z1+z2+z3=3z0(z1+z2+z3)2=9z02⇒z12+z22+z32+2(z12+z22+z32)=9z02⇒z12+z22+z32=3z62
k=1∑3(zk−z0)2=(z1−z0)2+(z2−z0)2+(z3−z0)2=z12+z22+z32+3z02−2(z1+z2+z3)z0=6z02−6z02=0
If z1,z2,z3∈C are the vertices of an equilateral triangle, whose centroid is z0, then k=1∑3(zk−z0)2 is equal to
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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