Menu Close

If-1-a-1-a-2-a-n-1-are-n-th-roots-of-unity-the-find-the-value-of-1-1-1-1-1-a-1-1-1-a-2-1-1-a-n-1-n-is-odd-number-

Tinku Tara 0 Comments
Pin




Question Number 52087 by prakash jain last updated on 03/Jan/19
If 1,a_1 ,a_2 ,...,a_(n−1) are n^(th) roots of unity the find the value of (1/(1+1))+(1/(1+a_1 ))+(1/(1+a_2 ))+..+(1/(1+a_(n−1) )) n is odd number
$$\mathrm{If}\:\mathrm{1},{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…,{a}_{{n}−\mathrm{1}} \:\mathrm{are}\:{n}^{{th}} \:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{unity}\:\mathrm{the}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{2}} }+..+\frac{\mathrm{1}}{\mathrm{1}+{a}_{{n}−\mathrm{1}} } \\ $$$${n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{number} \\ $$
Commented by maxmathsup by imad last updated on 03/Jan/19
let S_n =Σ_(k=0) ^(n−1) (1/(1+a_k )) with a_k =e^((i2kπ)/n) we have 1+a_k =1+cos(((2kπ)/n))+isin(((2kπ)/n)) =2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n)) =2cos(((kπ)/n))e^(i((kπ)/n)) ⇒(1/(1+a_k )) =(1/(2cos(((kπ)/n)))) e^(−((ikπ)/n)) =(1/(2cos(((kπ)/n)))){cos(((kπ)/n))−isin(((kπ)/n))}=(1/2) −(i/2)tan(((kπ)/n))⇒ S_n =(n/2) −(i/2) Σ_(k=0) ^(n−1) tan(((kπ)/n)) rest to find Σ_(k=0) ^(n−1) tan(((kπ)/n))...
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{a}_{{k}} }\:\:{with}\:{a}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:\:{we}\:{have}\:\mathrm{1}+{a}_{{k}} =\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right) \\ $$$$=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)+\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\:=\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{{k}\pi}{{n}}} \:\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{a}_{{k}} }\:=\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right)}\:{e}^{−\frac{{ik}\pi}{{n}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right)}\left\{{cos}\left(\frac{{k}\pi}{{n}}\right)−{isin}\left(\frac{{k}\pi}{{n}}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}{tan}\left(\frac{{k}\pi}{{n}}\right)\Rightarrow \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{tan}\left(\frac{{k}\pi}{{n}}\right)\:{rest}\:{to}\:{find}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{tan}\left(\frac{{k}\pi}{{n}}\right)… \\ $$
Answered by mr W last updated on 03/Jan/19
z^n =1 z=cos θ+i sin θ z^n =cos nθ+i sin nθ=1 ⇒nθ=2kπ, k=0,1,2,...,n−1 ⇒θ=((2kπ)/n) ⇒a_k =cos ((2kπ)/n)+i sin ((2kπ)/n) (1/(1+a_k ))=(1/(1+cos ((2kπ)/n)+i sin ((2kπ)/n))) =((1+cos ((2kπ)/n)−i sin ((2kπ)/n))/((1+cos ((2kπ)/n))^2 +(sin ((2kπ)/n))^2 )) =((1+cos ((2kπ)/n)−i sin ((2kπ)/n))/(2(1+cos ((2kπ)/n)))) =(1/2)[1−i((sin ((2kπ)/n))/(1+cos ((2kπ)/n)))] =(1/2)[1−i((sin ((2kπ)/n))/(2cos^2 ((kπ)/n)))] =(1/2)[1−i((2sin ((kπ)/n) cos ((kπ)/n))/(2cos^2 ((kπ)/n)))] =(1/2)[1−itan ((kπ)/n)] Σ_(k=0) ^(n−1) (1/(1+a_k ))=(1/2)[n−iΣ_(k=0) ^(n−1) tan ((kπ)/n)] tan (((n−1)π)/n)=tan (π−(π/n))=−tan (π/n) tan (((n−2)π)/n)=tan (π−((2π)/n))=−tan ((2π)/n) ... ⇒Σ_(k=0) ^(n−1) tan ((kπ)/n)=0 (since n=odd) ⇒Σ_(k=0) ^(n−1) (1/(1+a_k ))=(n/2)
$${z}^{{n}} =\mathrm{1} \\ $$$${z}=\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta \\ $$$${z}^{{n}} =\mathrm{cos}\:{n}\theta+{i}\:\mathrm{sin}\:{n}\theta=\mathrm{1} \\ $$$$\Rightarrow{n}\theta=\mathrm{2}{k}\pi,\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},…,{n}−\mathrm{1} \\ $$$$\Rightarrow\theta=\frac{\mathrm{2}{k}\pi}{{n}} \\ $$$$\Rightarrow{a}_{{k}} =\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}_{{k}} }=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}} \\ $$$$=\frac{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}−{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}}{\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}\right)^{\mathrm{2}} +\left(\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}−{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{i}\frac{\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}}{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{{n}}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{i}\frac{\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{{n}}}{\mathrm{2cos}^{\mathrm{2}} \:\frac{{k}\pi}{{n}}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{i}\frac{\mathrm{2sin}\:\frac{{k}\pi}{{n}}\:\mathrm{cos}\:\frac{{k}\pi}{{n}}}{\mathrm{2cos}^{\mathrm{2}} \:\frac{{k}\pi}{{n}}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{i}\mathrm{tan}\:\frac{{k}\pi}{{n}}\right] \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+{a}_{{k}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[{n}−{i}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{tan}\:\frac{{k}\pi}{{n}}\right] \\ $$$$\mathrm{tan}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}=\mathrm{tan}\:\left(\pi−\frac{\pi}{{n}}\right)=−\mathrm{tan}\:\frac{\pi}{{n}} \\ $$$$\mathrm{tan}\:\frac{\left({n}−\mathrm{2}\right)\pi}{{n}}=\mathrm{tan}\:\left(\pi−\frac{\mathrm{2}\pi}{{n}}\right)=−\mathrm{tan}\:\frac{\mathrm{2}\pi}{{n}} \\ $$$$… \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{tan}\:\frac{{k}\pi}{{n}}=\mathrm{0}\:\:\:\:\left({since}\:{n}={odd}\right) \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+{a}_{{k}} }=\frac{{n}}{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 03/Jan/19
Thanks.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *