If $1, a_{1}, a_{2}, \ldots \ldots . a_{n-1}$ are the $n$ roots of unity then $\left(1-a_{1}\right)\

Question

If $1, a_{1}, a_{2}, \ldots \ldots . a_{n-1}$ are the $n$ roots of unity then $\left(1-a_{1}ight)\left(1-a_{2}ight) \ldots . .\left(1-a_{n-1}ight)=A 	imes n$. Find A

Professor PC Thomas Classes · Answer

Let $x$ be an $n^{	ext {th }}$ root of unity.\ $\Rightarrow \quad(1)^{1 / n}=x \Rightarrow x^{n}-1=0$\ Since the roots are $1, a_{1}, a_{2} \ldots \ldots . . a_{n-1}$, we have $x^{n}-1=(x-1)\left(x-a_{1}ight)\left(x-a_{2}ight) \ldots \ldots .\left(x-a_{n-1}ight)$\ or $\frac{x^{n}-1}{x-1}=\left(x-a_{1}ight)\left(x-a_{2}ight) \ldots \ldots \ldots . .\left(x-a_{n-1}ight)$\ $\Rightarrow \quad x^{n-1}+x^{n-2}+\ldots .+x+1$\ $=\left(x-a_{1}ight)\left(x-a_{2}ight) \ldots\left(x-a_{n-1}ight)$\ Put $x=1$, then\ $\Rightarrow \quad n=\left(1-a_{1}ight)\left(1-a_{2}ight) \ldots \ldots .\left(1-a_{n-1}ight)$\