If $a_{1}, a_{2}, . . . . a_{n - 1}$ are $n$ th roots of unity then prove that
(i) $(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n$
(ii) $1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0$
(iii) $\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}$

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Solution

$a_{1}, a_{2}, a_{3}, \dots . a_{n - 1} \to n^{t h}$ roots of unity.
(i) $(x^{n} -) = (x - a_{1}) (x - a_{2}) \dots . . (x - a_{n - 1}) (x - 1)$
$\Rightarrow$ $\frac{x^{n} - 1}{x - 1} = (x - a_{1}) \dots \dots (x - a_{n - 1})$
$\Rightarrow 1 + x + x^{2} + \dots . + x^{n - 1} = (x - a_{1}) (x - a_{2}) \dots . . (x - a_{n - 1})$
Put $x = a$
$(1 - a_{1}) (1 - a_{2}) \dots . . (1 - a_{n - 1}) = 1 + \dots \dots . = n$
(ii) $x^{n} - 1 = 0$
Sum of roots $= 0$
$\Rightarrow 1 + a_{1} + a_{2} + \dots . a_{n - 1} = 0$
(iii) $\frac{x^{n} - 1}{x - 1} = (x - a_{1}) \dots . . (x - a_{n - 1})$
Taking log,
$log (n^{n} - 1) - log (x - 1)$
$= log (x - a_{1}) + log (x - a_{2}) + \dots . + log (x - a_{n - 1})$
Differentiation w.r.t: x
$\frac{n x^{n - 1}}{x^{n} - 1} - \frac{1}{x - 1}$
$= \frac{1}{x - a_{1}} + \frac{1}{x - a_{2}} + \dots + \frac{1}{x - a_{n - 1}}$
Put $x = 2$
$\frac{n \cdot 2^{n - 1}}{2^{n} - 1} - 1 = \frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + \dots . . + \frac{1}{2 - a_{n - 1}}$
Hence proved.

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