Roots of Unity
Let ω be a complex number that is not a real number such that ω 8 = 1 . Evaluate 3 1 k = 1 ∑ 2 0 1 6 ω 6 k 4 − ω 5 k 4 + ω 4 k 4 − ω 3 k 4 + ω 2 k 4 − ω k 4 + 1 ω 6 k 4 + ω 5 k 4 + ω 4 k 4 + ω 3 k 4 + ω 2 k 4 + ω k 4 + 1 .
The answer is 2016.
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1 solution
If ω = − 1 , then for k = 1 , ω 6 k 4 − ω 5 k 4 + ω 4 k 4 − ω 3 k 4 + ω 2 k 4 − ω k 4 + 1 ω 6 k 4 + ω 5 k 4 + ω 4 k 4 + ω 3 k 4 + ω 2 k 4 + ω k 4 + 1 = ( − 1 ) 6 − ( − 1 ) 5 + ( − 1 ) 4 − ( − 1 ) 3 + ( − 1 ) 2 − ( − 1 ) + 1 ( − 1 ) 6 + ( − 1 ) 5 + ( − 1 ) 4 + ( − 1 ) 3 + ( − 1 ) 2 + ( − 1 ) + 1 = 7 1 .
I was surprised that it worked for ω being a fourth root of unity as well; I thought when you wrote ω 8 = 1 , you intended that it worked for an eighth root of unity only and not a fourth root as well.
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Yeah, I just didn't want people to plug in i , and − i , and getting an answer.
Simplify the summand to ω 7 k 4 + 1 ω 7 k 4 − 1 ⋅ ω k 4 − 1 ω k 4 + 1 . Note that k 4 ≡ 0 or 1 ( m o d 8 ) . If k is odd, the expression is equal to ω 1 + 1 ω 1 − 1 ⋅ ω − 1 ω + 1 = − 1 . If k is even, the expression is equal to 7 . Thus the answer is 3 1 ( 1 0 0 8 ⋅ 7 − 1 0 0 8 ) = 3 1 ( 1 0 0 8 ) ( 6 ) = 2 ( 1 0 0 8 ) = 2 0 1 6 . Please tell me if there are any errors.