Let 2=a+ib ∵ z4−|z|4=4iz2 ⇒ (a2−b2)+2abi−(a2+b2)=4i(a+ib) ⇒ b2=2b and ab = 2a So z2=0 or z2=λ+2i,λ∈R Let z = x + iy ⇒ x2−y2+2xy=λ+2i ⇒ xy = 1 Locus of z is a rectangular hyperbola xy = 1 If Re(z1)>0 and Re(z2)<0 then Min. distance between z1 and z2 is the line segment joining two vertices of this hyperbola. |z1−z2|minnnn=√(1+1)2+(1+1)2=√8 ⇒|z1−z2|min2=8