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Question Number 52086 by prakash jain last updated on 03/Jan/19

If 1,a_(1,) a_2 ,...,a_(n−1) are n^(th) roots of unity, then prove that. (1+a_1 )(1+a_2 )..(1+a_(n−1) )= n if n is even 0 if n is odd

$$\mathrm{If}\:\mathrm{1},{a}_{\mathrm{1},} {a}_{\mathrm{2}} ,...,{a}_{{n}−\mathrm{1}} \:\mathrm{are}\:{n}^{{th}} \:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{unity},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}. \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)..\left(\mathrm{1}+{a}_{{n}−\mathrm{1}} \right)= \\ $$$${n}\:\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{even} \\ $$$$\mathrm{0}\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{odd} \\ $$

Commented by mr W last updated on 03/Jan/19

should it be =n if n is odd =0 if n is even?

$${should}\:{it}\:{be}\: \\ $$$$={n}\:{if}\:{n}\:{is}\:{odd} \\ $$$$=\mathrm{0}\:{if}\:{n}\:{is}\:{even}? \\ $$

Commented by mr W last updated on 03/Jan/19

please check the question sir: n=3: a_1 =ω=−(1/2)+((√3)/2)i a_2 =ω^2 =−(1/2)−((√3)/2) (1+a_1 )(1+a_2 )=((1/2)−((√3)/2)i)((1/2)+((√3)/2)i) =(1/4)+(3/4) =1≠n=3

$${please}\:{check}\:{the}\:{question}\:{sir}: \\ $$$${n}=\mathrm{3}: \\ $$$${a}_{\mathrm{1}} =\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${a}_{\mathrm{2}} =\omega^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\mathrm{1}\neq{n}=\mathrm{3} \\ $$

Commented by prakash jain last updated on 03/Jan/19

You are correct. The question should 1 for n odd. And 0 for n even

Answered by mr W last updated on 04/Jan/19

see question above a_k =cos ((2kπ)/n)+i sin ((2kπ)/n), k=0,1,2,...,n−1 1+a_k =1+cos ((2kπ)/n)+i sin ((2kπ)/n) 1+a_k =2cos^2 ((kπ)/n)+i 2sin ((kπ)/n)cos ((kπ)/n) 1+a_k =2cos ((kπ)/n)(cos ((kπ)/n)+i sin ((kπ)/n)) P=Π_(k=1) ^(n−1) (1+a_k )=2^(n−1) Π_(k=1) ^(n−1) cos ((kπ)/n) Π_(k=1) ^(n−1) (cos ((kπ)/n)+i sin ((kπ)/n)) if n=even: at k=(n/2): cos ((kπ)/n)=cos (π/2)=0 ⇒P=0 if n=odd: cos (((n−1)π)/n)=cos (π−(π/n))=−cos (π/n) sin (((n−1)π)/n)=sin (π−(π/n))=sin (π/n) [cos (((n−1)π)/n)+i sin (((n−1)π)/n)][cos (π/n)+i sin (π/n)] =[−cos (π/n)+i sin (π/n)][cos (π/n)+i sin (π/n)] =−[cos (π/n)−i sin (π/n)][cos (π/n)+i sin (π/n)] =−[cos^2 (π/n)+sin^2 (π/n)] =−1 ⇒Π_(k=1) ^(n−1) (cos ((kπ)/n)+i sin ((kπ)/n))=(−1)^((n−1)/2) =1 cos (π/n) cos (((n−1)π)/n)=−cos^2 (π/n) cos ((2π)/n) cos (((n−2)π)/n)=−cos^2 ((2π)/n) P=2^(n−1) cos^2 (π/n) cos^2 ((2π)/n) cos^2 ((3π)/n) ... cos^2 (((n−1)π)/(2n)) P=2^(n−1) [cos (π/n) cos ((2π)/n) cos ((3π)/n) ... cos (((n−1)π)/(2n))]^2 ......????

$${see}\:{question}\:{above} \\ $$$${a}_{{k}} =\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}},\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1} \\ $$$$\mathrm{1}+{a}_{{k}} =\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}} \\ $$$$\mathrm{1}+{a}_{{k}} =\mathrm{2cos}^{\mathrm{2}} \:\frac{{k}\pi}{{n}}+{i}\:\mathrm{2sin}\:\frac{{k}\pi}{{n}}\mathrm{cos}\:\frac{{k}\pi}{{n}} \\ $$$$\mathrm{1}+{a}_{{k}} =\mathrm{2cos}\:\frac{{k}\pi}{{n}}\left(\mathrm{cos}\:\frac{{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{{k}\pi}{{n}}\right) \\ $$$${P}=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+{a}_{{k}} \right)=\mathrm{2}^{{n}−\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{cos}\:\frac{{k}\pi}{{n}}\:\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{cos}\:\frac{{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{{k}\pi}{{n}}\right) \\ $$$$ \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{n}}=\boldsymbol{{even}}: \\ $$$${at}\:{k}=\frac{{n}}{\mathrm{2}}: \\ $$$$\mathrm{cos}\:\frac{{k}\pi}{{n}}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{P}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{n}}=\boldsymbol{{odd}}: \\ $$$$\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}=\mathrm{cos}\:\left(\pi−\frac{\pi}{{n}}\right)=−\mathrm{cos}\:\frac{\pi}{{n}} \\ $$$$\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}=\mathrm{sin}\:\left(\pi−\frac{\pi}{{n}}\right)=\mathrm{sin}\:\frac{\pi}{{n}} \\ $$$$\left[\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}\right]\left[\mathrm{cos}\:\frac{\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\pi}{{n}}\right] \\ $$$$=\left[−\mathrm{cos}\:\frac{\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\pi}{{n}}\right]\left[\mathrm{cos}\:\frac{\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\pi}{{n}}\right] \\ $$$$=−\left[\mathrm{cos}\:\frac{\pi}{{n}}−{i}\:\mathrm{sin}\:\frac{\pi}{{n}}\right]\left[\mathrm{cos}\:\frac{\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\pi}{{n}}\right] \\ $$$$=−\left[\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{{n}}+\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{{n}}\right] \\ $$$$=−\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{cos}\:\frac{{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{{k}\pi}{{n}}\right)=\left(−\mathrm{1}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{cos}\:\frac{\pi}{{n}}\:\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}=−\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{{n}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{2}\pi}{{n}}\:\mathrm{cos}\:\frac{\left({n}−\mathrm{2}\right)\pi}{{n}}=−\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{2}\pi}{{n}} \\ $$$${P}=\mathrm{2}^{{n}−\mathrm{1}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{{n}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{2}\pi}{{n}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{{n}}\:...\:\mathrm{cos}^{\mathrm{2}} \:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}} \\ $$$${P}=\mathrm{2}^{{n}−\mathrm{1}} \left[\mathrm{cos}\:\frac{\pi}{{n}}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{{n}}\:\mathrm{cos}\:\frac{\mathrm{3}\pi}{{n}}\:...\:\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right]^{\mathrm{2}} \\ $$$$......???? \\ $$

Commented by prakash jain last updated on 03/Jan/19

I was able to prove for odd. (1+x)(1+x^2 )...(1+x^(n−1) ) =Σ_(k=0) ^((n(n+1))/2) x^k =((x^(1+((n(n+1))/2)) −1)/(x−1)) x is first root of unity. ((x.(x^n )^((n+1)/2) −1)/(x−1))=((x−1)/(x−1))=1

$$\mathrm{I}\:\mathrm{was}\:\mathrm{able}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{for}\:\mathrm{odd}. \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)...\left(\mathrm{1}+{x}^{{n}−\mathrm{1}} \right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} {\sum}}{x}^{{k}} =\frac{{x}^{\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} −\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${x}\:\mathrm{is}\:\mathrm{first}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$$$\frac{{x}.\left({x}^{{n}} \right)^{\left({n}+\mathrm{1}\right)/\mathrm{2}} −\mathrm{1}}{{x}−\mathrm{1}}=\frac{{x}−\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1} \\ $$

Commented by mr W last updated on 03/Jan/19

thanks sir! i have to study in detail. i can′t get this result yet.

$${thanks}\:{sir}! \\ $$$${i}\:{have}\:{to}\:{study}\:{in}\:{detail}.\:{i}\:{can}'{t}\:{get}\:{this}\:{result}\:{yet}. \\ $$

Commented by mr W last updated on 04/Jan/19

I studied your workings and unterstand now. should the third line not be =Σ_(k=0) ^((n(n−1))/2) x^k =((x^(1+((n(n−1))/2)) −1)/(x−1)) ? why must n be odd? i can′t see this restriction in your workings.

$${I}\:{studied}\:{your}\:{workings}\:{and}\:{unterstand} \\ $$$${now}. \\ $$$${should}\:{the}\:{third}\:{line}\:{not}\:{be} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} {\sum}}{x}^{{k}} =\frac{{x}^{\mathrm{1}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} −\mathrm{1}}{{x}−\mathrm{1}}\:? \\ $$$${why}\:{must}\:{n}\:{be}\:{odd}?\:{i}\:{can}'{t}\:{see}\:{this} \\ $$$${restriction}\:{in}\:{your}\:{workings}. \\ $$

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