If a  and b  are real numbers between 0and 1 such that the points z1=a+i,z2=1+bi  and z3=0  from an equilateral triangle, then:

Since the triangle with vertices z1=a+i,z2=1+bi  and z3=0  is equilateral, we havez12+z22+z32=z1z2+z2z3+z3z1⇒     (a+i )2+(1+ib)2+0=(a+i)(1+ib)+0+0⇒        a2−b2+2i(a+b)=a−b+i(1+ab)Equating real and imaginary parts,            a2−b2=a−b                                          ……(1)and      2(a+b)=1+ab                                     …..(2)from Eq. (i), (a−b)[(a+b)−1]=0⇒                either a=b  or a+b=1 Taking a=b , we get from Eq. (ii)      4a=1+a2  or a2−4a+1=0∴          a=4±16−42=2±3Since 0