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


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 52087 by prakash jain last updated on 03/Jan/19

$If 1, a_{1}, a_{2}, . . ., a_{n - 1} are n^{t h} roots of$ $unity the find the value of$ $\frac{1}{1 + 1} + \frac{1}{1 + a_{1}} + \frac{1}{1 + a_{2}} + . . + \frac{1}{1 + a_{n - 1}}$ $n is odd number$
Commented by maxmathsup by imad last updated on 03/Jan/19
$let S_n =Σ_(k=0) ^(n−1) (1/(1+a_k )) with a_k =e^((i2kπ)/n) we have 1+a_k =1+cos(((2kπ)/n))+isin(((2kπ)/n)) =2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n)) =2cos(((kπ)/n))e^(i((kπ)/n)) ⇒(1/(1+a_k )) =(1/(2cos(((kπ)/n)))) e^(−((ikπ)/n)) =(1/(2cos(((kπ)/n)))){cos(((kπ)/n))−isin(((kπ)/n))}=(1/2) −(i/2)tan(((kπ)/n))⇒ S_n =(n/2) −(i/2) Σ_(k=0) ^(n−1) tan(((kπ)/n)) rest to find Σ_(k=0) ^(n−1) tan(((kπ)/n))...$
$l e t S_{n} = \sum_{k = 0}^{n - 1} \frac{1}{1 + a_{k}} w i t h a_{k} = e^{\frac{i 2 k π}{n}} w e h a v e 1 + a_{k} = 1 + c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})$ $= 2 {c o s}^{2} (\frac{k π}{n}) + 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n}) = 2 c o s (\frac{k π}{n}) e^{i \frac{k π}{n}} \Rightarrow \frac{1}{1 + a_{k}} = \frac{1}{2 c o s (\frac{k π}{n})} e^{- \frac{i k π}{n}}$ $= \frac{1}{2 c o s (\frac{k π}{n})} {c o s (\frac{k π}{n}) - i s i n (\frac{k π}{n})} = \frac{1}{2} - \frac{i}{2} t a n (\frac{k π}{n}) \Rightarrow$ $S_{n} = \frac{n}{2} - \frac{i}{2} \sum_{k = 0}^{n - 1} t a n (\frac{k π}{n}) r e s t t o f i n d \sum_{k = 0}^{n - 1} t a n (\frac{k π}{n}) . . .$
	Answered by mr W last updated on 03/Jan/19

	$z^{n} = 1$ $z = \cos θ + i \sin θ$ $z^{n} = \cos n θ + i \sin n θ = 1$ $\Rightarrow n θ = 2 k π, k = 0, 1, 2, . . ., n - 1$ $\Rightarrow θ = \frac{2 k π}{n}$ $\Rightarrow a_{k} = \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n}$ $\frac{1}{1 + a_{k}} = \frac{1}{1 + \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n}}$ $= \frac{1 + \cos \frac{2 k π}{n} - i \sin \frac{2 k π}{n}}{{(1 + \cos \frac{2 k π}{n})}^{2} + {(\sin \frac{2 k π}{n})}^{2}}$ $= \frac{1 + \cos \frac{2 k π}{n} - i \sin \frac{2 k π}{n}}{2 (1 + \cos \frac{2 k π}{n})}$ $= \frac{1}{2} [1 - i \frac{\sin \frac{2 k π}{n}}{1 + \cos \frac{2 k π}{n}}]$ $= \frac{1}{2} [1 - i \frac{\sin \frac{2 k π}{n}}{{2 \cos}^{2} \frac{k π}{n}}]$ $= \frac{1}{2} [1 - i \frac{2 \sin \frac{k π}{n} \cos \frac{k π}{n}}{{2 \cos}^{2} \frac{k π}{n}}]$ $= \frac{1}{2} [1 - i \tan \frac{k π}{n}]$ $\underset{k = 0}{\sum^{n - 1}} \frac{1}{1 + a_{k}} = \frac{1}{2} [n - i \underset{k = 0}{\sum^{n - 1}} \tan \frac{k π}{n}]$ $\tan \frac{(n - 1) π}{n} = \tan (π - \frac{π}{n}) = - \tan \frac{π}{n}$ $\tan \frac{(n - 2) π}{n} = \tan (π - \frac{2 π}{n}) = - \tan \frac{2 π}{n}$ $. . .$ $\Rightarrow \underset{k = 0}{\sum^{n - 1}} \tan \frac{k π}{n} = 0 (s i n c e n = o d d)$ $\Rightarrow \underset{k = 0}{\sum^{n - 1}} \frac{1}{1 + a_{k}} = \frac{n}{2}$
	Commented by prakash jain last updated on 03/Jan/19
	Thanks.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum