Given, α‌least ‌=p αmax=q Equation given is z+α|z−1|+2i=0;z∈C and i=√−1 Let z=x+iy Then, z+α|z−1|+2i=0 ⇒x+iy+α√(x−1)2+y2+2i=0 ⇒(x+α√(x−1)2+y2)+i(y+2)=0 ∴‌‌y+2=0 and x+α√(x−1)2+y2=0 ‌‌y=−2 and x2=α2(x2+1−2x+y2) ‌‌x2=α2(x2−2x+5) ‌‌α2=‌