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Let w be a cube root of unity which is not equal to 1. Then the number of distinct elements in the set {(1+w+w2+w3+...+wn)^m : m,n=1,2,3,...} is ?

Let w be a cube root of unity which is not equal to 1. Then the number of distinct elements in the set
{(1+w+w2+w3+...+wn)^m : m,n=1,2,3,...} is ?

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1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
6 years ago
Take m, n = 1,2,3 only

for other values the same thing will repeat

So total nine cases will be formed (3*3)

Now out of these nine few may be duplicate so you need to remove then

I hope its clear to you

thanks

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