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Let $n$ be a positive integer, $z_k = e^{\pi i \frac{2k+1}{n}}\;$ for $ 1 \le k \le n$ (complex roots of $-1$), and $r$ a positive real number. Find a closed form of the sum $$\left(\sum_{l=1}^{n}z_{k}^{n-l}r^{l-1}\right)\left(\sum_{l=1}^{n}\overline{z_{k}}^{n-l}r^{l-1}\right)$$

When developing the sum, I tried to use the fact that $z_k\overline{z_k}=1$ and $\sum_{k=1}^{n}z_k=1$, but still cannot find a way to sum up the good terms to close it.

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    Where does $r$ come into the problem? – 2017-01-30
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    Sorry, I edited my post. – 2017-01-30

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Hint. One may apply the standard geometric sum evaluation $$ \sum_{l=1}^nq^l=q\:\frac{1-q^{n}}{1-q},\quad q \neq1, $$ giving $$ \sum_{l=1}^{n}z_{k}^{n-l}r^{l-1}=z_{k}^{n-1}\:\sum_{l=1}^{n}\left(\frac{r}{z_k}\right)^{l-1}=z_{k}^{n-1}\:\frac{r}{z_k}\:\frac{1-\frac{r^{n}}{z^{n}_k}}{1-\frac{r}{z_k}}=\frac{r}{z_k}\cdot\frac{z^{n}_k-r}{z_k-r}. $$ I hope you can take it from here.

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    I think there is a typo in your hint, because $\sum_{l=1}^{n}z_{k}^{n-l}r^{l-1}=\frac{r}{z_k}\frac{z_k^n-r^n}{z_k-r}$. Now, with this I get that the same is $\frac{r^2(1+r^2)^2}{1+r^2-2r\cos\left(\frac{1+2k}{n}\pi\right)}$. Is this true ? Thanks. – 2017-01-30