I'm stuck on a complex number problem relating to the geometric series in polar form. By considering that: $$1+\operatorname{cis}{\theta}+\operatorname{cis}{2\theta}+\operatorname{cis}{3\theta}+....+\operatorname{cis}{n\theta}$$ as a geometric series, find: $$\sum_{r=0}^n \operatorname{cos}{r\theta}$$ I was thinking along the lines of finding the sum of the polar form series, and that I was assuming that the sum of the series would be: $$\operatorname{cis}{\frac{n(n+1)}{2}}{\theta}$$ Thus, if that is the answer then finding the sum of $\operatorname{cos}{r\theta}$ becomes very easy. But I doubt that that is the answer, so any help would be much appreciated!!

  • $\begingroup$ if $z \ne 1$ : $\sum_{r=0}^n z^r = \frac{1-z^{n+1}}{1-z}$. $\endgroup$ – reuns Mar 20 '17 at 0:48
  • $\begingroup$ I'm confused? How does the left hand side equate to the right hand side? $\endgroup$ – Maths Matador Mar 20 '17 at 1:12

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