Show that Π_(k=1) ^n (a−e^((2ikπ)/n) )(a−e^(−((2ikπ)/n)) )=(a^n −1)^2


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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

	$Consider the equation of degree n :$ $x^{n} - 1 = 0 () \Leftrightarrow x^{n} = 1 \Leftrightarrow x =^{n} \sqrt{1}$ $= {(\cos 2 k π + isin 2 k π)}^{\frac{1}{n}} = \cos \frac{2 k π}{n} + isin \frac{2 k π}{n}$ $for k = 1, 2, . . ., n . Thus, the equation ()$ $has n roots which each of them be a$ $n - th root of unit . We denote the that$ $roots to be δ_{1}, δ_{2}, . . ., δ_{n} . In addition, we$ $see that α_{k} = \cos \frac{2 k π}{n} - isin \frac{2 k π}{n} (k = \overset{―}{1 . . n})$ $are also n different n - th roots of unit$ $but α_{i} and δ_{i} are two conjugate number$ $together means α_{i} δ_{i} = 1 and so we have$ $δ_{1} . δ_{2} . . . . δ_{n} = α_{1} . α_{2} . . . . α_{n} = 1 . Then we have$ $\underset{k = 1}{\prod^{n}} (a - e^{\frac{2 k π i}{n}}) (a - e^{\frac{- 2 k π i}{n}}), (e^{\frac{2 k π i}{n}} = \cos \frac{2 k π}{n} + isin \frac{2 k π}{n}, (e^{\frac{- 2 k π i}{n}} = \cos \frac{2 k π}{n} - isin \frac{2 k π}{n})$ $= \underset{k = 1}{\overset{n}{Π}} (a - δ_{k}) (a - α_{k}) = \underset{k = 1}{\overset{n}{Π}} (a - δ_{k}) \underset{k = 1}{\overset{n}{Π}} (a - α_{k})$ $= [a^{n} - a^{n - 1} (\underset{k = 1}{\overset{n}{Σ}} δ_{k}) + a^{n - 2} (\underset{i \neq j}{Σ} δ_{i} δ_{j}) - . . . + a \underset{i_{1} \neq i_{2} \neq . . . \neq i_{n - 1}}{Σ} (δ_{i_{1}} δ_{i_{2}} . . . δ_{i_{n - 1}}) - δ_{1} δ_{2} . . δ_{n}]$ $= (a^{n} - 1) (becau se δ_{k} (k = \overset{―}{1 . . . n}) be n$ $roots of the equation : x^{n} - 1 = 0, so all$ $sums of form Σ δ_{k}, Σ δ_{i} δ_{j}, . . . . Σ (δ_{i_{1}} δ_{i_{2}} . . . δ_{i_{n - 1}})$ $equal to zero - {Vieta}^{'} s theorem)$ $Similarly, we have$ $\underset{k = 1}{\prod^{n}} (a - α_{k}) = (a^{n} - 1)$ $Consequently, \underset{k = 1}{\prod^{n}} (a - e^{\frac{2 k π i}{n}}) (a - e^{\frac{- 2 k π i}{n}})$ $= {(a^{n} - 1)}^{2} (q . e . d)$
	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

	$Thank you, but only in public situation$ $, Sir$
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