If z0 is the centroid of the triangle then z0=z1+z2+z33
⇒3z0=z1+z2+z3
To find z21+z22+z23
we have (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
or a2+b2+c2=(a+b+c)2−2(ab+bc+ca)
∴z21+z22+z23=(z1+z2+z3)2−2(z1z2+z2z3+z3z1)
⇒z21+z22+z23=(3z0)2−2(z1z2+z2z3+z3z1)
⇒z21+z22+z23=9z02−2(z1z2+z2z3+z3z1) ..........(1)
In an equilateral triangle all the sides are equal.
∴AB=BC=CA=a
∵∠ABC=π3
From coni method,
z1−z2z3−z2=aaeiπ3 ....(2)
and ∠BAC=π3
From coni method,
z3−z1z2−z1=aaeiπ3 ....(3)
From equations (2) and (3), we get
z1−z2z3−z2=z3−z1z2−z1
⇒(z1−z2)(z1−z2)=(z3−z2)(z3−z1)
⇒z21+z22+z23=z1z2+z2z3+z3z1 on simplification
Substituting for z1z2+z2z3+z3z1=z21+z22+z23 in eqn(1) we get
⇒z21+z22+z23=9z02−2(z1z2+z2z3+z3z1)
⇒z21+z22+z23=9z02−2(z21+z22+z23)
⇒3(z21+z22+z23)=9z02
∴z21+z22+z23=3z02