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Sum Series using De Moivres Theorem

Struggling on a question at the moment. Appreciate any help.
Screenshot_20210420-225912__01.jpg
I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)
(edited 3 years ago)
Original post by Toast210
Struggling on a question at the moment. Appreciate any help.
Screenshot_20210420-225912__01.jpg
I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)


Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of eiθe^{i\theta} instead of 12eiθ\frac{1}{2}e^{i\theta}, and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.
Reply 2
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Toast210
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Original post by ghostwalker
Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of eiθe^{i\theta} instead of 12eiθ\frac{1}{2}e^{i\theta}, and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.

IMG_20210420_234153__01.jpg
Original post by Toast210
IMG_20210420_234153__01.jpg


As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be 12neinθ\frac{1}{2^n}e^{in\theta}
Reply 4
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Toast210
OP
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Original post by ghostwalker
As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be 12neinθ\frac{1}{2^n}e^{in\theta}

Could you explain why?
Original post by Toast210
Could you explain why?

Because the sum of geo sequence result contains an r^n in there ... so what is (1/2 e^itheta)^n ?
Original post by Toast210
Could you explain why?


As part of the formula for a finite sum of the geometric series the common ratio is raised to the power of n. In this case the common ratio is 12eiθ\frac{1}{2}e^{i\theta}, so raising that to the power of n gives us (12eiθ)n=12neinθ\displaystyle\left(\frac{1}{2}e^{i\theta}\right)^n=\frac{1}{2^n}e^{in\theta}

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