$w_1$, $w_2$, ..., $w_{n-1}$ are the complex roots not equal to $1$ of $z^n-1=0$, where $n$ is odd.
Given: $\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}$. Show: $$\sum_{k=1}^{n-1} \frac{1-\bar{w_k}}{1+w_k} = n$$
I am struggling with this number proof.
I tried substituting in the given and got this: $$\sum_{k=1}^{n-1} \frac{\bar{w_k}-\bar{w_k}^2}{1+\bar{w_k}} = n$$
On
Since $n$ is odd, the only real solution to $x^n=1$ is $1$, so all the $w_i$ can be paired into $\frac{n-1}2$ complex conjugate pairs. For each pair $w,\overline w$ we can use the given relation to reduce it to $2-w-\overline w$, so the expression becomes $$2\frac{n-1}2-\sum_iw_i$$ Now the sum of $n$th roots of unity including $1$ is $0$, so the $w_i$ sum to $-1$ and the expression reduces to $n$.
Take $X=\sum_{k} \frac{1-conj(w_k)}{1+(w_k)}$. Then using the forumula u mentioned, it is clear that $X+conj(X)=\sum_k (2-w_k-conj(w_k)) + (2-1-1)$ since $w_k$ are roots of $z^n-1=z^n - (\sum_{k=1}^{n-1} w_k +1 )z^{n-1}...-1$, we have $\sum_k w_k +1 = 0$. Hence $X+conj(X)=2n$. Since $w_1,...,w_{n-1}$ are closed under conjugation i.e., $conj(w_i)=w_j$ for some $j$ we have $X=conj(X)$. Hence $X=n$.