Complex no. cube root of unity

\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 $\bf{m,n=\left\{1,2,3,...\right\}}}$.

6 Answers

1708
man111 singh ·

Nice Epsilon,Ketan , XyZ

http://www.askiitians.com/forums/Discuss-with-Askiitians-Tutors/33/47291/cubic-root-of-unity.htm

11
epsilon ·

if 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
Ketan Chandak ·

@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....

1
xYz ·

is the answer 7

i am getting
{0,1,-1,ω,-ω,(1+ω),-(1+ω)} as the 7 distinct elements of the range space

1057
Ketan Chandak ·

@xYz....i m also getting these 7 only...

11
epsilon ·

@ketan bhaiya u r right..I forgot the case of ()m...
sorry!!

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