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Question Number 107498 by Ar Brandon last updated on 11/Aug/20

Show that

\underset{k = 1}{\prod^{n}} (a - e^{\frac{2 i k π}{n}}) (a - e^{- \frac{2 i k π}{n}}) = {(a^{n} - 1)}^{2}

Answered by 1549442205PVT last updated on 11/Aug/20

Consider the equation of degree n :

x^{n} - 1 = 0 (*) \Leftrightarrow x^{n} = 1 \Leftrightarrow x =^{n} \sqrt{1}

= {(\cos 2 k π + isin 2 k π)}^{\frac{1}{n}} = \cos \frac{2 k π}{n} + isin \frac{2 k π}{n}

for k = 1, 2, \dots, n . Thus, the equation (*)

has n roots which each of them be a

n - th root of unit . We denote the that

roots to be δ_{1}, δ_{2}, \dots, δ_{n} . In addition, we

see that α_{k} = \cos \frac{2 k π}{n} - isin \frac{2 k π}{n} (k = \overset{―}{1 . . n})

are also n different n - th roots of unit

but α_{i} and δ_{i} are two conjugate number

together means α_{i} δ_{i} = 1 and so we have

δ_{1} . δ_{2} \dots . δ_{n} = α_{1} . α_{2} \dots . α_{n} = 1 . Then we have

\underset{k = 1}{\prod^{n}} (a - e^{\frac{2 k π i}{n}}) (a - e^{\frac{- 2 k π i}{n}}), (e^{\frac{2 k π i}{n}} = \cos \frac{2 k π}{n} + isin \frac{2 k π}{n}, (e^{\frac{- 2 k π i}{n}} = \cos \frac{2 k π}{n} - isin \frac{2 k π}{n})

= \underset{k = 1}{\overset{n}{Π}} (a - δ_{k}) (a - α_{k}) = \underset{k = 1}{\overset{n}{Π}} (a - δ_{k}) \underset{k = 1}{\overset{n}{Π}} (a - α_{k})

= [a^{n} - a^{n - 1} (\underset{k = 1}{\overset{n}{Σ}} δ_{k}) + a^{n - 2} (\underset{i \neq j}{Σ} δ_{i} δ_{j}) - \dots + a \underset{i_{1} \neq i_{2} \neq \dots \neq i_{n - 1}}{Σ} (δ_{i_{1}} δ_{i_{2}} \dots δ_{i_{n - 1}}) - δ_{1} δ_{2} . . δ_{n}]

= (a^{n} - 1) (becau se δ_{k} (k = \overset{―}{1 \dots n}) be n

roots of the equation : x^{n} - 1 = 0, so all

sums of form Σ δ_{k}, Σ δ_{i} δ_{j}, \dots . Σ (δ_{i_{1}} δ_{i_{2}} \dots δ_{i_{n - 1}})

equal to zero - {Vieta}^{'} s theorem)

Similarly, we have

\underset{k = 1}{\prod^{n}} (a - α_{k}) = (a^{n} - 1)

Consequently, \underset{k = 1}{\prod^{n}} (a - e^{\frac{2 k π i}{n}}) (a - e^{\frac{- 2 k π i}{n}})

= {(a^{n} - 1)}^{2} (q . e . d)

Commented by Ar Brandon last updated on 11/Aug/20

Thanks so very much Sir �� May I write you in case of any difficulties.��

Commented by 1549442205PVT last updated on 13/Aug/20

Thank you, but only in public situation

, Sir