Complex Numbers 5 Question 17
18. If $1, a _1, a _2, \ldots, a _{n-1}$ are the $n$ roots of unity, then show that $\quad\left(1-a _1\right)\left(1-a _2\right)\left(1-a _3\right) \ldots\left(1-a _{n-1}\right)=n$
(1984, 2M)
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Solution:
- Since, $1, a _1, a _2, \ldots, a _{n-1}$ are $n$th roots of unity.
$\Rightarrow\left(x^{n}-1\right)=(x-1)\left(x-a _1\right)\left(x-a _2\right) \ldots\left(x-a _{n-1}\right)$
$\Rightarrow \quad \frac{x^{n}-1}{x-1}=\left(x-a _1\right)\left(x-a _2\right) \ldots .\left(x-a _{n-1}\right)$
$\Rightarrow x^{n-1}+x^{n-2}+\ldots .+x^{2}+x+1$ $=\left(x-a _1\right)\left(x-a _2\right) \ldots .\left(x-a _{n-1}\right)$ $\because \frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+\ldots+x+1$
On putting $x=1$, we get $1+1+\ldots n$ times
$=\left(1-a _1\right)\left(1-a _2\right) \ldots\left(1-a _{n-1}\right)$ $\Rightarrow \quad\left(1-a _1\right)\left(1-a _2\right) \ldots\left(1-a _{n-1}\right)=n$
