Q50 / 160 · AP EAMCET 2016 · Mathematics

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

  • A
    0
  • B
    $n-1$
  • Correct answer
    C
    $n$
  • D
    1

Solution

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 $\begin{aligned} & (1+1+\ldots .+n \text { times }) \\ & =\left(1-z_1\right)\left(1-z_2\right)+\ldots .\left(1-z_{n-1}\right) \\ & n=\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right) \end{aligned}$

Hence, the correct answer is option: 'C'.

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