<s>[INST] How find the maximum of the complex number inequality 
Let $z_{i}(i=1,2,\cdots,n)$ be complex numbers such that $|z_{i}|\le 1,z_{1}+z_{2}+\cdots+z_{n}=0$. Define
$$f_{n}(z_{1},z_{2},\cdots,z_{n})=|z^3_{1}+z^3_{2}+\cdots+z^3_{n}|$$

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*If $n=4$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.

*For any positive $n$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.


I have attempted the $n=2,3$ cases, here is my try:
If $n=2$, then $f_{2}(z_1, z_2)=|z_1^3+(-z_1)^{3}|=0$.
If $n=3$, use the identity $a^3+b^3+c^3=3abc$ if $a+b+c=0$, so
$$f_{3}(z_{1},z_{2},z_{3})=3|z_{1}z_{2}z_{3}|\le 3$$
If $z_{1}=1,z_{2}=w,z_{3}=w^2$ is maximum, where $w^3=1,w\neq 1$
For the $n=4$ case, use $$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
then
$$f_{4}(z_{1},z_{2},z_{3},z_{4})=3|(z_{1}+z_{2})(z_{2}+z_{3})(z_{3}+z_{1})|$$but I can't get any farther.
 [/INST] Here's the asymptotic answer:  $n - \max f_n$ converges to zero! Justification will be broken up into mod $3$ cases.
(Sorry, I don't have a more exact answer and this was too long to write as a comment. )

Suppose $n=3k$, then $z_{k} = \exp(k \cdot 2\pi  i/3)$ gives $f_n = n$.

Suppose $n=3k+1$, the intuition is to only slightly change the previous and rely on the fact that $\cos'(0) = 0$.
Define $\Delta_n$ to be the smallest positive number such that:
$$-2k \cos(2\pi/3 + \Delta_n) = k+1$$
Choose:
$$z_{1},\ldots,z_{k} = \exp(2\pi/3 + \Delta_n)$$
$$z_{k+1},\ldots,z_{2k} = \exp(4\pi/3 - \Delta_n)$$
$$z_{2k+1},\ldots,z_{n} = 1$$
By construction, $f_n = 2k \cos(3\Delta_n) + k+1$ and the constraint is satisfied.
Clearly $\Delta_n \rightarrow 0$, so a first order Taylor approximation gives:
$$-2k \big(\cos(2\pi/3) + \Delta_n \cos'(2\pi/3) + O(\Delta_{n}^{2})\big) = k+1$$
$$\Rightarrow  \Delta_n = (2k\sin(2\pi/3))^{-1} + O(k^{-2})$$
Then,
$$f_n = 2k \big(1 + 3\Delta_n\cos'(0) + O(\Delta_{n}^{2})\big) + k+1$$
$$\Rightarrow f_n = 3k+1 + O(k^{-1})$$
$$\Rightarrow n - f_n = O(k^{-1})$$

Suppose $n=3k+2$, this is the exact same intuition.
Define $\Gamma_n$ to be the smallest positive number such that:
$$-2(k+1) \cos(2\pi/3 + \Gamma_n) = k$$
Choose:
$$z_{1},\ldots,z_{k+1} = \exp(2\pi/3 - \Gamma_n)$$
$$z_{k+2},\ldots,z_{2k+2} = \exp(4\pi/3 + \Gamma_n)$$
$$z_{2k+3},\ldots,z_{n} = 1$$
By construction, $f_n = 2(k+1) \cos(3\Gamma_n) + k$ and the constraint is satisfied.
Clearly $\Gamma_n \rightarrow 0$, so a first order Taylor approximation gives:
$$-2(k+1) \big(\cos(2\pi/3) + \Gamma_n \cos'(2\pi/3) + O(\Gamma_{n}^{2})\big) = k$$
$$\Rightarrow  \Gamma_n = (2k\sin(2\pi/3))^{-1} + O(k^{-2})$$
Then,
$$f_n = 2(k+1) \big(1 + 3\Gamma_n\cos'(0) + O(\Gamma_{n}^{2})\big) + k$$
$$\Rightarrow f_n = 3k+2 + O(k^{-1})$$
$$\Rightarrow n - f_n = O(k^{-1})$$
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