Possible Values of Real Part of Complex Number

To find the possible values of $x$, we start by expressing the complex number $z$ in terms of its real and imaginary parts, $z = x + iy$, where $x, y \in \mathbb{R}$ and $y \neq 0$.

We are given the condition: $\text{Im}\left(\frac{az+b}{z+1}\right) = y$

Substitute $z = x + iy$ into the expression $\frac{az+b}{z+1}$: $\frac{az+b}{z+1} = \frac{a(x+iy) + b}{(x+iy) + 1} = \frac{(ax + b) + iay}{(x + 1) + iy}$

To find the imaginary part of this expression, we rationalize the denominator by multiplying both the numerator and the denominator by the complex conjugate of the denominator, which is $(x + 1) - iy$:

$\frac{az+b}{z+1} = \frac{[(ax + b) + iay][(x + 1) - iy]}{((x + 1) + iy)((x + 1) - iy)}$

The denominator becomes: $(x + 1)^2 + y^2$

Now, expand the numerator: $\text{Numerator} = [(ax + b)(x + 1) + ay^2] + i [ay(x + 1) - y(ax + b)]$

Thus, the imaginary part of $\frac{az+b}{z+1}$ is given by: $\text{Im}\left(\frac{az+b}{z+1}\right) = \frac{ay(x + 1) - y(ax + b)}{(x + 1)^2 + y^2}$

Simplify the numerator of the imaginary part: $ay(x + 1) - y(ax + b) = y [a(x + 1) - (ax + b)] = y [ax + a - ax - b] = y(a - b)$

We are given that $a - b = 1$. Substituting this into the expression, we get: $\text{Im}\left(\frac{az+b}{z+1}\right) = \frac{y(1)}{(x + 1)^2 + y^2} = \frac{y}{(x + 1)^2 + y^2}$

Given that $\text{Im}\left(\frac{az+b}{z+1}\right) = y$, we have: $\frac{y}{(x + 1)^2 + y^2} = y$

Since $y \neq 0$, we can divide both sides by $y$: $\frac{1}{(x + 1)^2 + y^2} = 1$

$(x + 1)^2 + y^2 = 1$

Rearranging the equation to solve for $x$: $(x + 1)^2 = 1 - y^2$

Taking the square root on both sides: $x + 1 = \pm\sqrt{1 - y^2}$

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

Therefore, the possible values of $x$ are: $x = -1 + \sqrt{1 - y^2} \quad \text{and} \quad x = -1 - \sqrt{1 - y^2}$

Hence, the correct options are A and B.