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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 $$f_{n}(z_{1},z_{2},\cdots,z_{n})=|z^3_{1}+z^3_{2}+\cdots+z^3_{n}|$$

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

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    $\begingroup$ Just one correction: The $n = 2$ case has maximum 0 as $z_1 = -z_2$ and hence the cubes cancel. $\endgroup$ – Azhao17 Jul 8 '20 at 0:29
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    $\begingroup$ Evidences suggest that $\max f_4=3$ with a maximum point $\{z_1,z_2,z_3,z_4\}=\{1,\omega,\omega^2,0\}$ (where $\omega^2+\omega+1=0$). $\endgroup$ – Sangchul Lee Jul 8 '20 at 1:32
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    $\begingroup$ If $n = 3 k$ is a multiple of 3, then there is an easy solution. Note that $|\sum_i z_i^3| \le \sum_i |z_i^3| \le n$. Arrange the set $\{z_i\}$ in $n/3$ triples and choose, for the $z_i$-values in each of these triples, complex numbers $1,w,w^2$ with $w = \exp (j 2 \pi/3)$. Then $\sum_i z_i = 0$ and indeed $\sum_i z_i^3 = n$, so my first inequality is tight and the claim proved. Further, setting $z_i = 0$ in excess of $n = 3 k$ (see Sangchul Lee's comment), we can immediately say that $|\sum_i z_i^3| \le M$ with $M \ge 3 (n \; {\rm{div}} \; 3)$, where this doesn't yet constitute a bound. $\endgroup$ – Andreas Jul 8 '20 at 11:27
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    $\begingroup$ For $n=5$ there is $f_5(1, a, a, \bar a, \bar a) = 3.75$ with $a=-1/4+i \sqrt{15}/4$, which looks like the maximal value from numeric simulation. $\endgroup$ – Martin R Jul 8 '20 at 14:43
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    $\begingroup$ ... So for all $n=3k+2$ the maximum is at least $3k + 0.75$. $\endgroup$ – Martin R Jul 8 '20 at 14:52
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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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    $\begingroup$ Thanks,can you prove $n=4?$ Thanks $\endgroup$ – geromty Jul 15 '20 at 2:13
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    $\begingroup$ I think SangchulLee is right. I'm playing around with Lagrange to prove it, but I've nothing at the moment. $\endgroup$ – Sherwin Lott Jul 16 '20 at 3:14
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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.

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    $\begingroup$ n=4 is from china contest organized by a high school student,so Maybe have without derivative method,and the (2) is me add it. $\endgroup$ – geromty Mar 11 at 4:17
  • $\begingroup$ I see that you posted this question on mathoverflow but it was deemed off-topic. Might I suggest you try posting on artofproblemsolving.com/community and then report back? $\endgroup$ – user293794 Mar 11 at 17:25
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We have:

  • If $(z^*_1,z^*_2,\dots,z^*_n)\,$ is the solution of the given task, then $\max(|z^*_1|,|z^*_2|,\dots,|z^*_n|)=1.\,$

  • If $(z^*_1,z^*_2,\dots,z^*_n)\,$ is the solution of the given task, then $(z^*_1e^{i\varphi},z^*_2e^{i\varphi},\dots,z^*_ne^{i\varphi})\,$ is the solution too.
    This allows to assume $$z_n=-1,\quad z_1+z_2+\dots z_{n-1}=1.\tag1$$

Сondition $(1)$ means that geometrically complex numbers are displayed as segments of a closed polyline.

On the other hand, $z^3_n=-1,$ and the triple $$z_k = \bar z_k - e^{\large-2ik\LARGE^\pi\!/\!_3}, \quad z^3_1=z^3_2=z^3_3 = -1,\quad z_1+z_2 = 1\tag2$$

gives the stable optimal solution, if $\;\mathbf{n=3}.\;$ In this way, we have $$|s^\,_{3k+p}|\ge 3k+|s^\,_p|,\quad |s^\,_{k+1}|\ge |s_k|, \quad |s^\,_{3k}|=3k.\tag3$$


In the case $\;\color{brown}{\mathbf {n=4}}\;$ one can get $$s_4=z_1^3+z_2^3+z_3^3 -1 = (z_1+z_2+z_3)^3-3(z_1+z_2)(z_2+z_3)(z_3+z_1)-1,$$ $$s_4 = -3(z_1+z_2)(1-z_1)(1-z_2).$$ Then, in the notation $$z_1=p+iq,\quad z_2=u+iv,$$ $$|s_4|^2=f(p,q,u,v)=9((1-p)^2+q^2)((1-u)^2+v^2)\big((p+u)^2 +(q+v)^2\big),\tag4$$ where $$0\le p^2+q^2\le1,\quad0\le u^2+v^2\le1,\quad 0\le (p+u-1)^2+(q+v)^2\le1\tag5$$ The inner stationary points of the $\;f(p,q,u,v)\;$ can be defined from the system $\;f'_p=f'_q=f'_u=f'_v=0.\;$ Then \begin{cases} (p-1)\big((p+u)^2 +(q+v)^2\big)+(p+u)((p-1)^2+q^2)=0\\ q\big((p+u)^2 +(q+v)^2\big)+(q+v)((p-1)^2+q^2)=0\\ (u-1)\big((p+u)^2 +(q+v)^2\big)+(p+u)((u-1)^2+v^2)=0\\ v\big((p+q-1)^2 +(q+v)^2\big)+(q+v)((u-1)^2+v^2)=0, \end{cases} $\qquad\qquad\left|(p-1)\times[1]+q\times[2],\quad q\times[1]-(p-1)\times[2]\dots\Large\mathstrut\right|$

\begin{cases} \big((p+u)^2 +(q+v)^2\big)+(p-1)(p+u)+q(q+v)=0\\ q(p+u-1)-(p-1)(q+v)=0\\ \big((p+u)^2 +(q+v)^2\big)+(u-1)(p+u)+v(q+v)=0\\ v(p+u)-(u-1)(q+v)=0, \end{cases}

\begin{cases} (2p+u-1)(p+u)+(2q+v)(q+v)=0\\ (p+2u-2)(p+u)+(q+2v)(q+v)=0\\ q(p+u)-(p-1)(q+v)=0\\ v(p+u)-(u-1)(q+v)=0, \end{cases}

$$(p+u-1)^2+(q+v)^2=0.$$ Thus, the greatest value of $\;f\;$ takes place on the edges of the area $(5).$

Case $\;\mathbf{(p+u-1)^2+(q+v)^2 = 1,\; (p-1)^2+q^2\in (0,1),\; (u-1)^2+v^2\in(0,1)}.\;$

Applying Lagrange multipliers method, we can find the greatest value of the function $$\begin{align} &g(p,q,u,v,\lambda) = 18(p+u)\big((p-1)^2+q^2\big)\big((u-1)^2+v^2\big)\\[4pt] &+\lambda\big((p+u)^2+(q+v)^2-2(p+u))\big), \end{align}\tag6$$ wherein the stationary points are defined via the system $\;g'_p=g'_q=g'_u=g'_v=g'_\lambda=0,\;$ or \begin{cases} 18\big((u-1)^2+v^2\big)\big((p-1)^2+q^2+2(p+u)(p-1)\big) + 2\lambda(p+u-1)=0\\ 18q(p+u)\big((u-1)^2+v^2\big) + \lambda(q+v)=0\\ 18\big((p-1)^2+q^2\big)\big((u-1)^2+v^2+2(p+u)(u-1)\big) + 2\lambda(p+u-1)=0\\ 18v(p+u)\big((p-1)^2+q^2\big) + \lambda(q+v)=0\\ (p+u)^2+(q+v)^2=2(p+u), \end{cases}

\begin{cases} (p-1)((u-1)^2+v^2)=(u-1)((p-1)^2+q^2)\\ q((u-1)^2+v^2)=v((p-1)^2+q^2)\\ (q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\ (p+u)^2+(q+v)^2=2(p+u), \end{cases}

$\qquad\qquad\left|(p-1)\times[1]+q\times[2],\quad (u-1)\times[1]+v\times[2]\dots\Large\mathstrut\right|$

\begin{cases} (u-1)^2+v^2 =(p-1)(u-1)+qv\\ (p-1)^2+q^2 =(p-1)(u-1)+qv\\ (q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\ (p+u)^2+(q+v)^2=2(p+u), \end{cases}

\begin{cases} (u-p)^2+(v-q)^2=0\\ (p+u)^2+(q+v)^2=2(p+u),\\ (q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\ \end{cases} $$u=p,\quad v=q,\quad p^2+q^2=p,\quad (p-1)^2+4p(p-1)+q^2 = 2p(2p-1),$$ $$u=p=\dfrac13,\quad v=q=\dfrac{\sqrt2}3,\quad g=\dfrac{16}3,\quad |s_4| = \dfrac4{\sqrt3}.$$

Case $\;\mathbf{(p+u-1)^2+(q+v)^2 = 0,\; (p-1)^2+q^2\in (0,1),\; (u-1)^2+v^2\in(0,1)}.\;$

We can find the greatest value of the function $$\begin{align} &h(p,q)=f(p,q,1-p,v) = 9\big((p-1)^2+q^2\big)\big(p^2+q^2\big)\\[4pt] \end{align}\tag7$$ under the conditions $$(p-1)^2+q^2\le1,\quad p^2+q^2\le 1.\tag8$$ Since the goal function increases by $\;q,\;$ then its greatest value achieves under the condition $$q^2=\begin{cases} 1-p^2,\quad\text{if}\quad p\in[0,\,^1\!/_2]\\[4pt] 1-(1-p)^2,\quad\text{if}\quad p\in[\,^1\!/_2,1]. \end{cases}$$

Then $$\max\limits_{\large p\in[0,\,^1\!/_2]} h(p,\sqrt{1-p^2}) = \max\limits_{\large p\in[0,\,^1\!/_2]} 9\big(1+(1-p)^2-p^2\big) = 9\quad\text{at}\quad p=\,^1\!/_2,$$ $$\max\limits_{\large p\in[\,^1\!/_2,1]} h(p,\sqrt{1-(1-p)^2}) = \max\limits_{\large p\in[\,^1\!/_2,1]} 9\big(1+p^2-(1-p)^2\big) = 9\quad\text{at}\quad p=\,^1\!/_2,$$ with the known solution for $\;n=3.\;$

Therefore, $\color{brown}{\mathbf{|s_4|=|s_3|= 3.}}$


$\color{brown}{\textbf{Splitting strategy.}}$

Let us find $\;|s_2(t)|,\;$ where $$s_2=\min\big(z_1^3+(t-z_1)^3\big)\in\mathbb R,\quad i\in\mathbb R,\quad s_2<0.$$ This leads to the minimizing task for $$f_2(t,p,q)= (t-p-iq)^3+(p+iq)^3 = t\big(3(p+iq)^2-3t(p+iq)+t^2\big),$$ which is real if $\;p=\frac12t.\;$

Therefore,. $$g_2(t,q)=f_2\left(t,\dfrac t2,q\right) = (p-iq)^3+(p+iq)^3 = 2p^3-6pq^2 = \frac14t^3-3q^2t,$$ $$\min g_2(t,q)=f_2\left(t,\sqrt{1-\dfrac14t^2}\right) = -3t+t^3.$$

Then $$|s^\,_3|\ge|s^\,_2(1)-1|=3,$$ $$|s^\,_5|\ge\left|2s^\,_2\left(\dfrac12\right)-1\right|=\dfrac{15}4 > 3,$$ $$|s^\,_7|\ge\left|3s^\,_2\left(\dfrac13\right)-1\right|=\dfrac{35}9,$$ $$|s^\,_{7}|\ge\left|2s^\,_2\left(\dfrac32\right)-3\right|=\dfrac{21}4,$$ $$|s^\,_8|\ge\left|3s^\,_2\left(\dfrac23\right)-2\right|=\dfrac{64}9,$$ $$|s^\,_{10}|\ge\left|4s^\,_2\left(\dfrac12\right)-2\right|=\dfrac{15}2,$$ $$|s^\,_{10}|\ge\left|3s^\,_2\left(\dfrac43\right)-4\right|=\dfrac{80}9,$$ $$|s^\,_{13}|\ge\left|5s^\,_2\left(\dfrac35\right)-3\right|=\dfrac{273}{25},$$ $$|s^\,_{13}|\ge\left|4s^\,_2\left(\dfrac54\right)-5\right|=\dfrac{195}{16}>12\dots$$ $$|s^\,_{35}|\ge\left|12s^\,_2\left(\dfrac{11}{12}\right)-11\right|=\dfrac{5005}{144}=34\dfrac{109}{144}>34\dfrac34\dots$$

Obtained solutions allow to improve the primary estimations $(3).$

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