\hspace{-16}$ Let $\bf{\omega}$ be a cube root of unity which is not equal to $1$.\\\\ Then the number of distinct elements in the set $\bf{{(1+\omega+\omega^2+\omega^3+...+\omega^n)^m}$\\\\ Where $\bf{m,n=\left\{1,2,3,...\right\}}}$.
1708Nice Epsilon,Ketan , XyZ
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11if n to be a multiple of 3n there will be 2 distinct elements..
if n is a multiple of 3n+1, there will be 3 distinct elements..
if n is a multiple of 3n+2, there will be 1 distinct elements..
1057@epsilon....remember that 1+\omega +\omega ^{2}=0
so when n is a multiple of 3 there will be 2 terms but one will be zero....so 1 term only...which will be (omega) and now omegam can be (omega), (omega)2 or 1 depending on the form of m....
when n is of form 3n+1 then there will be 2 terms in the bracket and again no of distinct terms would depend on the form of m now...
when n is of form 3n+2 then there will be only one term which will always be 0 and 0m will always be zero....
1is the answer 7
i am getting
{0,1,-1,ω,-ω,(1+ω),-(1+ω)} as the 7 distinct elements of the range space