$z_1^2+z_2^2+z_3^2=3z_0^2$ if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre

Question

Prove that, if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre, then $z_1^2+z_2^2+z_3^2=3z_0^2$

My Attempt $$ z_0=\frac{z_1+z_2+z_3}{3}\implies 3z_0=z_1+z_2+z_3\\ 9z_0^2=z_1^2+z_2^2+z_3^2+2z_1z_2+2z_2z_3+2z_3z_1 $$

How do I proceed further and complete the proof ?

Answer 1

Here, we clearly know (by rotation theorem):
$$\frac{z_2-z_0}{z_1-z_0} = \frac{z_3-z_0}{z_2-z_0} = \frac{z_1-z_0}{z_3-z_0}$$

On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$

Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$

Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$

Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$

Answer 2

Let $$z_1=z_0+r(\cos\theta+i\sin\theta),$$ $$z_2=z_0+r(\cos(120^{\circ}+\theta)+i\sin(120^{\circ}+\theta))$$ and $$z_3=z_0+r(\cos(240^{\circ}+\theta)+i\sin(240^{\circ}+\theta)).$$ Id est, it's enough to prove that $$\cos2\theta+\cos(240^{\circ}+2\theta)+\cos(480^{\circ}+2\theta)=0,$$ $$\sin2\theta+\sin(240^{\circ}+2\theta)+\sin(480^{\circ}+2\theta)=0,$$ $$\cos\theta+\cos(120^{\circ}+\theta)+\cos(240^{\circ}+\theta)=0$$ and $$\sin\theta+\sin(120^{\circ}+\theta)+\sin(240^{\circ}+\theta)=0.$$ Can you end it now?