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Question Number 21293 by Tinkutara last updated on 19/Sep/17

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.

$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{such} \\ $$$$\mathrm{that}\:\frac{{n}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{let}\:\alpha_{\mathrm{0}} ,\:\alpha_{\mathrm{1}} ,\:....,\:\alpha_{{n}−\mathrm{1}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{of}\:\mathrm{order}\:{n}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}\:+\:{b}\alpha_{{k}} ^{\mathrm{2}} \right)\:=\:\left({a}^{\frac{{n}}{\mathrm{2}}} \:+\:{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{complex}\:\mathrm{numbers}\:{a}\:\mathrm{and}\:{b}. \\ $$

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) ) Putting x=i(√(a/b)), we get i^n ((a/b))^(n/2) −1=(i(√(a/b))−α_0 )(i(√(a/b))−α_1 )...(i(√(a/b))−α_(n−1) ) i^n a^(n/2) −b^(n/2) =(i(√a)−(√b)α_0 )(i(√a)−(√b)α_1 )...(i(√a)−(√b)α_(n−1) ) ...(1) Similarly putting x=−i(√(a/b)), we get (−i^n )a^(n/2) −b^(n/2) =(−i(√a)−(√b)α_0 )(−i(√a)−(√b)α_1 )...(−i(√a)−(√b)α_(n−1) ) i^n a^(n/2) −b^(n/2) =(i(√a)+(√b)α_0 )(i(√a)+(√b)α_1 )...(i(√a)+(√b)α_(n−1) ) ...(2) Multiplying (1) and (2), we get (i^n a^(n/2) −b^(n/2) )^2 =(a+bα_0 ^2 )(a+bα_1 ^2 )...(a+bα_(n−1) ^2 )=Π_(k=0) ^(n−1) (a+bα_k ^2 ) Since n=2m, where m is odd, ∴ i^(2m) =−1. So, (a^(n/2) +b^(n/2) )^2 =Π_(k=0) ^(n−1) (a+bα_k ^2 )

$${x}^{{n}} −\mathrm{1}=\left({x}−\alpha_{\mathrm{0}} \right)\left({x}−\alpha_{\mathrm{1}} \right)\left({x}−\alpha_{\mathrm{2}} \right)...\left({x}−\alpha_{{n}−\mathrm{1}} \right) \\ $$$${Putting}\:{x}={i}\sqrt{\frac{{a}}{{b}}},\:{we}\:{get} \\ $$$${i}^{{n}} \left(\frac{{a}}{{b}}\right)^{\frac{{n}}{\mathrm{2}}} −\mathrm{1}=\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{\mathrm{0}} \right)\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{\mathrm{1}} \right)...\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{{n}−\mathrm{1}} \right) \\ $$$${i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{1}} \right)...\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right)\:...\left(\mathrm{1}\right) \\ $$$${Similarly}\:{putting}\:{x}=−{i}\sqrt{\frac{{a}}{{b}}},\:{we}\:{get} \\ $$$$\left(−{i}^{{n}} \right){a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{1}} \right)...\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right) \\ $$$${i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{\mathrm{1}} \right)...\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right)\:...\left(\mathrm{2}\right) \\ $$$${Multiplying}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right),\:{we}\:{get} \\ $$$$\left({i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} =\left({a}+{b}\alpha_{\mathrm{0}} ^{\mathrm{2}} \right)\left({a}+{b}\alpha_{\mathrm{1}} ^{\mathrm{2}} \right)...\left({a}+{b}\alpha_{{n}−\mathrm{1}} ^{\mathrm{2}} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}+{b}\alpha_{{k}} ^{\mathrm{2}} \right) \\ $$$${Since}\:{n}=\mathrm{2}{m},\:{where}\:{m}\:{is}\:{odd},\:\therefore\:{i}^{\mathrm{2}{m}} =−\mathrm{1}.\:{So}, \\ $$$$\left({a}^{\frac{{n}}{\mathrm{2}}} +{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}+{b}\alpha_{{k}} ^{\mathrm{2}} \right) \\ $$

Commented by Tinkutara last updated on 24/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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