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Prove that $\displaystyle\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)=0$ for $n\in\mathbb{N},n>1$

I'm thinking at a demonstration by induction, as base case $n=2$

$$\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)=\sum_{k=0}^{1}\cos\left(\frac{2\pi k}{n}+\phi\right)=\cos\phi+\cos(\pi+\phi)=\cos\phi-\cos\phi=0$$ but I don't have any idea to inductive step, how can someone prove this?

I'm thinking use this to proof that $\displaystyle\sum_{k=0}^{n-1}\alpha^{k}=0$ where $\alpha=\exp\left(\frac{2\pi}{n}\iota\right)$, thanks for any help.

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marked as duplicate by lab bhattacharjee trigonometry Apr 21 '16 at 2:05

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    $\begingroup$ You probably mean $=0$ ? $\endgroup$ – Yves Daoust Apr 20 '16 at 22:29
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    $\begingroup$ Just use the sum of geometric series, and that $\alpha^n=1$. $\endgroup$ – Berci Apr 20 '16 at 22:35
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    $\begingroup$ Roots of unity maybe? $\endgroup$ – Aritra Das Apr 20 '16 at 22:36
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Your sum is the real part of

$$\sum_{k=0}^{n-1}e^{i(2\pi k/n+\phi)}=e^{i\phi}\sum_{k=0}^{n-1}\left(e^{i2\pi /n}\right)^k=e^{i\phi}\frac{e^{in2\pi/n}-1}{e^{i2\pi/n}-1}=0.$$

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  • $\begingroup$ and so simple that I do not know how I did not think about it :c $\endgroup$ – cand Apr 20 '16 at 22:35
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$$ \begin{align} \sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right) &=\mathrm{Re}\left(\sum_{k=0}^{n-1}e^{\frac{2\pi ik}n+i\phi}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\sum_{k=0}^{n-1}e^{\frac{2\pi ik}n}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\frac{1-e^{\frac{2\pi in}n}}{1-e^{\frac{2\pi i}n}}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\frac{1-e^{2\pi i}}{1-e^{\frac{2\pi i}n}}\right)\\[6pt] &=\mathrm{Re}(0)\\[12pt] &=0 \end{align} $$ As long as $n\ge2$, $1-e^{\frac{2\pi i}n}\ne0$.

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Since the sum of the $n$-th roots of unity is zero, we have: $$ \sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}\right) = \sum_{k=0}^{n-1}\sin\left(\frac{2\pi k}{n}\right) = 0.\tag{1}$$ For a given $n$, the function: $$ f_n(\phi) = \sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right) $$ is quite trivially an analytic function (as the sum of $n$ analytic functions), but due to $(1)$ we have $f_n(0)=f_n'(0)=f_n''(0)=\ldots=0$, hence $\color{red}{f_n\equiv 0}$.

As an alternative, for every $n$ we have that $f_n(\phi)$ is a solution of the differential equation $f''(\phi)+f(\phi)=0$ fulfilling the constraints $f(0)=f'(0)=0$. The solutions are unique and global by the Cauchy-Lipschitz theorem, hence the same conclusion as above holds.

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    $\begingroup$ By the angle addition formula, you directly have $f_n(\phi)=C\cos(\theta)-S\sin(\theta)$. $\endgroup$ – Yves Daoust Apr 20 '16 at 22:54
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    $\begingroup$ @YvesDaoust: right for sure, good point. $\endgroup$ – Jack D'Aurizio Apr 20 '16 at 22:55

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