Inequality with complex numbers

The constant $\frac{1}{4 \sqrt{2}}$ can be replaced by $\frac{1}{\pi}$, which is the best possible constant independent of $n$.

Let $R(z) = \max(0, \text{Re}(z))$. Choose $\theta \in [0,2\pi]$ to maximize $F(z_1,\ldots, z_n,\theta) = \sum_{j=1}^n R(e^{i\theta} z_j)$. Note that for any complex number $z$, $$\frac{1}{2\pi} \int_0^{2 \pi} R(e^{i \theta} z) \ d\theta = \frac{|z|}{2 \pi} \int_{0}^\pi \sin \theta \ d\theta = \frac{|z|}{\pi}$$ The maximal value of $F(z_1,\ldots,z_n,\theta)$ is at least the average value for $\theta \in [0,2\pi]$, namely $\frac{1}{\pi} \sum_{j=1}^n |z_j|$. Now note that if $J = \{j: R(e^{i\theta} z_j) > 0\}$, $\left|\sum_{j \in J} z_j\right| \ge \text{Re} \sum_{j \in J} e^{i \theta} z_j = F(z_1,\ldots,z_n,\theta)$.

To see that this estimate is best possible, consider cases where $n$ is large and the $z_n$ are the $n$'th roots of unity.

For any complex $|w|=1$, $w\cdot z\le|z|$. Therefore, $$ \sum_{k=1}^n z_k\cdot w\le\left|\sum_{k=1}^n z_k\right|\tag{1} $$ If we consider complex numbers that are contained in a wedge with angle $\theta$, then we have that by letting $w$ be the unit complex number in the middle of the wedge $$ \begin{align} \left|\sum_{k=1}^n z_k\right| &\ge\sum_{k=1}^n z_k\cdot w\\ &\ge\sum_{k=1}^n |z_k|\cos(\theta/2)\tag{2} \end{align} $$ because the angle between $z_k$ and $w$ is at most $\theta/2$.

Note that for a given angle $\theta$, we can find a wedge $W$ of angle $\theta$ that $$ \sum_{z_k\in W} |z_k|\ge\frac{\theta}{2\pi}\sum_{k=1}^n |z_k|\tag{3} $$ that is, there must be a wedge that has at least the average of all wedges.

Putting together $(2)$ and $(3)$, we have $$ \begin{align} \left|\sum_{z_k\in W} z_k\right| &\ge\cos(\theta/2)\sum_{z_k\in W} |z_k|\\ &\ge\frac{\theta}{2\pi}\cos(\theta/2)\sum_{k=1}^n |z_k|\tag{4} \end{align} $$ The maximum of $\dfrac{\theta}{2\pi}\cos(\theta/2)$ is $0.1786$ when $\theta$ is $1.720667$. However, using $\theta=\pi/2$, we get $\dfrac{\theta}{2\pi}\cos(\theta/2)=\frac{1}{4\sqrt{2}}$. Plugging this into $(4)$, we get $$ \left|\sum_{z_k\in W} z_k\right|\ge\frac{1}{4\sqrt{2}}\sum_{k=1}^n |z_k|\tag{5} $$

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*}