If z_1,z_2,z_3 are the vertices of an equilateral triangle…

If $z_1,z_2,z_3$ are the vertices of an equilateral triangle, verify that $\ z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$.

Step 1 · Center at the centroid

Let $c=\dfrac{z_1+z_2+z_3}3$ and $w_k=z_k-c$, so $w_1+w_2+w_3=0$. For an equilateral triangle centered at its centroid, the vertices are related by $120^\circ$ rotation: $\ w_2=\omega w_1,\ w_3=\omega^2w_1$, where $\omega=e^{2\pi i/3}$ satisfies $\omega^3=1$ and $1+\omega+\omega^2=0$.

Step 2 · Evaluate both sides for the centered triangle

$$ w_1^2+w_2^2+w_3^2=w_1^2(1+\omega^2+\omega^4)=w_1^2(1+\omega^2+\omega)=w_1^2\cdot0=0 $$

(using $\omega^4=\omega$ since $\omega^3=1$). Similarly:

$$ w_1w_2+w_2w_3+w_3w_1=w_1^2(\omega+\omega^3+\omega^2)=w_1^2(\omega+1+\omega^2)=w_1^2\cdot0=0 $$

Both sides vanish for the centered triangle.

Step 3 · Translate back to $z_k=w_k+c$

$$ \sum_kz_k^2=\sum_k(w_k+c)^2=\sum_kw_k^2+2c\sum_kw_k+3c^2=0+0+3c^2=3c^2 $$

$$ \sum_{k<j}z_kz_j=\sum_{k<j}(w_k+c)(w_j+c)=\sum_{k<j}w_kw_j+c\sum_{k}(3-1)w_k+3c^2 $$

More directly: $\sum_{k<j}(w_k+c)(w_j+c)=\sum_{k<j}w_kw_j+2c\sum_kw_k+3c^2=0+0+3c^2=3c^2$ (using that each $w_k$ appears in exactly $2$ of the $3$ pairs).

Step 4 · Conclusion

$$ z_1^2+z_2^2+z_3^2=3c^2=z_1z_2+z_2z_3+z_3z_1 \qquad\Longrightarrow\qquad \boxed{z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1} $$

for any equilateral triangle, centered at the origin or not.

Equivalent form

The identity is equivalent to $\ (z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$, which directly encodes that the three side-vectors of an equilateral triangle are related by $120^\circ$ rotations.

Verification

$z_1=0,\ z_2=1,\ z_3=\frac12+i\frac{\sqrt3}2$ (equilateral, side $1$): $\ \sum z_k^2=0+1+\left(\frac12+i\frac{\sqrt3}2\right)^2=1+\left(-\frac12+i\frac{\sqrt3}2\right)=\frac12+i\frac{\sqrt3}2$. And $\sum z_kz_j=0\cdot1+1\cdot\left(\frac12+i\frac{\sqrt3}2\right)+\left(\frac12+i\frac{\sqrt3}2\right)\cdot0=\frac12+i\frac{\sqrt3}2$. Equal ✓.