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For a positive integer n, evaluate the sum S = sum over k = 1 to 2n of [ sin(2*pi*k/(2n+1)) - i*cos(2*pi*k/(2n+1)) ], where i = sqrt(-1).

1. i
2. -i
3. 0
4. 1

Correct answer: i

Solution

Writing the term as -i*e^(i*2*pi*k/(2n+1)), the sum telescopes via roots of unity: the cosines sum to -1 and the sines sum to 0, giving S = -i*(-1) = i.

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