If \(1, \alpha_1, \alpha_2, \ldots ., \alpha_{n-1}\) are the \(n\)th roots of unity and \(n\) is an even…
If \(1, \alpha_1, \alpha_2, \ldots ., \alpha_{n-1}\) are the \(n\)th roots of unity and \(n\) is an even natural number, then \(\left(l+\alpha_1\right)\left(l+\alpha_2\right) \ldots .\left(l+\alpha_{n-1}\right)=\)
1
-1
0
2
Solution
Since, 1, \(\alpha_1, \alpha_2 \ldots \ldots . . . . ., \alpha_{n-1}\) are the \(n\)th roots of unity, then \(x^n-1=(x-1)\left(x-\alpha_1\right)\left(x-\alpha_2\right)\) \(\left(x-\alpha_{n-1}\right)\)
\(\Rightarrow \quad\left(x-\alpha_1\right)\left(x-\alpha_2\right) \ldots \ldots\left(x-\alpha_{n-1}\right)=\frac{x^n-1}{x-1}\)
On putting \(x=-1\), we get
\(\begin{aligned}
& \left(-1-\alpha_1\right)\left(-1-\alpha_2\right) \ldots \ldots . . .\left(-1-\alpha_{n-1}\right)=\frac{(-1)^n-1}{-1-1} \\
& \left(1+\alpha_1\right)\left(1+\alpha_2\right) \ldots \ldots . . .\left(1+\alpha_{n-1}\right)=0
\end{aligned}\)
[as \(n\) is an even natural number] Hence, option (c) is correct.