An alternative method to find $\sum_{k=1}^{2n-1} | \beta ^k - 1|$ 
Let $\beta \in \mathbb{C}$ such that $\beta ^n = 1$ but $ \beta ^k
 \neq 1$, $\forall k=1,2,\cdots, n-1$. Find the value of $$
 \sum_{k=1}^{2n-1} | \beta ^k - 1|.$$

I came across such a question when I read a book about complex analysis. I would say that I find the value by brute force:
Note that $$\sum_{k=1}^{2n-1} | \beta ^k - 1| =\sum_{k=1}^{n-1} | \beta ^k - 1|+\sum_{k=1}^{n} | \beta ^{k+n-1} - 1|$$ with the appropriate change of variable in the second summation of the RHS.
However, $\beta ^n =1$ and when $k=1$, we have $| \beta ^{k+n-1} - 1|=0$. Hence, the above becomes $$\sum_{k=1}^{2n-1} | \beta ^k - 1| =\sum_{k=1}^{n-1} | \beta ^k - 1|+\sum_{k=2}^{n} | \beta ^{k-1} - 1|.$$ Again, by appropriate change of variable, we have $$\sum_{k=1}^{2n-1} | \beta ^k - 1| =\sum_{k=1}^{n-1} | \beta ^k - 1|+\sum_{k=1}^{n-1} | \beta ^k - 1|=2\sum_{k=1}^{n-1} | \beta ^k - 1|.$$ It remains to find $\sum_{k=}^{n-1} | \beta ^k - 1|$. After writing $\beta$ as $e^{\frac{2 \pi i}{n}}$ and finding the norm explicitly, I find that (well, after quite a while) $$\sum_{k=1}^{n-1} | \beta ^k - 1|=2 \cot \frac{\pi}{2n}$$ and the problem is solved.
However, I am not satisfied with this solution. I would like to invite all of you to give an alternative method for solving this problem.
I have in my mind some ideas. One of them would be transforming the question to a geometric problem. If we draw the $\beta ^k -1$ out on the complex plane, it can be seen that the problem is actually asking for the sum of the lengths of the diagonals of a regular $n$-gon.   Some would suggest that this sum would be found by 'complex analysis' (just search sum of length in MSE) but this is exactly what I want to avoid. I would appreciate any help here.
 A: As you seem to be wanting some guidance and not a complete response, I will give you a push in the right direction, but not provide full proofs. 
Consider the diagram below, which shows an $n$ sided polygon of unit side length with vertices labelled $X_k$, which will correspond to our $\beta^k$. The diagonal $X_nX_k$ is denoted by $d_k$. 
$\hskip.9in$
Your sum $$\sum_{k=1}^{n-1}|\beta^k-1|$$ corresponds to the sum $$\sum_{k=1}^{n-1}X_nX_{k}=\sum_{k=1}^{n-1}d_k$$ 
I will leave it to you to show that $d_k=\dfrac{\sin \left ( \dfrac{k\pi}{n} \right )}{\sin \left ( \dfrac{\pi}{n} \right )}$.
Now, it can be easily shown that $$\sum_{k=1}^{n-1} \sin \left ( \frac{k\pi}{n} \right ) = \cot \left ( \frac{\pi}{2n} \right ) $$ by considering the real part of the geometric series $$\sum_{k=0}^{n-1}z^{2k}$$ From here you can show that $$\sum_{k=1}^{n-1}d_k= \frac{1}{2 \sin ^2 \left ( \frac{\pi}{2n} \right )}$$ But recall that our polygon has unit side length. Scale the result above accordingly and you have your required sum.
