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?
Best Answer
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 $\max|Q|=8$.