Animated Solution for Mathematics - Complex Numbers: Let $S=\{z=x+iy:\frac{2z-3i}{4z+2i}\text{ is a real number}\}$. Then which of the following is NOT correct?

Welcome, fellow explorers of the mathematical universe. Today, we are diving into a classic JEE Advanced problem that bridges the gap between algebra and geometry.

We are tasked with finding the locus of a complex number $z$ such that the expression $W=\frac{2z-3i}{4z+2i}$ is a real number. This is not just a calculation; it is a journey into the Argand plane, where every equation tells a story of position and constraint.

To begin, we define our complex number $z$ in its most fundamental form: $z=x+iy$, where $x$ and $y$ are real numbers. Our goal is to find the set $S$ of all such $z$ that satisfy the condition that $W$ is real.

We substitute $z=x+iy$ into our expression:

By grouping the real and imaginary parts, we transform the expression into a more manageable structure:

Now, we face a complex number in the denominator. To separate the real and imaginary parts of $W$, we must rationalize.

We multiply the numerator and the denominator by the complex conjugate of the denominator, which is $4x-i(4y+2)$. This step is our mathematical scalpel; it cuts through the complexity, leaving us with a purely real denominator.

The expression becomes:

For $W$ to be a real number, its imaginary part must vanish. After rationalization, the denominator is a real number, so we only need to focus on the imaginary part of the numerator.

By expanding the numerator and extracting the imaginary component, we set it to zero:

Let us expand this carefully. We get $8xy-12x-(8xy+4x)=0$.

Notice the elegance here: the $8xy$ terms cancel out, leaving us with $-16x=0$, which simplifies beautifully to $x=0$. This tells us that the locus of $z$ is the imaginary axis, or the $y$-axis, in the Argand plane.

However, we must not be hasty. In the world of JEE, the most dangerous traps are the ones we create by ignoring domain constraints.

The original expression has a denominator, and that denominator cannot be zero. We must ensure that $4z + 2i eq 0$.

Substituting $z=x+iy$, we find that $4x + i(4y + 2) eq 0$. Since we already know $x=0$, the real part is already zero. Thus, we must ensure the imaginary part is not zero: $4y + 2 eq 0$, which implies $y eq -\frac{1}{2}$.

We have discovered that the set $S$ is the entire imaginary axis ($x=0$), but with a single, crucial point excluded: $(0,-\frac{1}{2})$. This is the "hole" in our locus.

By visualizing the Argand plane and respecting the domain, we have conquered this problem with precision and insight. Keep practicing, and remember: every equation has a geometric soul waiting to be revealed.