It is given that $$z = x + iy$$ satisfies $${\mathop{\rm Im}\nolimits} \left( {{{az + b} \over {z + 1}}} \right) = y$$.

Therefore,

$${\mathop{\rm Im}\nolimits} \left( {{{a(x + iy) + b} \over {(x + iy) + 1}}} \right) = y$$

$$ \Rightarrow Im\left( {{{ax + iay + b} \over {x + iy + 1}}} \right) = y$$

Rationalizing the above equation and multiplying and dividing LHS by (x + 1 $$-$$ iy), we get

$${\mathop{\rm Im}\nolimits} \left( {{{(ax + iay + b)} \over {(x + iy + 1)}} \times {{(x + 1 - iy)} \over {(x + 1 - iy)}}} \right) = y$$

Using $${a^2} - {b^2} = (a + b)(a - b)$$, we get

$${\mathop{\rm Im}\nolimits} \left( {{{(ax + iay + b)} \over {{{(x + 1)}^2} - {{(iy)}^2}}}} \right) = y$$

$${\mathop{\rm Im}\nolimits} \left( {{{a{x^2} + ax - iayx + iaxy - {i^2}a{y^2} + bx + b - iby} \over {{{(x + 1)}^2} + {y^2}}}} \right) = y$$ (as $${i^2} = - 1$$)

$${\mathop{\rm Im}\nolimits} \left( {{{(a{x^2} + ax + a{y^2} + bx + b) + i(axy - ayx + ay - by)} \over {{{(x + 1)}^2} + {y^2}}}} \right) = y$$

Rearranging LHS, we get

$$ \Rightarrow {{ay - by} \over {{{(x + 1)}^2} + {y^2}}} = y$$ (as imaginary value in bracket is coefficient of i)

$$ \Rightarrow y(a - b) = y({(x + 1)^2} + {y^2}) \Rightarrow (a - b) = {(x + 1)^2} + {y^2}$$

It is given that a $$-$$ b = 1 and y $$\ne$$ 0

$$ \Rightarrow 1 = {(x + 1)^2} + {y^2}$$

$$ \Rightarrow 1 - {y^2} = {(x + 1)^2} \Rightarrow (x + 1) = \pm \sqrt {1 - {y^2}} $$ (as $${x^2} = b \Rightarrow x = \pm \sqrt 6 $$)

$$ \Rightarrow 1 = - 1 \pm \sqrt {1 - {y^2}} $$

or, $$x = - 1 + \sqrt {1 - {y^2}} $$ and $$x = - 1 - \sqrt {1 - {y^2}} $$