Step-by-Step Solution:

1. Set up the expression: We are given a complex number z = x + iy and an equation involving it: ${\mathop{\rm Im} \nolimits} \left( {{{az + b} \over {z + 1}}} \right) = y$ We are also given that a, b, x, y are real numbers, a - b = 1, and $y \neq 0$.

2. Substitute z = x + iy: Let the given complex expression be $W = {{az + b} \over {z + 1}}$. Substitute z = x + iy into the numerator and the denominator: Numerator: az + b = a(x + iy) + b = (ax + b) + i(ay) Denominator: z + 1 = (x + iy) + 1 = (x + 1) + iy

   So, the expression becomes: $W = {{(ax + b) + i(ay)} \over {(x + 1) + iy}}$

3. Find the imaginary part of W: To find the imaginary part, we rationalize the expression by multiplying the numerator and denominator by the conjugate of the denominator, which is (x + 1) - iy. $W = {{(ax + b) + i(ay)} \over {(x + 1) + iy}} \times {{(x + 1) - iy} \over {(x + 1) - iy}}$ The denominator becomes: $((x + 1) + iy)((x + 1) - iy) = (x + 1)^2 - (iy)^2 = (x + 1)^2 + y^2$ The numerator becomes: $((ax + b) + i(ay))((x + 1) - iy) = (ax + b)(x + 1) - i(ax + b)y + i(ay)(x + 1) - i^2(ay)y$ $= [(ax + b)(x + 1) + ay^2] + i[ay(x + 1) - y(ax + b)]$ The imaginary part of the numerator is ay(x + 1) - y(ax + b). $ay(x + 1) - y(ax + b) = axy + ay - axy - by = y(a - b)$ So, the imaginary part of W is: ${\mathop{\rm Im} \nolimits} (W) = {{y(a - b)} \over {(x + 1)^2 + y^2}}$

4. Apply the given conditions: The problem states that ${\mathop{\rm Im} \nolimits} (W) = y$. $y = {{y(a - b)} \over {(x + 1)^2 + y^2}}$ Since we are given $y \neq 0$, we can divide both sides by y: $1 = {{(a - b)} \over {(x + 1)^2 + y^2}}$ We are also given a - b = 1. Substituting this into the equation: $1 = {{1} \over {(x + 1)^2 + y^2}}$

5. Solve for x: From the equation above, we get: $(x + 1)^2 + y^2 = 1$ We need to find the possible values of x. Let's rearrange the equation to solve for x: $(x + 1)^2 = 1 - y^2$ For x to be a real number, the right side must be non-negative, i.e., $1 - y^2 \ge 0$, which means $y^2 \le 1$. Since we also know $y \neq 0$, we have the condition $0 < y^2 \le 1$. Taking the square root of both sides: $x + 1 = \pm \sqrt{1 - y^2}$ $x = -1 \pm \sqrt{1 - y^2}$ This gives two possible forms for the value of x:

   • $x_1 = -1 + \sqrt{1 - y^2}$
   • $x_2 = -1 - \sqrt{1 - y^2}$

6. Compare with the options: Let's check which of the given options match these forms.

   • A: $1 - \sqrt {1 + {y^2}}$ - Does not match.
   • B: $- 1 - \sqrt {1 - {y^2}}$ - Matches $x_2$.
   • C: $1 + \sqrt {1 + {y^2}}$ - Does not match.
   • D: $- 1 + \sqrt {1 - {y^2}}$ - Matches $x_1$.

   Therefore, the possible values of x are given by options B and D. For any y such that $0 < y^2 \le 1$, these expressions give a valid real value for x satisfying the given condition.

Conclusion:

The possible values of x are $x = -1 - \sqrt{1 - y^2}$ and $x = -1 + \sqrt{1 - y^2}$. These correspond to options B and D.