複數不等式組

tommywong

artofproblemsolving.com/community/c7h2942222_complex_inequality

Let $z_1,z_2,z_3,z_4,z_5 \in \mathbb{C}$
and satisfy,
$$\begin{cases} |z_1|\leq 1\\  |z_2|\leq 1\\ \bigl |2z_3-(z_1+z_2)\bigr|\leq \bigl |z_1-z_2\bigr| \\  \\ \color{red}{\bigl |2z_4-(z_2+z_3)\bigr|\leq \bigl |z_2-z_3\bigr|}  \\ \\ \bigl |2z_5-(z_3+z_4)\bigr|\leq \bigl |z_3-z_4\bigr|\end{cases}$$
Find the maximum value of $\bigl|z_5\bigr|$.

---

Let $z_1,z_2,z_3,z_4,z_5 \in \mathbb{C}$
and satisfy,
$$\begin{cases} |z_1|\leq 1\\  |z_2|\leq 1\\ \bigl |2z_3-(z_1+z_2)\bigr|\leq \bigl |z_1-z_2\bigr| \\  \\ \color{green}{\bigl |2z_4-(z_1+z_2)\bigr|\leq \bigl |z_1-z_2\bigr|}  \\ \\ \bigl |2z_5-(z_3+z_4)\bigr|\leq \bigl |z_3-z_4\bigr|\end{cases}$$
Find the maximum value of $\bigl|z_5\bigr|$.

kuing

改写成平几，可能感兴趣的人会多一些

给定单位圆 `\odot O`。
两点 `A_1`, `A_2` 在 `\odot O` 内（含边界，下同）。
点 `A_3` 在以 `A_1A_2` 为直径的圆内；
点 `A_4` 在以 `A_2A_3` 为直径的圆内；
点 `A_5` 在以 `A_3A_4` 为直径的圆内。
求 `OA_5` 的最大值。

我不会。

从直观上不难看出，取最大值时一定各点都是在各自所限制的圆的边界上，不会在内部。

tommywong

Mathskidd話第4行不等式寫錯咗，我修改番題目。
應該係$\sqrt{3}$，但唔知做法係點。
$z_1=\left(\sqrt{\dfrac{1}{3}}+i\sqrt{\dfrac{2}{3}}\right)e^{i\theta}$
$z_2=\left(\sqrt{\dfrac{1}{3}}-i\sqrt{\dfrac{2}{3}}\right)e^{i\theta}$
$z_3=\left(2\sqrt{\dfrac{1}{3}}+i\sqrt{\dfrac{1}{3}}\right)e^{i\theta}$
$z_4=\left(2\sqrt{\dfrac{1}{3}}-i\sqrt{\dfrac{1}{3}}\right)e^{i\theta}$
$z_5=\sqrt{3}e^{i\theta}$

tommywong

$\dfrac{e^{i\alpha}+e^{i\beta}}{2}
=\dfrac{\cos\alpha+\cos\beta}{2}+i\dfrac{\sin\alpha+\sin\beta}{2}
=\cos\dfrac{\alpha-\beta}{2}e^{i\dfrac{\alpha+\beta}{2}}$

$\dfrac{e^{i\alpha}-e^{i\beta}}{2}
=\dfrac{\cos\alpha-\cos\beta}{2}+i\dfrac{\sin\alpha-\sin\beta}{2}
=\sin\dfrac{\beta-\alpha}{2}e^{i\dfrac{\pi-(\alpha+\beta)}{2}}$

$|\dfrac{e^{i\alpha}-e^{i\beta}}{2}|
=|\sin\dfrac{\alpha-\beta}{2}|$

假設取最大值時各點在圓的邊界上
Let $z_1=e^{i\theta_1},~z_2=e^{i\theta_2}$
$z_3=\dfrac{z_1+z_2}{2}+|\dfrac{z_1-z_2}{2}| e^{i\theta_3}$
$z_4=\dfrac{z_1+z_2}{2}+|\dfrac{z_1-z_2}{2}| e^{i\theta_4}$
$z_5=\dfrac{z_3+z_4}{2}+|\dfrac{z_3-z_4}{2}| e^{i\theta}$

$\dfrac{z_1+z_2}{2}=\cos\dfrac{\theta_1-\theta_2}{2}e^{i\dfrac{\theta_1+\theta_2}{2}}$
$|\dfrac{z_1-z_2}{2}|=|\sin\dfrac{\theta_1-\theta_2}{2}|$

$z_3=\cos\dfrac{\theta_1-\theta_2}{2}e^{i\dfrac{\theta_1+\theta_2}{2}}+|\sin\dfrac{\theta_1-\theta_2}{2}| e^{i\theta_3}$
$z_4=\cos\dfrac{\theta_1-\theta_2}{2}e^{i\dfrac{\theta_1+\theta_2}{2}}+|\sin\dfrac{\theta_1-\theta_2}{2}| e^{i\theta_4}$

$\dfrac{z_3+z_4}{2}=\cos\dfrac{\theta_1-\theta_2}{2}
e^{i\dfrac{\theta_1+\theta_2}{2}}
+|\sin\dfrac{\theta_1-\theta_2}{2}|
\dfrac{e^{i\theta_3}+e^{i\theta_4}}{2}$
$=\cos\dfrac{\theta_1-\theta_2}{2}
e^{i\dfrac{\theta_1+\theta_2}{2}}
+|\sin\dfrac{\theta_1-\theta_2}{2}|
\cos\dfrac{\theta_3-\theta_4}{2}
e^{i\dfrac{\theta_3+\theta_4}{2}}$

$|\dfrac{z_3-z_4}{2}|=|\sin\dfrac{\theta_1-\theta_2}{2}|
|\dfrac{e^{i\theta_3}-e^{i\theta_4}}{2}|
=|\sin\dfrac{\theta_1-\theta_2}{2}||\sin\dfrac{\theta_3-\theta_4}{2}|$

$z_5=\cos\dfrac{\theta_1-\theta_2}{2}
e^{i\dfrac{\theta_1+\theta_2}{2}}
+|\sin\dfrac{\theta_1-\theta_2}{2}|
\cos\dfrac{\theta_3-\theta_4}{2}
e^{i\dfrac{\theta_3+\theta_4}{2}}
+|\sin\dfrac{\theta_1-\theta_2}{2}||\sin\dfrac{\theta_3-\theta_4}{2}| e^{i\theta}$

$|z_5|\le |\cos\dfrac{\theta_1-\theta_2}{2}|
+|\sin\dfrac{\theta_1-\theta_2}{2}|
|\cos\dfrac{\theta_3-\theta_4}{2}|
+|\sin\dfrac{\theta_1-\theta_2}{2}||\sin\dfrac{\theta_3-\theta_4}{2}|$
$=|\cos\dfrac{\theta_1-\theta_2}{2}|
+|\sin\dfrac{\theta_1-\theta_2}{2}|
\left(|\cos\dfrac{\theta_3-\theta_4}{2}|
+|\sin\dfrac{\theta_3-\theta_4}{2}|\right)$
$\le |\cos\dfrac{\theta_1-\theta_2}{2}|
+\sqrt{2}|\sin\dfrac{\theta_1-\theta_2}{2}|
\le \sqrt{3}$

取等條件為
$\begin{cases}\theta_1+\theta_2=\theta_3+\theta_4=2\theta\\
\dfrac{\theta_3-\theta_4}{2}=\arccos\dfrac{1}{\sqrt{2}}=\dfrac{\pi}{4}\\
\dfrac{\theta_1-\theta_2}{2}=\arccos\dfrac{1}{\sqrt{3}}\end{cases}$

$\begin{cases}\theta_1=\theta+\arccos\dfrac{1}{\sqrt{3}}\\
\theta_2=\theta-\arccos\dfrac{1}{\sqrt{3}}\\
\theta_3=\theta+\dfrac{\pi}{4}\\
\theta_4=\theta-\dfrac{\pi}{4}\end{cases}$

$z_1=\left(\sqrt{\dfrac{1}{3}}+i\sqrt{\dfrac{2}{3}}\right)e^{i\theta}$
$z_2=\left(\sqrt{\dfrac{1}{3}}-i\sqrt{\dfrac{2}{3}}\right)e^{i\theta}$
$z_3=\left(2\sqrt{\dfrac{1}{3}}+i\sqrt{\dfrac{1}{3}}\right)e^{i\theta}$
$z_4=\left(2\sqrt{\dfrac{1}{3}}-i\sqrt{\dfrac{1}{3}}\right)e^{i\theta}$
$z_5=\sqrt{3}e^{i\theta}$