We will show that for any $a\in \mathbb{C}$, there exists at most one point on a line of this form such that $\frac{1+z^{23}}{az^{64}}\in \mathbb{R}$. Let $z=cr$. This is equivalent to \begin{align*} &\frac{1+z^{23}}{az^{64}} = \overline{\frac{1+z^{23}}{az^{64}}} = \frac{1+\overline{z}^{23}}{\overline{a} \overline{z}^{64}} \\ &\iff (1+(cr)^{23})\overline{a}c^{64}\overline{r}^{64}=(1+\overline{z}^{23})(ac^{64}r^{64})\\ &\iff(1+(cr)^{23})\overline{a}\overline{r}^{64}=(1+c^{23}\overline{r}^{23})(ar^{64})\\ &\iff \overline{a}\overline{r}^{64} - ar^{64} = c^{23}( \overline{r}^{23}ar^{64} - r^{23}\overline{a}\overline{r}^{64})\\ \end{align*} Since $23$ is odd, there exists exactly one real solution $c$ to $$c^{23} = \frac{\overline{a}\overline{r}^{64} - ar^{64}}{\overline{r}^{23}ar^{64} - r^{23}\overline{a}\overline{r}^{64}} = \frac{-Im(ar^{64})}{|r|^{46} Im(ar^{41})}$$ We consider all lines of the form $$\{ (cr, c\in \mathbb{R}^+) | r=e^{ti} \text{ for some } t\in [\theta - \pi/20, \theta + \pi/20]\}$$ (WLOG $|r|=1$, or the argument doesn't change if we replace $r$ with $\frac{r}{|r|}$.) On each line $cr$ (except when $ar^{41}\in \mathbb{R}$), we know there exists exactly one $c\in \mathbb{R}$ such that $z=cr$. We are seeking a point on one of these lines satisfying $\frac{1+(cr)^{23}}{a(cr)^{64}}=-1$ (and $c>0$). Let $f(r) = \frac{1+z(r)^{23}}{az(r)^{64}}$ where $z(r)$ is the unique complex number such that $z(r)/r\in \mathbb{R}, \frac{1+z(r)^{23}}{az(r)^{64}}\in \mathbb{R}$. The key is to show that $f$ is surjective (in $\mathbb{R}$). Note $f$ is continuous except where $|z(r)|=0$, which only happens when $Im(ar^{64})=0 \iff ar^{64}\in \mathbb{R}$. Case 1: There does not exist $r$ such that $ar^{64}\in \mathbb{R}$ and $r^{23} \in \{-1,1\}$. Color a point $r$ on the arc $e^{ti} (t\in [\theta - \pi/20, \theta + \pi/20])$ red if $ar^{64}$ is real, and silver if $ar^{41}$ is real. Claim: We are done if we can find consecutive points $r_1,s_1,r_2,s_2,r_3$ in this order such that their arguments are all within $[\theta-\pi/20, \theta+\pi/20]$, $r_1,r_2,r_3$ are red and $s_1,s_2$ are silver. If a point $s$ is silver, then when we approach it (from either direction of the unit circle), we can see that $c$ tends to infinity, so $\lim_{t\to s} f(t)=0$. If a point $r$ is red, then we can see that $c$ tends to zero, so $\lim_{t\to r} f(t) = \pm \infty$ where $\pm$ is $+$ if $Re(ar^{64})>0$ and $-$ if $Re(ar^{64})<0$. Note when a point travelling around a unit circle passes through a silver or red point, the sign of $c$ changes, and we need the sign of $c$ to be positive. Thus, the sign of $c$ is positive in either both minor arcs $(r_1,s_1), (r_2,s_2)$ or both minor arcs $(s_1,r_2),(s_2,r_3).$ In either case, when the point travels from $s_j$ to $r_j$ (or $r_{j+1}$), in at least one of the cases (when $j=1,2$) we have $Re(ar^{64})<0$, so we are done by intermediate value theorem; we have $f(r) < -1 < f(s)$. Similarly, we are also done if we can find consecutive points $s_1,r_1,r_2,s_2$ such that $s_j$'s are silver and $r_j$'s are red. (two cases; either okay point is between two reds, or between one of $s_1,r_1$ or $ s_2,r_2$) Since there exist at least 4 silver points and between two silver points there exists 1 or 2 red points we are done. The cases not covered in case 1 can be handled similarly.

notationsLet $S_{\theta}^R=S_{\theta}\cap\{z\in\mathbb{C}:|z|\le R\}$, $l_{\theta}=\{z\in\mathbb{C}:ze^{-i\theta}\in\mathbb{R}_{\ge0}\}$, $l_{\theta}^R=l_{\theta}\cap\{z\in\mathbb{C}:|z|\le R\}$, $\gamma_{\theta}^R=S_{\theta}\cap\{z\in\mathbb{C}:|z|=R\}$ We have $\partial S_{\theta}^R=l_{\theta-\frac{\pi}{20}}^R+\gamma_{\theta}^R-l_{\theta+\frac{\pi}{20}}^R$ Write $f(z)=1+z^{23}+az^{64}$ FTSOC. We have $\left|\operatorname{Re}\frac1{2\pi i}\int_{l_{\theta}^R}\dfrac{{\rm d}f(z)}{f(z)}\right|=\left|\operatorname{Re}\frac1{2\pi i}\int_0^R\dfrac{{\rm d}f(te^{i\theta})}{f(te^{i\theta})}\right|\le1~(\forall \theta)$ (This is the most important claim actually. It can be proven by noticing that there is one $l_{\theta'}$ that doesn't intersect with $f(l_{\theta}^R)$, but I don't know the details) But we also have $\lim_{R\to\infty} \frac1{2\pi i}\int_{\gamma_{\theta}^R}\dfrac{{\rm d}f(z)}{f(z)}=\frac{64}{20}>2$ which means $\lim_{R\to\infty} \operatorname{Re}\frac1{2\pi i}\int_{\partial S_{\theta}^R}\dfrac{{\rm d}f(z)}{f(z)}>0$ leading to $\exists R,\frac1{2\pi i}\int_{\partial S_{\theta}^R}\dfrac{{\rm d}f(z)}{f(z)}\neq0$ so we conclude $\exists z\in S_{\theta}^R\subset S_{\theta},f(z)=0$. $\blacksquare$