Find maximum and minimum of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$)

Question

How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ?

My try:

$$\begin{align}\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right| &\leq |z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \\ &\leq (|z_1|+|z_2|)^2 + (|z_2|+|z_3|)^2 + (|z_3|+|z_1|)^2 \\ &\leq 2^2+2^2+2^2 \leq 12\end{align}$$

However the answer given is $8$. Where am I going wrong and how to do it correctly?

Answer 1

Let $$Q:=(z_0-z_1)^2+(z_1-z_2)^2+(z_2-z_0)^2\in{\mathbb C}\ .$$ The minimal value of $|Q|$ is of course $0$, which is attained when $z_0=z_1=z_2$, but also for an equilateral triangle. In order to determine $\max|Q|$ under the given constraints we may assume $$z_0=e^{it},\quad z_1=-e^{-i\alpha},\quad z_2=-e^{i\alpha}$$ with $t\in{\mathbb R}$ and $0\leq\alpha\leq{\pi\over3}$. Then $$\eqalign{Q&=(e^{it}+e^{i\alpha})^2+(e^{it}+e^{-i\alpha})^2+(2i\sin\alpha)^2 \cr &=2e^{2it}+4e^{it}\cos\alpha+8\cos^2\alpha-6\ . \cr}$$ Put $\cos\alpha=:p\in\bigl[{1\over2},1\bigr]$. Then $$|Q|\leq2+4p+|8p^2-6|\ .$$ If ${\sqrt{3}\over2}\leq p\leq1$ then $$|Q|\leq2+4p+8p^2-6=8\left(p+{1\over4}\right)^2-{9\over2}\leq{25\over2}-{9\over2}=8\ ,$$ and if ${1\over2}\leq p\leq{\sqrt{3}\over2}$ then $$|Q|\leq2+4p+6-8p^2={17\over2}-8\left(p-{1\over4}\right)^2\leq{17\over2}-{1\over2}=8\ .$$ On the other hand $z_0=1$, $z_1=z_2=-1$ gives $|Q|=8$, so that altogether we have proven that $\max|Q|=8$.

Answer 2

Given $$(z_1-z_2)+(z_2-z_3)+(z_3-z_1)=0$$ and $$\left|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2\right|=\\ \left|z_1^2-2z_1z_2+z_2^2+z_2^2-2z_2z_3+z_3^2+z_3^2-2z_3z_1+z_1^2\right|=\\ 2\left|z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(z_3-z_1)\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(-(z_1-z_2)-(z_2-z_3))\right|=\\ 2\left|(z_1-z_2)(z_1-z_3)+(z_2-z_3)^2)\right|=...$$ replacing $z_1=1$ $$...=2\left|(1-z_2)(1-z_3)+(z_2-z_3)^2)\right|=2\left|(1-z_2)(1-z_3)+(z_2-1+1-z_3)^2\right|=\\ 2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2+2(z_2-1)(1-z_3)\right|=\\ 2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2-2(1-z_2)(1-z_3)\right|=\\ 2\left|(1-z_2)^2+(1-z_3)^2-(1-z_2)(1-z_3)\right|=...$$ which is $$...=2\left|\frac{(1-z_2)^3+(1-z_3)^3}{1-z_2+1-z_3}\right|=...$$ using law of sines ... $$...=2\left|\frac{2^3\sin^3{\alpha}+2^3\sin^3{\beta}}{2\sin{\alpha}+2\sin{\beta}}\right|=8\left|\frac{\sin^3{\alpha}+\sin^3{\beta}}{\sin{\alpha}+\sin{\beta}}\right|\leq ...\tag{1}$$ both $\alpha, \beta \in (0,\pi)$ (corner cases can be treated individually), which means $$0<\sin{\alpha}\leq 1,0<\sin{\beta}\leq 1$$ or $$0<\sin^3{\alpha}\leq \sin{\alpha}<1,0<\sin^3{\beta}\leq \sin{\beta}<1$$ thus $$0<\sin^3{\alpha} + \sin^3{\beta} \leq \sin{\alpha} + \sin{\beta}$$ and, continuing (1) $$...\leq 8$$

Answer 3

Supplement to @rtybase's very nice answer.

WLOG, assume that $z_1 = 1$.

When $z_2 = 1, z_3 = -1$, we have $\left|(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2\right| = 8$.

We claim that the maximum is $8$.

It suffices to prove that $$\left|(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2\right| \le 8.$$

Clearly, we only need to prove the case that $\mathrm{Re}(2 - z_2 - z_3) > 0$ (otherwise $z_2 = z_3 = 1$).

We have \begin{align*} &\left|(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2\right|\\ ={}& \left|(1 - z_2)^2 + (1 - z_3 - (1 - z_2))^2 + (1 - z_3)^2\right|\\ ={}& 2\left|(1 - z_2)^2 - (1-z_2)(1-z_3) + (1 - z_3)^2\right|\\ ={}& 2\cdot \left|\frac{(1 - z_2)^3 + (1 - z_3)^3}{1 - z_2 + 1 - z_3}\right|\\ \le{}& 2\cdot \frac{|1-z_2|^3 + |1 - z_3|^3}{\mathrm{Re}(2 - z_2 - z_3)} \end{align*} where we use $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, and $|(1 - z_2)^3 + (1 - z_3)^3|\le |1-z_2|^3 + |1 - z_3|^3$ by triangle inequality, and $|u| \ge \mathrm{Re}(u)$, and $\mathrm{Re}(2 - z_2 - z_3) > 0$.

Let $z_2 = \cos (2\alpha) + \mathrm{i}\sin(2\alpha)$ and $z_3 = \cos (2\beta) + \mathrm{i}\sin(2\beta)$. We have $|1-z_2| = 2|\sin \alpha|$, and $|1 - z_3| = 2|\sin \beta|$, and $\mathrm{Re}(2 - z_2 - z_3) = 2\sin^2\alpha + 2\sin^2\beta$. We have $$ 2\cdot \frac{|1-z_2|^3 + |1 - z_3|^3}{\mathrm{Re}(2 - z_2 - z_3)} = 2\cdot \frac{8|\sin \alpha|^3 + 8|\sin \beta|^3}{2\sin^2\alpha + 2\sin^2\beta} \le 8 $$ where we use $|\sin\alpha|^3 \le \sin^2\alpha$ and $|\sin \beta|^3 \le \sin^2\beta$.

We are done.