TITLE: Complex Numbers and their relationship with higher Mathematics
QUESTION [8 upvotes]: Let $z_1, z_2, \cdots, z_n$ be complex numbers satisfying
$$|z_1|+|z_2|+\cdots +|z_n|=1.$$
Prove that there is a non-empty subset of $\{z_1,z_2,\cdots,z_n\}$ the sum of whose elements has modulus at least $1/4.$
It was a problem from the Chinese Mathematical Olympiad and hence it has an elementary and beautiful solution. But the author also remarked that the bound $1/4$ can actually be improved to $1/\pi$ using higher mathematics. I tried to do it myself but have failed, can anyone help? I learned basic complex analysis, thanks.

REPLY [7 votes]: For $x\in\Bbb R$ let $x^+$ be the positive part of $x$; that is, $x^+=x$ if $x\ge0$, $x^+=0$ if $x<0$. Calculate that for any fixed $\theta$ $$\frac1{2\pi}\int_0^{2\pi}(\cos(t+\theta))^+\,dt=\frac1{2\pi}\int_0^{2\pi}(\cos(t))^+\,dt=\frac1\pi.$$
For $z\in\Bbb C$ let $$\phi(z)=(\Re z)^+.$$It follows that $$\frac1{2\pi}\int_0^{2\pi}\phi(e^{it}z)\,dt=\frac1\pi|z|.$$(Write $z=re^{i\theta}$...)
So $$\frac1{2\pi}\int_0^{2\pi}\sum_j\phi(e^{it}z_j)\,dt=\frac1\pi.$$Hence there exists $t$ with $$\sum_j\phi(e^{it}z_j)\ge\frac1\pi.$$Let $S=\{j\,:\,\Re(e^{it}z_j)\ge0\}$. Then the previous inequality shows that $$\Re\left( e^{it}\sum_{j\in S}z_j\right)\ge\frac1\pi,$$hence $$\left|\sum_{j\in S}z_j\right|\ge\frac1\pi.$$