Prove that all solutions of $\frac{n}{2z} = \sum\limits_{i=1}^n \frac{1}{z-c_i}$ lie on the unit circle given that $|c_i| = 1$ for $1 \le i \le n.$
If $n=1,$ then $z = -c_1.$ If $n=2,$ then $z = \pm \sqrt{c_1 c_2}.$ Trying to prove the statement through brute force for $n \ge 3$ quickly becomes unfeasible. I tried a proof by contradiction by assuming that $|z| < 1$ or $|z| > 1$:
$\frac{n}{2|z|} = \left|\sum\limits_{i=1}^n \frac{1}{z-z_i}\right| \le \sum\limits_{i=1}^n \frac{1}{|z-z_i|} \le \frac{n}{|1-|z||}.$ If $|z| > 1,$ this gives $-1 \le |z|,$ which we already know. If $|z| < 1,$ this gives $|z| \ge 1/3,$ which is not helpful enough. Thus, we must take into account the argument of the LHS and RHS of the original equation. But assuming that a point lies off of the unit circle does not place any constraints on its argument, so we can't get a contradiction that way either. We must somehow consider the argument and magnitude of $z$ simultaneously. How do we do this? Any approaches, hints, or ideas? What would be the motivation behind these approaches? Is there something obvious I'm missing?
Note that $$\begin{align} \sum\limits_{i=1}^n \frac{1}{z-c_i} - \frac{n}{2z} &= \sum_{i=1}^n (\frac{1}{z-c_i} - \frac 1{2z}) \\ &= \frac{1}{2z}\sum_{i=1}^n \frac{z+c_i}{z-c_i}\\ &= \frac{1}{2z}\sum_{i=1}^n \frac{|z|^2-z\bar{c_i}+c_i\bar z -1}{|z-c_i|^2}. \end{align}$$
Since $-z\bar{c_i}+c_i\bar z$ has real part $0$, $\displaystyle \frac{n}{2z} = \sum\limits_{i=1}^n \frac{1}{z-c_i}$ implies $$(|z|^2-1) \sum_{i=1}^n \frac{1}{|z-c_i|^2}=0$$ hence $|z|=1$.