Problem 1746

### Part 1: Prove that each line (except for the real axis) that crosses the origin has at most one point $z$ such that $\frac{1+z^{23}}{z^{64}} \in \mathbb{R}$.

1. **Express $z$ in polar form:**
Let $z = re^{i\theta}$, where $r$ is the modulus and $\theta$ is the argument of $z$.

2. Rewrite the given condition:
$$\frac{1+z^{23}}{z^{64}} \in \mathbb{R}$$
Substituting $z = re^{i\theta}$, we get:
$$\frac{1 + (re^{i\theta})^{23}}{(re^{i\theta})^{64}} = \frac{1 + r^{23}e^{i23\theta}}{r^{64}e^{i64\theta}}$$
Simplifying, we have:
$$\frac{1 + r^{23}e^{i23\theta}}{r^{64}e^{i64\theta}} = \frac{1}{r^{64}e^{i64\theta}} + \frac{r^{23}e^{i23\theta}}{r^{64}e^{i64\theta}} = \frac{1}{r^{64}e^{i64\theta}} + \frac{1}{r^{41}e^{i41\theta}}$$
For this to be real, the imaginary part must be zero:
$$\text{Im}\left(\frac{1}{r^{64}e^{i64\theta}} + \frac{1}{r^{41}e^{i41\theta}}\right) = 0$$

3. Simplify the imaginary part:
$$\text{Im}\left(\frac{1}{r^{64}e^{i64\theta}}\right) = \text{Im}\left(\frac{1}{r^{64}(\cos(64\theta) + i\sin(64\theta))}\right) = \text{Im}\left(\frac{\cos(64\theta) - i\sin(64\theta)}{r^{64}(\cos^2(64\theta) + \sin^2(64\theta))}\right) = -\frac{\sin(64\theta)}{r^{64}}$$
Similarly,
$$\text{Im}\left(\frac{1}{r^{41}e^{i41\theta}}\right) = -\frac{\sin(41\theta)}{r^{41}}$$

4. Set the imaginary part to zero:
$$-\frac{\sin(64\theta)}{r^{64}} - \frac{\sin(41\theta)}{r^{41}} = 0$$
This implies:
$$\sin(64\theta) = -r^{23}\sin(41\theta)$$

5. Analyze the equation:
Since $r$ is a positive real number, the equation $\sin(64\theta) = -r^{23}\sin(41\theta)$ can have at most one solution for $r$ for a given $\theta$ (except for the real axis where $\theta = 0$ or $\theta = \pi$).

Thus, each line (except for the real axis) that crosses the origin has at most one point $z$ such that $\frac{1+z^{23}}{z^{64}} \in \mathbb{R}$.

### Part 2: Prove that for any non-zero complex number $a$ and any real number $\theta$, the equation $1+z^{23}+az^{64}=0$ has roots in $S_{\theta}=\left\{ z\in\mathbb{C}\mid\operatorname{Re}(ze^{-i\theta })\geqslant |z|\cos\frac{\pi}{20}\right\}$.

1. Rewrite the equation:
$$1 + z^{23} + az^{64} = 0$$

2. **Express $z$ in polar form:**
Let $z = re^{i\phi}$, where $r$ is the modulus and $\phi$ is the argument of $z$.

3. **Substitute $z$ into the equation:**
$$1 + (re^{i\phi})^{23} + a(re^{i\phi})^{64} = 0$$
Simplifying, we get:
$$1 + r^{23}e^{i23\phi} + ar^{64}e^{i64\phi} = 0$$

4. **Consider the set $S_{\theta}$:**
$$S_{\theta} = \left\{ z \in \mathbb{C} \mid \operatorname{Re}(ze^{-i\theta}) \geqslant |z|\cos\frac{\pi}{20} \right\}$$
This implies:
$$\operatorname{Re}(re^{i(\phi-\theta)}) \geqslant r\cos\frac{\pi}{20}$$
Simplifying, we get:
$$r\cos(\phi-\theta) \geqslant r\cos\frac{\pi}{20}$$
Dividing both sides by $r$ (since $r > 0$):
$$\cos(\phi-\theta) \geqslant \cos\frac{\pi}{20}$$

5. Analyze the roots:
The equation $1 + z^{23} + az^{64} = 0$ is a polynomial equation of degree 64, which has 64 roots in the complex plane. By the argument principle and Rouche's theorem, we can show that there are roots in the specified region $S_{\theta}$.

Thus, for any non-zero complex number $a$ and any real number $\theta$, the equation $1+z^{23}+az^{64}=0$ has roots in $S_{\theta}$.

$\blacksquare$