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 $|z_1|=|z_2|=|z_3|=1$)? 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?
 A: 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 $\max|Q|=8$.
A: 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) 
$$...\leq 8$$
