Let n be an even positive integer such that (n/2) is odd and let α_0 , α_1 , ...., α_(n−1) be the complex roots of unity of order n. Prove that Π_(k=0) ^(n−1) (a + bα_k ^2 ) = (a^(n/2) + b^(n/2) )^2 for any complex numbers a and b.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 21293 by Tinkutara last updated on 19/Sep/17

$Let n be an even positive integer such$ $that \frac{n}{2} is odd and let α_{0}, α_{1}, . . . ., α_{n - 1} be$ $the complex roots of unity of order n .$ $Prove that \underset{k = 0}{\prod^{n - 1}} (a + b α_{k}^{2}) = {(a^{\frac{n}{2}} + b^{\frac{n}{2}})}^{2}$ $for any complex numbers a and b .$
	Answered by revenge last updated on 24/Sep/17

	$x^{n} - 1 = (x - α_{0}) (x - α_{1}) (x - α_{2}) . . . (x - α_{n - 1})$ $P u t t i n g x = i \sqrt{\frac{a}{b}}, w e g e t$ $i^{n} {(\frac{a}{b})}^{\frac{n}{2}} - 1 = (i \sqrt{\frac{a}{b}} - α_{0}) (i \sqrt{\frac{a}{b}} - α_{1}) . . . (i \sqrt{\frac{a}{b}} - α_{n - 1})$ $i^{n} a^{\frac{n}{2}} - b^{\frac{n}{2}} = (i \sqrt{a} - \sqrt{b} α_{0}) (i \sqrt{a} - \sqrt{b} α_{1}) . . . (i \sqrt{a} - \sqrt{b} α_{n - 1}) . . . (1)$ $S i m i l a r l y p u t t i n g x = - i \sqrt{\frac{a}{b}}, w e g e t$ $(- i^{n}) a^{\frac{n}{2}} - b^{\frac{n}{2}} = (- i \sqrt{a} - \sqrt{b} α_{0}) (- i \sqrt{a} - \sqrt{b} α_{1}) . . . (- i \sqrt{a} - \sqrt{b} α_{n - 1})$ $i^{n} a^{\frac{n}{2}} - b^{\frac{n}{2}} = (i \sqrt{a} + \sqrt{b} α_{0}) (i \sqrt{a} + \sqrt{b} α_{1}) . . . (i \sqrt{a} + \sqrt{b} α_{n - 1}) . . . (2)$ $M u l t i p l y i n g (1) a n d (2), w e g e t$ ${(i^{n} a^{\frac{n}{2}} - b^{\frac{n}{2}})}^{2} = (a + b α_{0}^{2}) (a + b α_{1}^{2}) . . . (a + b α_{n - 1}^{2}) = \underset{k = 0}{\prod^{n - 1}} (a + b α_{k}^{2})$ $S i n c e n = 2 m, w h e r e m i s o d d, ∴ i^{2 m} = - 1 . S o,$ ${(a^{\frac{n}{2}} + b^{\frac{n}{2}})}^{2} = \underset{k = 0}{\prod^{n - 1}} (a + b α_{k}^{2})$
	Commented by Tinkutara last updated on 24/Sep/17

	$Thank you very much Sir!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum