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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 9 of 160

Marks: +1, -0

If

1, \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{n-1}

n^{\text{th}}

roots of unity then

\sum\limits_{1 \le i < j \le n-1} \alpha_i \alpha_j =

1
0
$-1$
$i$

Solution:

Let

z^n = 1

z^n - 1 = (z-1)(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\ldots(z-\alpha_{n-1})

Sum of

n

th roots of unity

= 0

1 + \alpha_1 + \alpha_2 + \alpha_3 + \ldots + \alpha_{n-1} = 0

\Rightarrow \alpha_1 + \alpha_2 + \alpha_3 + \ldots + \alpha_{n-1} = -1

Also, the coefficient of

z^{n-1}

is zero, so sum of product of zeroes taken two at a time

= 0

\Rightarrow \alpha_1 + \alpha_2 + \ldots + \alpha_{n-1} + \sum\limits_{1 \le i < j \le n-1} \alpha_i \alpha_j = 0

\Rightarrow -1 + \sum\limits_{1 \le i < j \le n-1} \alpha_i \alpha_j = 0

\therefore \sum\limits_{1 \le i < j \le n-1} \alpha_i \alpha_j = 1

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