If z1=2 and (1−i)z2+(1+i)z¯2=82 then the  minimum value of z1−z2 is

We have,z1=2z1 lies on the circle having center at origin and radius 2and (1−i)z2+(1+i)z¯2=82∴    (1−i)x2+iy2+(1+i)x2−iy2=82⇒    x2+y2=42So, z2 lies on the straight line x+y=42z1−z2min= shortest distance between circle and straight line = AB = OB- OA =4-2=2