1) Let a,b,x and y be real numbers such that a-b=1 and y≠ o. If the complex number z=x+iy satisfies Im(az+bz+1)=y , then which of the following is(are) possible value(s) of x? A) 1−√1+y2 B) −1−√1−y2 C) 1+√1+y2 D) −1+√1−y2 Answer: Option B,DExplanation:az+bz+1=ax+b+aiy(x+1)+iy =(ax+b+aiy)((x+1)−iy)(x+1)2+y2 ∵ Im(az+b)z+1)=−(ax+b)y+ay(x+1)(x+1)2+y2 ⇒ (a−b)y(x+1)2+y2=y ∵ a-b=1 ∵ (x+1)2+y2=1 ∵ x=−1±√1−y2