IIT JEE , AIEEE Forum - Mathematics , Algebra - what is the easiest way to solve these kinds of problem

hsbhatt

Since $\omega$ , $\omega^2$ ..., $\omega^{n-1}$ are roots of z^n = 1 .

$(z^n - 1) = (z-1)(z-\omega)(z-\omega^2)(z-\omega^3)...(z-\omega^{n-1})$

or $1+z+z^2+z^3+...+z^{n-1} = (z-\omega)(z-\omega^2)(z-\omega^3)...(z-\omega^{n-1})$

Now substitute z =1.

Hence $(1-\omega)(1-\omega^2)(1-\omega^3)...(1-\omega^{n-1}) = n$