Let $C=\{ z \in \mathbb{C} : |z|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, z_{240} \in C$ (not necessarily different) be $240$ complex numbers, satisfying the following two conditions: (1) For any open arc $\Gamma$ of length $\pi$ on $C$, there are at most $200$ of $j ~(1 \le j \le 240)$ such that $z_j \in \Gamma$. (2) For any open arc $\gamma$ of length $\pi/3$ on $C$, there are at most $120$ of $j ~(1 \le j \le 240)$ such that $z_j \in \gamma$. Find the maximum of $|z_1+z_2+\ldots+z_{240}|$.
Problem
Source: 2022 China TST, Test 1, P5
Tags: inequalities, complex numbers
Private-Name
28.03.2022 08:26
Let the $240$ plurals be $e^{\theta_1 i},e^{\theta_2 i}, \cdots ,e^{\theta_{240}i},0 \leqslant \theta_1 \leqslant \theta_2 \leqslant \cdots \leqslant \theta_{240} \leqslant 2 \pi ,z_k=e^{\theta_k i}$. And let $\omega_k=z_k+$$z_{k+40}+z_{k+80}+z_{k+120}+z_{k+160}+z_{k+200},(1\leqslant k \leqslant 40).$For the complex number $\omega_k=a_{0 k}+a_{1 k}+a_{2 k}+a_{3 k}+a_{4 k}+a_{5 k},(1\leqslant k \leqslant 40),a_{i j}=z_{40 i+j}$,it should satisfy the following $2$ conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on the unit circle, at most $5$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\Gamma$.
(2) For any open arc $\gamma$ of length $ \frac \pi 3$ on the unit circle, at most $3$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\gamma$.
So we have$|\omega|=|a_0+a_1+a_2+a_3+a_4+a_5| \leqslant |(a_0+a_5)+(a_1+a_4)+a_2+a_3|$
$\leqslant |2\cos(\frac \pi 2+\alpha)+2\cos(\frac \pi 6+\beta)+1+1|\leqslant 2+\sqrt3(\alpha,\beta\in[0,\frac\pi2])$
so$\displaystyle|\sum_{i=1}^{240}z_i|=|\sum_{i=1}^{40}\omega_i|\le40(2+\sqrt3).$
And when $z_1=z_2=\cdots=z_{40}=i,z_{41}=z_{42}=\cdots=z_{80}=-i,z_{81}=z_{82}=\cdots=$$z_{120}=\frac{\sqrt3}2+\frac 12i,$
$z_{121}=z_{122}=\cdots=z_{160}=\frac{\sqrt3}2-\frac 12i,z_{161}=z_{162}=\cdots=z_{240}=1$,we have$\displaystyle|\sum_{i=1}^{240}z_i|=80+40\sqrt3$
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szjzc2018
05.04.2022 05:25
Just notice that $z_1+z_2+...+z_{240}=\int_0^{2\pi}F(\theta)e^{i\theta}d\theta$,where F($\theta$) indicates the number of $z_i$s in the arc $(\theta-\frac{\pi}{6},\theta+\frac{\pi}{6})$ This intergal equals $\frac{1}{6} \int_0^{2\pi}(F(\theta)e^{i\theta}+F(\theta+\frac{\pi}{3})e^{i(\theta+\frac{\pi}{3})}+...F(\theta+\frac{5\pi}{3})e^{i(\theta+\frac{5\pi}{3})})d\theta$ WLOG $z_1+z_2+...+z_{240}$ belongs to R+. Ignoring the finite breaking point,all we have to notice is when there are 40 $i$s,40 $-i$s,80 $1$s,40 $w$s,40 $w^5$s($w=e^{\frac{i\pi}{3}}$),for all $\theta$,$Re((F(\theta)e^{i\theta}+F(\theta+\frac{\pi}{3})e^{i(\theta+\frac{\pi}{3})}+...F(\theta+\frac{5\pi}{3})e^{i(\theta+\frac{5\pi}{3})}))$are maximized. Thus the answer is $80+40\sqrt{3}$
mihaig
05.04.2022 05:52
Masterpieces
msc_helloworld
19.04.2022 11:12
1、the problem can be reduced to 6 complex numbers, satisfying the following two conditions: (3)For any open arc $\Gamma$ of length $\pi$ on $C$,there are at most 5 complex numbers such that they $\in \Gamma$. (4)For any open arc $\gamma$ of length $\pi/3$ on $C$,there are at most 3 complex numbers such that they $\in \gamma$. 2、Not too hard to guess the maximum is $sqrt(3)+2$, when they are 1 -1 i i $exp(i*pi/3)$ $exp(i*2pi/3)$ ,just notice that they should as close as possible at the unit circle, thus one is 1 and one is -1 should be a good choice, since condition (3) restrict they can not be more close. 3、Next step is the key: when you want to prove the length of the sum of 6 complex numbers(=Y) $<=$ $sqrt(3)+2$, we try to prove the length of the component of Y in any direction is equal or less than $\sqrt(3)+2$. To accomplish this , just discuss how many complex numbers in a half unit circle, which is easy to do so.(at first 4, then 5, then 3, then 2, then 1, then 0). the answer is $80+40\sqrt(3)$, Done.