, 2025

Solution:

We show that no solutions $z$ not on the unit circle can exist. First, we eliminate $|z| > 1$.

Claim 1. For all $j$ and $|z| > 1$, the real part of $\frac{z + \omega_{j}}{z - \omega_{j}}$ is positive.

Proof. We use geometry. Note that $\omega_{j}$ and $-\omega_{j}$ are antipodes on the unit circle. Since $z$ lies outside the unit circle, it follows that $\angle \omega_{j}z(-\omega_{j})$ is acute. But this means the complex number $\frac{z + \omega_{j}}{z - \omega_{j}}$ lies strictly in the first or fourth quadrant of the complex plane and thus has positive real part, as desired. $\square$

It is then clear that whenever $|z| > 1$, the sum $\sum_{j = 1}^{n} \frac{z + \omega_{j}}{z - \omega_{j}}$ has positive real part and thus cannot be $0$. The case where $|z| < 1$ is analogous, except $\angle \omega_{j}z(-\omega_{j})$ is obtuse instead, so $\frac{z + \omega_{j}}{z - \omega_{j}}$ has negative real part for all $j$. Therefore, all solutions $z$ to the original equation must satisfy $|z| = 1$.

Solution:

We show more generally that for any positive integers $k$, $a_{1}$, $\ldots$, $a_{k}$, and distinct $\omega_{j}$ on the unit circle, the equation
$$\sum_{j = 1}^{k} a_{j} \left(\frac{z + \omega_{j}}{z - \omega_{j}}\right) = 0$$
has $k$ distinct solutions on the unit circle. The original problem then follows upon consolidating duplicate $\omega_{j}$'s. Without loss of generality, assume that $\omega_{1}$, $\ldots$, $\omega_{k}$ are in this order going clockwise around the unit circle.

Claim 2. There is a solution on the (clockwise) arc from $\omega_{j}$ to $\omega_{j + 1}$ for all $j$ (where $\omega_{k + 1} = \omega_{1}$).

Proof. First, $\omega_{j}$ and $-\omega_{j}$ are antipodes on the unit circle, so if $z$ is on the unit circle, $\angle \omega_{j}z(-\omega_{j}) = 90^{\circ}$. This means $\frac{z + \omega_{j}}{z - \omega_{j}}$ is purely imaginary. Now consider the imaginary part of the left hand side of the equation, which is a real and continuous function on the arc strictly between $\omega_{j}$ and $\omega_{j + 1}$ for each $j$. In particular, as $z$ approaches $\omega_{j + 1}$ from the clockwise direction, this function approaches $\infty$. On the other hand, as $z$ approaches $\omega_{j}$ from the counterclockwise direction, this function approaches $-\infty$. By the Intermediate Value Theorem, there must be a solution on this arc, as desired. $\square$

It follows that there are at least $k$ solutions on the unit circle. But the equation is equivalent to a polynomial of degree $k$. Hence, there are exactly $k$ solutions, all of which lie on the unit circle.