Prove that: (1) In the complex plane, each line (except for the real axis) that crosses the origin has at most one point ${z}$, satisfy $$\frac {1+z^{23}}{z^{64}}\in\mathbb R.$$(2) For any non-zero complex number ${a}$ and any real number $\theta$, the equation $1+z^{23}+az^{64}=0$ has roots in $$S_{\theta}=\left\{ z\in\mathbb C\mid\operatorname{Re}(ze^{-i\theta })\geqslant |z|\cos\frac{\pi}{20}\right\}.$$Proposed by Yijun Yao
Problem
Source: 2023 China TST Problem 6
Tags: algebra, China TST
pco
14.03.2023 20:07
EthanWYX2009 wrote: Prove that: (1) In the complex plane, each line (except for the real axis) that crosses the origin has at most one point ${z}$, satisfy $\frac {1+z^{23}}{z^{64}}\in\mathbb R$. Setting $z=\rho e^{i\theta}$ with $\rho\ge 0$ and $\theta\in[0,2\pi)$ : $\frac{1+z^{23}}{z^{64}}\in\mathbb R$ $\iff$ $\rho>0$ and $\sin 64\theta+\rho^{23}\sin 41\theta=0$ and so : Either $\sin 41\theta=\sin 64\theta=0$ and so $\theta=k\pi$ (since $41$ and $64$ are coprime) and so line is x-axis. Either $\rho^{23}=-\frac{\sin 64\theta}{\sin 41\theta}$ and so at most one solution for $(\theta,\theta+\pi)$ (since $\rho>0$)
CANBANKAN
15.03.2023 01:41
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.
a22886
19.03.2023 10:35
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 $\exists z\in S_{\theta}^R\subset S_{\theta},f(z)=0$. $\blacksquare$