TITLE: Inequality with complex numbers
QUESTION [23 upvotes]: Consider the following problem.
Fix $n \in \mathbb N$. Prove that for every set of complex numbers $\{z_i\}_{1\le i \le n}$, there is a subset $J\subset \{1,\dots , n\}$ such that 
$$\left|\sum_{j\in J} z_j\right|\ge \frac{1}{4\sqrt 2} \sum_{k=1}^n |z_k|.$$
I believe I proven have a stronger statement. Is this proof correct, and if so, what is the optimal constant?
My proof. Consider all the $z_i$ with positive real part. Call the real part of the sum of these numbers $X^+$. In a similar way, form $X^-$, $Y^+$, and $Y^-$. Without loss of generality, let $X^+$ have the greatest magnitude of these.
Note that because $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|\ge |z|$, we have
$$ \left(\sum_{k=1}^n  |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| \right) \ge \sum_{k=1}^n |z_k|.$$
But note that $\sum \limits_{k=1}^n  |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| = X^+ + |X^-|+ Y^+ +|Y^-|$, so we have 
$$ 4X^+ \ge \sum_{k=1}^n |z_k|.$$ By choosing $J$ to be the set of complex number with positive real part, this proves a stronger statement, because the factor of $1/\sqrt 2$ isn't needed.

REPLY [2 votes]: Let $z_k = r_k e^{i\varphi_k}$, $k\in[n] = \{1, 2, \dots, n \}$.
\begin{align*}
\max_{I\subseteq [n]}\left|\sum_{k\in I}z_k \right| &\ge
\frac{1}{2\pi}\int_0^{2\pi}\left|\sum_{{\rm Im}(z_ke^{ix})\ge 0} z_k e^{ix}\right|\,dx
\\&\ge \frac{1}{2\pi}\int_0^{2\pi} {\rm Im} \sum_{{\rm Im}(z_ke^{ix})\ge 0} z_k e^{ix} \,dx
\\& = \sum_{k=1}^n \frac{1}{2\pi}\int_{0\le \varphi_k + x\le \pi} {\rm Im}(r_ke^{i(\varphi_k + x)})\,dx
\\& = \sum_{k=1}^n \frac{r_k}{2\pi}\int_{0\le \varphi_k + x\le \pi} \sin(\varphi_k + x)\,dx
\\& = \sum_{k=1}^n \frac{r_k}{2\pi} \cdot 2
\\& = \frac{1}{\pi}\sum_{k=1}^n |z_k|
\end{align*}