If z1z2∈C,z12+z22∈R,z1z12−3z22=2 and z23z12−z22=11, then the value of z12+z22 is

We have,z1z12−3z22=2          (1)z23z12−z22=11        (2)Multiplying (2) by i and adding it to (1), we getz13−3z22z1+i3z12z2−z23=2+11i⇒ z1+iz23=2+11i       (3)Multiplying (2) by i and subtracting it from (1), we getz13−3z22z1−i3z12z2−z23=2−11i⇒ z1−iz23=2−11i          (4)Multiplying (3) and (4), we getz12+z223=22−121i2=4+121=125⇒ z12+z22=5