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Recently I came across this summation: $$\sum_{k=0}^n\cos\Bigl(k\frac{2\pi}{n}\Bigr)$$ when I was trying to evaluate the following summation $$\sum_{k=0}^{n} z^k+z^{-k}$$ where $z \in \mathbb{C}, z=\operatorname{cis}(\frac{2\pi}{n})$

But I got stuck... Could anyone give me a tip ?

Thanks in advance!

Bernard
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Ramon Coche
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2 Answers2

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$\sum\limits_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$

$\sum\limits_{k=0}^n z^{-k}=z\frac{1-z^{-(n+1)}}{z-1}$

Is that where you started.

herb steinberg
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  • Thank you for the explanation this helped me a lot I couldn't see these 2 geometric progressions now I'm able to move on! – Ramon Coche Jun 11 '21 at 22:16
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Hint:

$\cos\dfrac{2k\pi}n$ is the real part of $\:\Bigl(\mathrm e^{\tfrac{2i\pi}n}\Bigr)^{\!k}$.

Bernard
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  • A better hint is to write $z = \Bigl(\mathrm e^{\tfrac{2i\pi}n}\Bigr)$ and then go to the OP, where we know about $\sum(z^k+z^{-k})$, but we maybe do not know that it is $\left(\sum z^k\right)+\left(\sum z^{-k}\right)$... – dan_fulea Jun 11 '21 at 21:23
  • Thanks a lot for the hint! I'll try to understand a bit better about Euler's notation for complex numbers – Ramon Coche Jun 11 '21 at 22:15
  • Note that, for the same price, considering the imaginary part, you obtain the sum of the sines. – Bernard Jun 11 '21 at 22:17