Question

The three vertices of a triangle are represented by the complex numbers,$0,z_1$ and $z_2$. If triangle is equilateral, then show that ${z_1}^2+{z_2}^2=z_1{z}_2$. Further if $z_0$ is circumcentre then prove that ${z_1}^2+{z_2}^2=3{z_0}^2$.

Answer 1

Since the triangle is equilateral

$\therefore |z_1-0|=|z_2-z_1|=|0-z_2|$

$⟹|z_1{|}^2=|z_2-z_1{|}^2=|z_2{|}^2$

$⟹z_1{\overline{z}}_1=(z_2-z_1)({\overline{z}}_2-{\overline{z}}_1)=z_2{\overline{z}}_2$

Now $z_1{\overline{z}}_1=z_2{\overline{z}}_2⟹\frac{z_1}{z_2}=\frac{{\overline{z}}_2}{{\overline{z}}_1}$

Also $z_2{\overline{z}}_2=(z_2-z_1)({\overline{z}}_2-{\overline{z}}_1)=(z_2-z_1)(z_1-z_2)\frac{{\overline{z}}_2}{z_1}$

$\therefore z_1{z}_2=-z_2^2-z_1^2+2z_1{z}_2⟹z_1^2+z_2^2=z_1{z}_2$

$z_0$ is circumcentre

$z_1^2+z_2^2+0^2=3z_0^2$

$⟹z_1^2+z_2^2=3z_0^2$