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-

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

If 1, a_{1}, a_{2}, \dots, 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

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}) \dots

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, \dots, 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}

\dots

\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.