Let $\alpha_k=1,2,...,10$ be the roots of the unity of order 11, $\alpha_k \neq 1$.

Question

Let $\alpha_k=1,2,...,10$ be the roots of the unity of order 11, $\alpha_k \neq 1$. Then compute the following sum:

$$\sum_{k=1}^{10}\frac{1-\overline{\alpha_k}}{1+\alpha_k}.$$

So what I did was I thought the roots o unity must be the roots of the polynomial:

$$f=x^{10}+x^{9}+x^{8}+...+1=0$$

so then $f=x^{11}-1=0\to x^{11}=1$ and found that

$$\alpha_k=(\cos(\frac{2k\pi}{11})+i\sin(\frac{2k\pi}{11}).$$

But just by putting them into the formula is not really pleasant to solve.. is there any other way than just calculating the sum "as is"?

Answer 1

Consider the general case.

Let $\alpha_1,\dots,\alpha_{n-1}$ be the $n$-th roots of unity different from $1$, where $n$ is odd to avoid division by zero.

Note that $\alpha$ is an $n$-th root of unity iff $\overline{\alpha}$ is an $n$-th root of unity.

Pair the terms as follows: $$ \frac{1-\overline{\alpha_k}}{1+\alpha_k}+ \frac{1-\alpha_k}{1+\overline{\alpha_k}} = 2-\alpha_k-\overline{\alpha_k} $$ because $\overline{\alpha_k}=1/\alpha$.

Sum both sides to get $$ 2\sum_{k=1}^{n-1}\frac{1-\overline{\alpha_k}}{1+\alpha_k}= 2(n-1)-\sum_{k=1}^{n-1}\alpha_k-\sum_{k=1}^{n-1}\overline{\alpha_k} =2(n-1)+2 =2n $$ because $1+\sum_{k=1}^{n-1}\alpha_k=0$ since there is no $x^{n-1}$ term in $x^n-1=0$.