If z2+z+1=0,  where z  is a complex number, then the value of (z+1z)2+(z2+1z2)2+(z3+1z3)2+......+(z6+1z6)2  is

z2+z+1=0  ⇒z=ω  or ω2  since 1+ω+ω2=0So, z+1z=ω+ω2=−1,  z2+1z2=ω2+ω=−1,  z3+1z3=ω3+ω3=2     since ω3=1z4+1z4=−1,  z5+1z5=−1  and z6+1z6=2∴  The given sum=1+1+4+1+1+4=12