Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2241 Upvotes
Verified Answer
The correct answer is:
$n$
We have given that, $1, z_1, z_2, z_3, \ldots . z_{n-1}$ are the $n$th root of unity.
$\begin{aligned}
& \therefore \quad z^n-1=(z-1)\left(z-z_1\right)\left(z-z_2\right), \ldots .,\left(z-z_{n-1}\right) \\
& (z-1)\left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right) \\
& =(z-1)\left(z-z_1\right)+\ldots . .\left(z-z_{n-1}\right) \\
& \left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right)=\left(z-z_1\right) \\
& \left(z-z_1\right)=\left(z-z_{n-1}\right)
\end{aligned}$
Put $z=1$, we get
$\begin{aligned}
& (1+1+\ldots .+n \text { times }) \\
& =\left(1-z_1\right)\left(1-z_2\right)+\ldots .\left(1-z_{n-1}\right) \\
& n=\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right)
\end{aligned}$
$\begin{aligned}
& \therefore \quad z^n-1=(z-1)\left(z-z_1\right)\left(z-z_2\right), \ldots .,\left(z-z_{n-1}\right) \\
& (z-1)\left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right) \\
& =(z-1)\left(z-z_1\right)+\ldots . .\left(z-z_{n-1}\right) \\
& \left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right)=\left(z-z_1\right) \\
& \left(z-z_1\right)=\left(z-z_{n-1}\right)
\end{aligned}$
Put $z=1$, we get
$\begin{aligned}
& (1+1+\ldots .+n \text { times }) \\
& =\left(1-z_1\right)\left(1-z_2\right)+\ldots .\left(1-z_{n-1}\right) \\
& n=\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right)
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.