Let $(z_1,\ldots z_n)\subseteq \mathbb{C}.$ Does $$ \sum_{j=1\,,\,j\neq k}^n \frac{z_k}{z_k-z_j}=\sum_{j=1\,,\,j\neq k'}^n \frac{z_{k'}}{z_{k'}-z_j} $$ for all $k,k'\in\{1,...,n\}\,$, implies that the $(z_j)_{j=1,...,n}$ are the n-th roots of $z:=z_1^n$ ?
For example,
for $n=2$, we can verify that $z_1=-z_2$ .
For $n=3$, we can also verify that the $(z_j)_{j=1,2,3}$ can be obtained from $z_k\,,$ by a multiplication with $-\frac12+i\frac{\sqrt{3}}{2}.$
The same goes for $n=4$, but with multiplication by $i$...
How can we prove it for all $n\in\mathbb{N}$ ?
Yes, the answer is affirmative. I'm using a bit of calculus here.
Let $f(z):=\prod_{j=1}^n(z-z_j)$. Then $f'(z)/f(z)=\sum_{j=1}^n 1/(z-z_j)$ and, for $1\leqslant k\leqslant n$, $$\sum_{j\neq k}\frac{z_k}{z_k-z_j}=z_k\lim_{z\to z_k}\left(\frac{f'(z)}{f(z)}-\frac1{z-z_k}\right)\underset{\text{L'Hôpital}}{\phantom{\big[}=\phantom{\big]}}\frac{z_kf''(z_k)}{2f'(z_k)}.$$
We're given that this quantity doesn't depend on $k$. Then $g(z_k)/f'(z_k)=\lambda$ doesn't depend on $k$ as well, where $g(z):=zf''(z)-(n-1)f'(z)$ has degree less than $n-1$.
Thus, $g(z_k)-\lambda f'(z_k)=0$ for $1\leqslant k\leqslant n$, which implies $g(z)=\lambda f'(z)$ identically. This is impossible if $\lambda\neq 0$ (the degrees don't match), hence $zf''(z)=(n-1)f'(z)$.
Now solve this ODE...