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.
 A: 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,
$$f_n = 2(k+1) \big(1 + 3\Gamma_n\cos'(0) + O(\Gamma_{n}^{2})\big) + k$$
$$\Rightarrow f_n = 3k+2 + O(k^{-1})$$
$$\Rightarrow n - f_n = O(k^{-1})$$
A: Here is a hacky solution for the $n=4$ case. I'm sure it is not what the problem-writer intended, but it should do the trick. By the way, where is this problem from?
Anyway, using the idea in the OP, I will prove that the max of $|(z_1+z_2)(z_2+z_3)(z_3+z_1)|$ is $1$. To do this, first note that WLOG we may suppose that $z_4=-1$ because given any $4$ points that lie in the interior of the circle satisfying $z_1+z_2+z_3+z_4=0$, we may dilate them until one touches the boundary without changing their sum while increasing $f_n$, and then we may rotate until that point is $-1$. Then, using the equation $z_1+z_2+z_3=1$, we may rewrite
$$|(z_1+z_2)(z_2+z_3)(z_3+z_1)|=|(z_1-1)(z_2-1)(z_3-1)|=|z_1-1||z_2-1||z_3-1|$$
Now, by AM-GM, $|z_1-1||z_2-1||z_3-1|\leq (\frac{|z_1-1|+|z_2-1|+|z_3-1|}{3})^3$ so it suffices to show that $g(z_1,z_2,z_3)=|z_1-1|+|z_2-1|+|z_3-1|\leq 3$. For this we use some geometric reasoning, and then calculus.
Claim: We may assume that $z_1$ and $z_2$ lie on the boundary of the circle.
Pf: Suppose $z_1,z_2,$ and $z_3$ all lie in the interior. Then we may slide $z_1$ in the direction $z_1-1$ while adjusting $z_2$ so that $z_1+z_2+z_3=1$ still holds until either $z_1$ or $z_2$ hits the boundary, all the while not decreasing $g(z_1,z_2,z_3)$. The point is that adding $\Delta(z_1-1)$ for $\Delta>0$ to $z_1$ increases $|z_1-1|$ by $\Delta$ because it adds in exactly the right direction while subtracting at most $\Delta(z_1-1)$ from $z_2$ decreases $|z_2-1|$ by at most $\Delta$. I can make this more precise if necessary, but anyway by choosing $\Delta$ minimal so that either $|z_1+\Delta(z_1-1)|=1$ or $|z_2-\Delta(z_1-1)|=1$, we may assume WLOG that either $z_1$ or $z_2$ has norm $1$. Repeating this argument with the remaining $2$ points in the interior, we prove the claim.
The above claim reduces the problem to a optimization in two parameters: maximize $h(x,y)=|e^{ix}-1|+|e^{iy}-1|+|e^{ix}+e^{iy}|$, where we have used that $z_3-1=-z_1-z_2$, subject to the constraint that $|z_3|=|1-e^{ix}-e^{iy}|\leq1$. For this, we use calculus:
$$\partial_xh(x,y)=\frac{\sin(x)}{\sqrt{2-2\cos(x)}}-\frac{\sin(x-y)}{\sqrt{2+2\cos(x-y)}}\\
\partial_yh(x,y)=\frac{\sin(y)}{\sqrt{2-2\cos(y)}}+\frac{\sin(x-y)}{\sqrt{2+2\cos(x-y)}}$$
To avoid the singularities of the derivative of the absolute value function, we want to optimize in a region away from $x=0$, $y=0$, or $x-y=\pi$. I leave it to you to verify that we can't have a maximum that satisfies the constraint in these regimes. So sweeping issues of differentiability under the rug, we solve $\partial_xh(x,y)=\partial_yh(x,y)=0$. Adding the two equations, we find that
$$\frac{\sin(x)}{\sqrt{1-\cos(x)}}=-\frac{\sin(y)}{\sqrt{1-\cos(y)}}$$
and note that we may write
$$\frac{\sin(x)}{\sqrt{1-1\cos(x)}}=\text{sgn}(\sin(x))\sqrt{1+\cos(x)}$$
which makes it easy to solve the above equation to find that $x=-y$. Plugging this into $\partial_xh(x,y)=0$, we will find that $\cos(x)=\cos(2x)$, which has solutions $x=0$ and $x=\pm \frac{2}{3}\pi$. Then you may check that $x=y=0$ is not the max and that $(x_0,y_0)=(\pm\frac{2}{3}\pi,\mp\frac{2}{3})$ is a local max, and therefore a global max. Plugging in, we see that $h(x_0,y_0)=3$ so we are done.
