If $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$ are the $n, n$th roots of unity, then $(1-\omega)(1-\omega)^{2} \ldots\left(1-\omega^{n-1}\right)$ is equal to
(A) 0
(B) 1
(C) $n$
(D) $n^{2}$Passage 3 Solution of Equations Certain types of algebraic equations can be solved with the help of De'Moivre's theorem
Equations of the type $p z^{n}+q=0:$ If $p z^{n}+q=0$, where $p$ and $q$ are complex numbers, and $p \neq 0$, then
$$
z^{n}=-q / p
$$
The roots of the given equation are, therefore, the $n$ values of $(-q / p)^{1 / n}$. For example, consider the equation $z^{7}+1=0$.$z^{7}+1=0 \Rightarrow z^{7}=-1=$ cis $(2 p+1) \pi$, where $p$ is an integer. Therefore, $z=\operatorname{cis}[(2 p+1) \pi 7], p=0,1, \ldots, 6$
On putting $p=0,1,2,3,4,5,6$, the roots are seen to be $\cos (\pi 7) \pm i \sin (\pi / 7), \cos (3 \pi / 7) \pm i \sin \left(3 \pi^{\prime} 7\right), \cos$
$(5 \pi / 7) \pm i \sin (5 \pi / 7),-1 .$
Equations of the type $p z^{2 n}+q z^{n}+r=0$, where $p, q$ and $r$ are complex numbers and $p \neq 0$.
$$
z^{n}=\frac{-q \pm \sqrt{q^{2}-4 p r}}{2 p}
$$
Denoting these values of $z^{n}$ by $\alpha$ and $\beta$, we have two equations $z^{n}=\alpha$ and $z^{n}=\beta$, each of which can be solved by the method given in the above example.
Equations of the type $a(p z+q)^{n}+b(r z+s)^{n}=0:$ The substitution $\frac{p z+q}{n+s}=w$ reduces the given equation to the form
$$
a w^{n}+b=0
$$
which can be solved by the method given above. If $w_{k}$ be a root of the equation (i), the corresponding root $z_{k}$ of the given equation is obtained by solving the equation
$$
\frac{p z_{k}+q}{r z_{k}+s}=w_{k}
$$