We are supposed to find an upper bound on :
$E=|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2$ where $z_1,z_2,z_3$ are unimodular complex numbers, i.e $|z_i|=1$.
Let us take the first term : $|z_1-z_2|^2$
$|z_1-z_2|^2=(z_1-z_2)(\bar{z_1}-\bar{z_2})=|z_1|^2+|z_2|^2-z_1\bar{z_2}-\bar{z_1}z_2=|z_1|^2+|z_2|^2-2Re(z_1\bar{z_2})$
Taking : $z_j=e^{i\theta_j}$, hence $z_1\bar{z_2}=e^{i\theta_1}e^{-i\theta_2}=e^{i(\theta_1-\theta_2)}=\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)$.
So, $Re(z_1\bar{z_2})=\cos(\theta_1-\theta_2)$.
Now, $\cos(\theta_1-\theta_2)\geq(-1)$ i.e. $Re(z_1\bar{z_2})\geq(-1)$ = $(-2)Re(z_1\bar{z_2})\leq2$, thus,
$|z_1|^2+|z_2|^2+(-2)Re(z_1\bar{z_2})\leq2 + |z_1|^2+|z_2|^2$
Hence,
$|z_1-z_2|^2\leq 4$
Similarly, for second and third term, so we have :
$E\leq 12$
But the solution says $9$, What did I do wrong ? Can any one help ?
Best Answer
Note that $$|\cos(\theta_1-\theta_2)+\cos(\theta_2-\theta_3)+\cos(\theta_3-\theta_1)|\le \frac {3}{2}$$ because $z_1,z_2$,and $z_3$ are on the unit circle.
That should solve your problem with your upper bound.