2

If $z_{1},z_{2},z_{3},z_{4}$ are $4$ points on a circle $|z| = 1$ such that $z_{1}+z_{2}+z_{3}+z_{4}=0\;,$

Then least value of expression $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{4}|^2+|z_{4}-z_{1}|^2$ is

$\bf{My\; Try::}$ Here $|z_{1}|=|z_{2}|=|z_{3}|=|z_{4}|=1$

So $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{4}|^2+|z_{4}-z_{1}|^2=2\left(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2-2\left[\Re(z_{1}\bar{z_{2}})+\Re(z_{2}\bar{z_{3}})+\Re(z_{3}\bar{z_{4}})+\Re(z_{4}\bar{z_{1}})\right]\right)$

So we get $$=8-4\left[\Re(z_{1}\bar{z_{2}})+\Re(z_{2}\bar{z_{3}})+\Re(z_{3}\bar{z_{4}})+\Re(z_{4}\bar{z_{1}})\right]$$ Now How can i solve after that , Help required, Thanks

Community
  • 1
juantheron
  • 53,015

1 Answers1

1

Firstly, in your last equation, shouldn't the second term have a $-2$ prefactor?

The condition $\sum z_i =0$ can be read as the centre of mass of the inscribed quadrilateral is the centre of the circle (the origin). I challenge you to find an example as such, that is not an inscribed rectangle!

Since they are all rectangles, the sum of squares of the sides is equal to $8$.

Chip
  • 1,199