Chinese IMO National Team Selection Test, 2023

(1) For a non-zero complex number $z$, let $z = r(\cos\theta + i\sin\theta)$ where $r > 0$ and $\theta \in [0, 2\pi)$, we have
$$f(z) := \frac{1+z^{23}}{z^{64}} = \frac{1}{r^{64}} (\cos(-64\theta) + i\sin(-64\theta) + r^{23}(\cos(-41\theta) + i\sin(-41\theta))).$$
Therefore,
$$\begin{align*} \frac{1+z^{23}}{z^{64}} \in \mathbb{R} &\Leftrightarrow \mathrm{Im} \frac{1+z^{23}}{z^{64}} = \frac{-1}{64} (\sin 64\theta + r^{23} \sin 41\theta) = 0 \\ &\Leftrightarrow \sin 64\theta + r^{23} \sin 41\theta = 0. \end{align*}$$
For $\theta \neq 0, \pi$, when $\sin 41\theta = 0$, we have $\sin 64\theta \neq 0$ and there are no solutions to the equation.
When $\sin 41\theta \neq 0$, we have
$$r = g_0(\theta) := \left( \frac{-\sin 64\theta}{\sin 41\theta} \right)^{1/23},$$
and for $r > 0$, $\sin 64\theta$ and $\sin 41\theta$ must have different signs. Noting that $g_0(\theta) = -g_0(\theta + \pi)$, we can conclude that the proposition holds.

For a fixed $\lambda \in [0, \pi)$, we consider the set of points on the complex plane
$$\Gamma_{\lambda} := \{z \mid f(z)e^{-i\lambda} \in \mathbb{R}\} = \{z \mid \sin(\lambda + 64\theta) + r^{23}\sin(\lambda + 41\theta) = 0\}.$$
The problem can be divided into two cases:
(1°) $\lambda + 64\theta \equiv \lambda + 41\theta \equiv 0 \pmod{\pi}$, then $\theta \equiv \frac{k\pi}{23} \pmod{\pi}$ and $\lambda \equiv \frac{5k\pi}{23} \pmod{\pi}$ hold for some integer $k$.
(2°) If the above equation does not hold, then similar to the discussion in the first question: There is at most one point $z$ on every line passing through the origin that lies in $\Gamma_{\lambda}$. (Specifically, there is at most one intersection point between any ray originating from the origin and $\Gamma_{\lambda}$)

$$g_{\lambda}(\theta) := \left( \frac{-\sin(\lambda + 64\theta)}{\sin(\lambda + 41\theta)} \right)^{1/23}.$$
As previously mentioned, we want to prove that when $z$ is in the sector corresponding to some interval of $\theta$, the set composed of all points in this interval that make $g_{\lambda}(\theta) > 0$,
$$(**) \quad -e^{-i\lambda} f(z) = - \frac{1}{(g_{\lambda}(\theta))^{64}} (\cos(\lambda + 64\theta) + r^{23} \cos(\lambda + 41\theta)) \text{ takes all real values.}$$
The $\theta$ that satisfies $g_{\lambda}(\theta) > 0$ is made up of several subintervals. At the endpoints of these intervals, there are two possible cases:
a) The $\theta$ at the endpoint satisfies $\lambda + 41\theta \equiv 0 \pmod{\pi}$, but $\lambda + 64\theta \not\equiv 0 \pmod{\pi}$. In this case, as $\theta$ approaches this endpoint within the interval, $g_{\lambda}(\theta) \to \infty$ and the corresponding $a \to 0$ ($\rho \to 0$);
b) The $\theta$ at the endpoint satisfies $\lambda + 64\theta \equiv 0 \pmod{\pi}$, but $\lambda + 41\theta \not\equiv 0 \pmod{\pi}$. In this case, as $\theta$ approaches this endpoint within the interval, $g_{\lambda}(\theta) \to 0$ and the corresponding $a \to \infty$ (when $\cos(\lambda + 64\theta)$ at the endpoint is $-1$, $\rho \to +\infty$; when it's 1, $\rho \to -\infty$).
Now we proceed to prove that for any interval of length $\frac{\pi}{10}$ (in terms of $\theta$), there exists a subinterval in which (***) holds.
1) If there exists $\theta_0$ in the interval (with an appropriate $\lambda$) satisfying the aforementioned condition 1°, then the distance from one of the endpoints of the interval (without loss of generality, assuming it to be the right endpoint) to $\theta_0$ is not less than $\frac{\pi}{20} > \frac{2\pi}{41} > \frac{3\pi}{64}$. Therefore, during the process of changing from $\theta_0$ to this endpoint, the following two situations may occur (also without loss of generality, assuming that $\sin(\lambda + 64\theta)$ takes positive values in $(\theta_0, \theta_0 + \frac{\pi}{64})$). One case is that

At this point, we observe that the piecewise continuous function $\rho$ (in terms of $\theta$) satisfies the following conditions:
$$\lim_{\theta \to (\theta_0 + \frac{\pi}{32})^{-}} \rho = \pm\infty, \quad \lim_{\theta \to (\theta_0 + \frac{3\pi}{64})^{+}} \rho = \mp\infty, \quad \rho(\theta_0 + \frac{\pi}{41}) = \rho(\theta_0 + \frac{2\pi}{41}) = 0.$$
Therefore, on the interval $[\theta_0 + \frac{\pi}{41}, \theta_0 + \frac{3\pi}{32}) \cup (\theta_0 + \frac{3\pi}{64}, \theta_0 + \frac{2\pi}{41}]$, $\rho$ can take on any real value.
The other case is that

At this point, we observe that $\rho$ can take on any real value on the interval $(\theta_0 + \frac{\pi}{32}, \theta_0 + \frac{3\pi}{64})$.
2) For the aforementioned case 2°, considering the values of $\theta$ (and $\lambda$) within this interval, we observe that it contains at least 4 zeros of $\sin(\lambda + 41\theta)$. Furthermore, noting that $\frac{3}{43} > \frac{4}{64}$, it follows that among these four zeros of $\sin(\lambda + 41\theta)$, there must exist two adjacent zeros, denoted as $\theta_1$ and $\theta_1 + \frac{\pi}{41}$, between which there are two zeros of $\sin(\lambda + 64\theta)$, namely $\theta_2$ and $\theta_2 + \frac{\pi}{64}$. Without loss of generality, let's assume that $\theta_1 + \frac{2\pi}{41}$ is also within the interval. In this case, either we have (assuming, without loss of generality, that $\sin(\lambda + 64\theta)$ takes positive values in $(\theta_1, \theta_2)$).

in which case one takes $[\theta_1, \theta_2) \cup (\theta_2 + \frac{\pi}{64}, \theta_1 + \frac{\pi}{41}]$ as the desired interval; or we have

in which case one takes $(\theta_2, \theta_2 + \frac{\pi}{64})$ as the desired interval.
In conclusion, the proposition is proven. □