Find sum $u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $

Question

I have $$ u_k = \cos\frac{2k\pi}{n} + i \sin\frac{2k\pi}{n}$$ And I should calculate: $$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$ But I have stucked:
Firstly I calculate $$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$ and put $$ \alpha_k = u_k \cdot u_{k+1} = ... = e^{\frac{i\pi(2k+1)}{n}} $$ and sum $$ \alpha_0 + ... + \alpha_{n-2} = ... = e^{\frac{i\pi}{n}} \cdot \frac{1-e^{\frac{2(n-1)i\pi}{n}}}{1-e^{\frac{2i\pi}{n}}}$$ and I don't know how ti finish that. If it comes to $$u_{n-1}u_0 = e^{i\pi} = -1 $$

Answer 1

Note that $$\alpha_k = u_{k}\cdot u_{k+1} = e^{i\frac{2\pi}{n}(2k+1)} = e^{i\frac{2\pi}{n}}\cdot e^{i\frac{4\pi}{n}k}.$$ Therefore, $$\sum_{k=0}^{n-2}{\alpha_k} = e^{i\frac{2\pi}{n}}\sum_{k=0}^{n-2}{\left(e^{i\frac{4\pi}{n}}\right)^{k}} = e^{i\frac{2\pi}{n}} \frac{1-e^{i\frac{4\pi(n-1)}{n}}}{1-e^{i\frac{4\pi}{n}}} = e^{i\frac{2\pi}{n}}\frac{1-e^{-i\frac{4\pi}{n}}}{1-e^{i\frac{4\pi}{n}}} = \frac{e^{i2\pi/n}-e^{-i2\pi/n}}{e^{i2\pi/n}(e^{-i2\pi/n}-e^{i2\pi/n})}=-e^{-i\frac{2\pi}{n}}.$$

Answer 2

Hint: The points $u_k$, and therefore $\alpha_k$, are distributed completely evenly along the unit circle.

Answer 3

Here's a way without geometric sums.

Note that $u_{i+1} = u_i e^{2i\pi/n}$ where $u_n:=u_0=1$, thus $\sum_{k=0}^{n-1} u_iu_{i+1} = e^{2i\pi/n} \sum_{k=0}^{n-1} u_i^2$.

By Newton's identities $ \sum_{k=0}^{n-1} u_i^2 = \left(\sum_{k=0}^{n-1} u_i\right)^2- \sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $\sum_{k=0}^{n-1} u_i=0$ and $\sum_{i,j}u_iu_j = 0$, thus $\sum_{k=0}^{n-1} u_i^2=0$.