Sums over roots of unity from a logarithmic derivative

1. Let $f(x)=x^n-1=\prod_k(x-\omega^k)$. By the product rule,
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^{n-1}\frac1{x-\omega^k}.$$
Since $f'(x)=nx^{n-1}$, the identity follows.

2. Take $x=2$:
$$\sum_{k=0}^{n-1}\frac1{2-\omega^k}=\frac{n\,2^{n-1}}{2^n-1}.$$
For $n=3$ this is $\frac{12}7\approx1.714$.

3. The terms $k$ and $n-k$ have $\omega^{n-k}=\omega^{-k}=\overline{\omega^k}$. For $z=\omega^k\ne1$ with $|z|=1$,
$$\frac1{1-z}+\frac1{1-\frac1z}=\frac1{1-z}+\frac z{z-1}=\frac{1-z}{1-z}=1.$$
So the terms $k=1,\dots,n-1$ can be grouped into pairs $\{k,n-k\}$ each summing to $1$. If $n$ is even, the term $k=\frac n2$ has $\omega^k=-1$ and equals $\frac12$, which is half of a "pair with itself". In all cases the total is $\frac{n-1}2$. $\blacksquare$

Lesson. A logarithmic derivative turns a product over roots into a sum over roots; pairing $z$ with $\bar z=1/z$ evaluates sums over the unit circle.