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

Have you ever looked at a complex expression and felt like it was a tangled knot of variables? In JEE Advanced, we often encounter problems that look like algebraic nightmares, but they are actually elegant geometric stories waiting to be told.

Today, we are going to unravel the mystery of the complex number $z=x+iy$ under the transformation $W=\frac{az+b}{z+1}$, given the constraint $a-b=1$. Let's walk through this journey together.

Our first instinct might be to panic at the sight of $a$, $b$, $x$, and $y$ all mixed together. But let's stay calm. We start by substituting $z=x+iy$ into our expression $W$.

By grouping the real and imaginary parts, we get:

This is the crucial first step. We have effectively separated the 'real' and 'imaginary' components of the numerator and denominator. It is like organizing your tools before starting a complex repair job.

Now, we face a complex number in the denominator. We cannot easily extract the imaginary part of a fraction like this. The standard 'JEE toolkit' move here is to rationalize the denominator.

We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $(x+1)-iy$:

When we multiply the denominator by its conjugate, we use the identity $(A+iB)(A-iB)=A^2+B^2$. Here, $A=(x+1)$ and $B=y$. The denominator becomes a purely real number: $(x+1{)}^2+y^2$.

We only need the imaginary part of $W$. We don't need to expand the entire numerator! We only care about the terms that result in an $i$.

These come from (Real $\times$ Imaginary) and (Imaginary $\times$ Real):

Let's expand this carefully:

Look at that! The $-axy$ and $+axy$ terms cancel out perfectly. We are left with $ay-by$, which is simply $y(a-b)$.

Since the problem tells us $a-b=1$, our numerator simplifies to just $y$. The entire imaginary part of $W$ is now:

We are told that $\text{Im}(W)=y$. Setting our result equal to $y$ gives us:

Since $y eq 0$, we can divide both sides by $y$ to get:

This is the equation of a circle in the complex plane! It is centered at $(-1,0)$ with a radius of $1$.

To find $x$, we simply solve for it:

And there you have it! We have arrived at the two possible values for $x$. It wasn't a nightmare; it was a beautiful, logical path. Keep this mindset for your next problem—look for the structure, use your tools, and trust the math.