2023 Chinese IMO National Team Selection Test

(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, $$\begin{array}{rl}\frac{1+z^{23}}{z^{64}}\in \mathbb{R}& \iff \mathrm{Im}\frac{1+z^{23}}{z^{64}}=\frac{-1}{64}(\sin 64\theta +r^{23}\sin 41\theta )=0\\ & \iff \sin 64\theta +r^{23}\sin 41\theta =0.\end{array}$$ For $\theta \ne 0,\pi$, when $\sin 41\theta =0$, we have $\sin 64\theta \ne 0$ and there are no solutions to the equation. When $\sin 41\theta \ne 0$, 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°) $\lambda +64\theta \equiv \lambda +41\theta \equiv 0(\mathrm{mod}\pi )$, then $\theta \equiv \frac{k\pi }{23}(\mathrm{mod}\pi )$ and $\lambda \equiv \frac{5k\pi }{23}(\mathrm{mod}\pi )$ 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 $g_{\lambda }(\theta )>0$, $$(\ast \ast )-e^{-i\lambda }f(z)=-\frac{1}{(g_{\lambda }(\theta ){)}^{64}}(\cos (\lambda +64\theta )+r^{23}\cos (\lambda +41\theta ))\text{ takes all real values.}$$ 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 $\lambda +41\theta \equiv 0(\mathrm{mod}\pi )$, but $\lambda +64\theta \not\equiv 0(\mathrm{mod}\pi )$. 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 $\lambda +64\theta \equiv 0(\mathrm{mod}\pi )$, but $\lambda +41\theta \not\equiv 0(\mathrm{mod}\pi )$. 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

	${\theta }_0$		${\theta }_0+\frac{\pi }{64}$		${\theta }_0+\frac{\pi }{41}$		${\theta }_0+\frac{\pi }{32}$		${\theta }_0+\frac{3\pi }{64}$		${\theta }_0+\frac{2\pi }{41}$
$\sin (\lambda +64\theta )$	0	+	0	-	-	-	0	+	0	-	-
$\sin (\lambda +41\theta )$	0	-	-	-	0	+	+	+	+	+	0
$\cos (\lambda +64\theta )$	$\pm 1$		$\mp 1$				$\pm 1$		$\mp 1$

At this point, we observe that the piecewise continuous function $\rho$ (in terms of $\theta$) satisfies the following conditions: $$\underset{\theta \to ({\theta }_0+\frac{\pi }{32}{)}^{-}}{\lim }\rho =\pm \infty ,\underset{\theta \to ({\theta }_0+\frac{3\pi }{64}{)}^{+}}{\lim }\rho =\mp \infty ,\rho ({\theta }_0+\frac{\pi }{41})=\rho ({\theta }_0+\frac{2\pi }{41})=0.$$ 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

	${\theta }_0$		${\theta }_0+\frac{\pi }{64}$		${\theta }_0+\frac{\pi }{41}$		${\theta }_0+\frac{\pi }{32}$		${\theta }_0+\frac{3\pi }{64}$		${\theta }_0+\frac{2\pi }{41}$
$\sin (\lambda +64\theta )$	0	+	0	-	-	-	0	+	0	-	-
$\sin (\lambda +41\theta )$	0	+	+	+	0	-	-	-	-	-	0
$\cos (\lambda +64\theta )$	$\pm 1$		$\mp 1$				$\pm 1$		$\mp 1$

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 $({\theta }_1,{\theta }_2)$).

	${\theta }_1$	${\theta }_2$	${\theta }_2+\frac{\pi }{64}$	${\theta }_1+\frac{\pi }{41}$	${\theta }_2+\frac{\pi }{32}$
$\sin (\lambda +64\theta )$	+	+	0	0	+
$\sin (\lambda +41\theta )$	0	-	-	-	0
$\cos (\lambda +64\theta )$		$\pm 1$	$\mp 1$		$\pm 1$

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

	${\theta }_1$	${\theta }_2$	${\theta }_2+\frac{\pi }{64}$	${\theta }_1+\frac{\pi }{41}$	${\theta }_2+\frac{\pi }{32}$
$\sin (\lambda +64\theta )$	+	+	0	0	+
$\sin (\lambda +41\theta )$	0	+	+	0	-
$\cos (\lambda +64\theta )$		$\pm 1$	$\mp 1$		$\pm 1$

in which case one takes $({\theta }_2,{\theta }_2+\frac{\pi }{64})$ as the desired interval. In conclusion, the proposition is proven. □