Put z=x+iy,z=x−iy ⇒z+2+i=x−iy+2+i ⇒z(z+2+i)=(x+iy)(x+2+i(1−y)) ⇒ Real part =x(x+2)−y(1−y) Imaginary part =x(1−y)+y(x+2) ⇒x2+y2+2x−y+2k=0...(1) and x+2y+3k=0...(2) eliminating x from (1) and (2) 5y2+(12k−5)y+9k2−4k=0 since y∈R use D≥0 (12k−5)2−4.5(9k2−4k)≥0 ⇒36k2+40k−25≤0 ⇒α+β=