If z1,z2 and z3 are three points lying on the circle z =2 tnimum value of z1+z22+z2+z32+z3+z12is
z1=z2=z3=2 Now ,z1+z2+z32≥0⇒ z1+z2+z3z¯1+z¯2+z¯3≥0
⇒ ∑z12+∑z1z¯2+z¯1z2≥0⇒ 2∑z12+∑z1z¯2+z¯1z2≥∑z12⇒ z1+z22+z2+z32+z3+z12≥12