Let a, b, x and y be real numbers such that $a - b = 1$ and $y \neq 0$. If the complex number $z = x + iy$ satisfies $\operatorname{Im}\left(\frac{az + b}{z + 1}\right) = y$, then which of the following is(are) possible value(s) of $x$ ?

Let a, b, x and y be real numbers such that a - b = 1 and y ≠ 0. If the complex number z = x + iy satisfies Im((az + b)/(z + 1)) = y, then which of the following is(are) possible value(s) of x ? $a,b,x,y\in\mathbb{R},\quad a-b=1,\quad y\ne 0$ $z=x+iy$ $\operatorname{Im}\!\left(\dfrac{az+b}{z+1}\right)=y,\qquad \text{find possible } x$ $(\text{1})\ -1+\sqrt{1-y^2}\quad(\text{2})\ -1-\sqrt{1-y^2}$ $(\text{3})\ 1+\sqrt{1+y^2}\quad(\text{4})\ 1-\sqrt{1+y^2}$ $w=\dfrac{az+b}{z+1}=\dfrac{(ax+b)+i\,ay}{(x+1)+i\,y}$ $w=\dfrac{[(ax+b)+i\,ay]\,[(x+1)-iy]}{(x+1)^2+y^2}$ $\operatorname{Im}(\text{num}) = -(ax+b)y + ay(x+1)$ $= y\big[-(ax+b)+a(x+1)\big] = y(a-b) = y$ $\operatorname{Im} w=\dfrac{y}{(x+1)^2+y^2}$ $\dfrac{y}{(x+1)^2+y^2}=y$ $y\ne 0\ \Rightarrow\ \dfrac{1}{(x+1)^2+y^2}=1$ $(x+1)^2+y^2=1$ $(x+1)^2+y^2=1$ $\text{locus of } z:\ \text{circle, centre }(-1,0),\ \text{radius }1$ $(x+1)^2=1-y^2$ $x+1=\pm\sqrt{1-y^2}$ $x=-1\pm\sqrt{1-y^2}$ $x=-1+\sqrt{1-y^2}\quad \text{or}\quad x=-1-\sqrt{1-y^2}$ $x=-1+\sqrt{1-y^2}\ \text{or}\ x=-1-\sqrt{1-y^2}$ $\boxed{\text{Options (1) and (2)}}$