Let $z_1, z_2, ..., z_n$ be complex numbers with the same, positive modulus. Prove that: $$\Re\left( \sum_{i=1}^n\sum_{j=1}^n\left( \frac{z_i}{z_j} \right)\right) = 0 \equiv \sum_{i=1}^nz_i = 0$$
We can rewrite $\Re\left( \sum_{i=1}^n\sum_{j=1}^n\left( \frac{z_i}{z_j} \right)\right) = 0$ as $\Re\left( \left(\sum_{i=1}^n z_i \right)\left(\sum_{i=1}^n\frac 1{z_j} \right)\right)$, therefore when $\sum_{i=1}^nz_i = 0$, we have $\Re\left( 0 \times\left(\sum_{i=1}^n\frac 1{z_j} \right)\right) = \Re\left( 0 \right)$ = 0. But how prove converse of this implication?
Let the common modulus be $r >0$. Then $\frac{1}{z_j} = \frac{\overline{z_j}}{r^2}$ and therefore $$ \sum_{i=1}^n\sum_{j=1}^n\left( \frac{z_i}{z_j} \right) = \left( \sum_{i=1}^n z_i \right) \left( \sum_{j=1}^n \frac{1}{z_j} \right) \\ = \left( \sum_{i=1}^n z_i \right) \left( \sum_{j=1}^n \frac{\overline{z_j}}{r^2} \right) = \frac{1}{r^2} \left| \sum_{i=1}^n z_i \right|^2 \, \in\Bbb R . $$ It follows that the expression on the left-hand side is a real number, and $$ \Re\left( \sum_{i=1}^n\sum_{j=1}^n\left( \frac{z_i}{z_j} \right)\right) = \sum_{i=1}^n\sum_{j=1}^n\left( \frac{z_i}{z_j} \right) = \frac{1}{r^2} \left| \sum_{i=1}^n z_i \right|^2 $$ which is zero exactly if $\sum_{i=1}^n z_i = 0$.