Let w = e^{i*pi/3} = cos(60 degrees) + i sin(60 degrees). Evaluate…

Question

Let w = e^{i*pi/3} = cos(60 degrees) + i sin(60 degrees). Evaluate the real part of the sum sum_{k=1}^{60} k * w^k, that is the value of sum_{k=1}^{60} k*cos(k*pi/3). Report the exact integer value.

Accepted Answer

Note w = e^{i*pi/3} is a primitive 6th root of unity, so w^6 = 1 and the powers w^k repeat with period 6. The required real sum is Re(sum_{k=1}^{60} k w^k). Group the 60 terms into 10 blocks of 6 consecutive indices. Within one block starting at index m, the real parts are weighted cosines of angles m*60, (m+1)*60, ... degrees. A cleaner route: let T = sum_{k=1}^{60} k w^k. Using sum_{k=1}^{N} k x^k = x(1 - (N+1)x^N + N x^{N+1})/(1 - x)^2 with x = w, N = 60. Since w^6 = 1 we have w^60 = 1 and w^61 = w. Thus the numerator x(1 - 61*w^60 + 60*w^61) = w(1 - 61 + 60w) = w(-60 + 60w) = 60w(w - 1), and (1 - w)^2 = (w - 1)^2, so T = 60w(w - 1)/(w - 1)^2 = 60w/(w - 1). With w = (1 + i*sqrt 3)/2, w - 1 = (-1 + i*sqrt 3)/2, so w/(w - 1) = (1 + i sqrt3)/(-1 + i sqrt3) = multiply by conjugate = (1 + i sqrt3)(-1 - i sqrt3)/4 = (-1 - i sqrt3 - i sqrt3 + 3)/4 = (2 - 2 i sqrt3)/4 = (1 - i sqrt3)/2. Hence T = 60*(1 - i sqrt3)/2 = 30 - 30 i sqrt3, whose real part is 30.
Final answer: 30

Final answer: 30

Let w = e^{i*pi/3} = cos(60 degrees) + i sin(60 degrees). Evaluate…

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