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 $|z_1|=|z_2|=|z_3|=1$)?

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 $\max|Q|=8$.

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) $$...\leq 8$$