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-n-1-n-if-n-is-even-0-if-n-is-odd-

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

If 1, a_{1,} a_{2}, \dots, 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, \dots, 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} \dots \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} \dots \cos \frac{(n - 1) π}{2 n}]}^{2}

\dots \dots ? ? ? ?

Commented by prakash jain last updated on 03/Jan/19

I was able to prove for odd .

(1 + x) (1 + x^{2}) \dots (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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