Croatian Mathematical Competitions

Let $z = re^{i\theta}$, where $r > 0$ and $\theta \in \mathbb{R}$. Then $\bar{z} = re^{-i\theta}$.

We have:
$$\frac{\bar{z}}{z} + \frac{z}{\bar{z}} = \frac{re^{-i\theta}}{re^{i\theta}} + \frac{re^{i\theta}}{re^{-i\theta}} = e^{-2i\theta} + e^{2i\theta} = 2\cos(2\theta)$$

We are told that $2\cos(2\theta)$ is a positive integer. The possible values for $2\cos(2\theta)$ are $1$ and $2$ (since $2\cos(2\theta) \leq 2$ and must be a positive integer).

Case 1: $2\cos(2\theta) = 2$

Then $\cos(2\theta) = 1 \implies 2\theta = 2\pi k$ for some integer $k$, so $\theta = \pi k$.

Thus, $z = re^{i\pi k} = r(-1)^k$. So $z$ is a nonzero real number.

Case 2: $2\cos(2\theta) = 1$

Then $\cos(2\theta) = \frac{1}{2} \implies 2\theta = \pm \frac{\pi}{3} + 2\pi k$ for some integer $k$.

So $\theta = \pm \frac{\pi}{6} + \pi k$.

Thus, $z = re^{i(\pm \frac{\pi}{6} + \pi k)}$ for $r > 0$ and integer $k$.

Therefore, all complex numbers $z \neq 0$ such that $z$ is a nonzero real number, or $z$ has argument $\pm \frac{\pi}{6}$ or $\pm \frac{7\pi}{6}$ (modulo $2\pi$), i.e., $z = re^{i\theta}$ where $\theta = \pi k$ or $\theta = \pm \frac{\pi}{6} + \pi k$ for integer $k$ and $r > 0$.