If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity, $\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right)$ is equal to

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Aicharya · Accepted Answer

We have given that, $1, z_1, z_2, z_3, \ldots . z_{n-1}$ are the $n$th root of unity. $\begin{aligned} & 	herefore \quad z^n-1=(z-1)\left(z-z_1ight)\left(z-z_2ight), \ldots .,\left(z-z_{n-1}ight) \ & (z-1)\left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1ight) \ & =(z-1)\left(z-z_1ight)+\ldots . .\left(z-z_{n-1}ight) \ & \left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1ight)=\left(z-z_1ight) \ & \left(z-z_1ight)=\left(z-z_{n-1}ight) \end{aligned}$ Put $z=1$, we get $\begin{aligned} & (1+1+\ldots .+n 	ext { times }) \ & =\left(1-z_1ight)\left(1-z_2ight)+\ldots .\left(1-z_{n-1}ight) \ & n=\left(1-z_1ight)\left(1-z_2ight) \ldots\left(1-z_{n-1}ight) \end{aligned}$

If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity, $\left(1-z_1\right)\left(1-z_2\right)…

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