$
	Commented by mr W last updated on 04/Jan/19

	$I s t u d i e d y o u r w o r k i n g s a n d u n t e r s t a n d$ $n o w .$ $s h o u l d t h e t h i r d l i n e n o t b e$ $= \underset{k = 0}{\sum^{\frac{n (n - 1)}{2}}} x^{k} = \frac{x^{1 + \frac{n (n - 1)}{2}} - 1}{x - 1} ?$ $w h y m u s t n b e o d d ? i {c a n}^{'} t s e e t h i s$ $r e s t r i c t i o n i n y o u r w o r k i n g s .$
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