(1) Bonus ‌∵z+z=2‌Re(z)‌‌‌ If ‌z=x+iy ‌⇒z+z=2x ‌z2+(z)2=2(x2−y2) ‌(az2+bz)+(az2+bz)=2a . . . (1) ‌(bz2+az)+(bz2+az)=2b . . . (2) ‌‌ add (1) and (2) ‌ ‌(a+b)z2+(a+b)z+(a+b)z2+(a+b)z=2(a+b) ‌(a+b)[z2+z+(z)2+z]=2(a+b) ‌‌ sub. (1) and (2) ‌ ‌(a−b)[z2−z+z2−z]=2(a−b) . . . (3) ‌z2+z2−z−z=2 . . . (4) ‌‌ Case I: If ‌a+b≠0 ‌‌ From (3) & (4) ‌ ‌2x+2(x2−y2)=2‌⇒x2−y2+x=1 . . . (5) 2(x2−y2)−2x=2‌⇒x2−y2−x=1 . . . (6) ‌(5)−(6) ‌2x=0⇒x=0 ‌‌ from (5) ‌y2=−1‌‌⇒‌ not possbible ‌ ‌∴‌ Ans ‌=0 Case II: If a+b=0 then infinite number of solution. So, the set X have infinite number of elements.