Let-n-be-an-even-positive-integer-such-that-n-2-is-odd-and-let-0-1-n-1-be-the-complex-roots-of-unity-of-order-n-Prove-that-k-0-n-1-a-b-k-2-a-n-2-b-n-2-2

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}, \dots ., 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}) \dots (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}) \dots (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}) \dots (i \sqrt{a} - \sqrt{b} α_{n - 1}) \dots (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}) \dots (- 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}) \dots (i \sqrt{a} + \sqrt{b} α_{n - 1}) \dots (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}) \dots (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!