Animated Solution for Mathematics - Complex Numbers: If $S=\left\{z\in \mathbb{C}:\frac{z-i}{z+2i}\in \mathbb{R}\right\}$, then :

Welcome, fellow explorers of the complex plane! Today, we are going to peel back the layers of a classic JEE Advanced problem. We are tasked with finding the locus of a complex number $z$ that satisfies a very specific, elegant condition: the ratio $\frac{z-i}{z+2i}$ must be a purely real number.

At first glance, this might look like a daunting algebraic mess. However, in the complex plane, a number is purely real if it sits right on the horizontal axis.

If a ratio of two complex numbers is real, it means the angle of the numerator and the angle of the denominator must be the same (or differ by $\pi$). This is the geometric soul of the problem.

To solve this, we need to break $z$ down into its fundamental components. Let $z=x+iy$, where $x$ and $y$ are real numbers. Our expression becomes:

Let's group the real and imaginary parts. The numerator is $x+i(y-1)$, and the denominator is $x+i(y+2)$.

To make sense of this, we need to rationalize the denominator. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $x-i(y+2)$.

When we multiply the denominator by its conjugate, the denominator becomes:

This is a purely real, positive value (provided $z eq -2i$). This is the moment where we must be careful—we have a constraint!

The point $z=-2i$ is strictly forbidden because it would make our denominator zero. Keep this in your back pocket; it is the "trap" that catches many students.

Now, look at the numerator. We have $(x+i(y-1))(x-i(y+2))$. We only care about the imaginary part, because for $w$ to be real, the imaginary part must be zero.

The imaginary part of this product is found by cross-multiplying:

Let's expand this: $-xy-2x+xy-x$. The $xy$ terms cancel out perfectly, leaving us with $-3x$.

The condition that the entire fraction is real boils down to the condition that its imaginary part is zero:

For this fraction to be zero, the numerator must be zero. Thus, $-3x=0$, which gives us $x=0$.

What does $x=0$ mean in the complex plane? It means the real part of $z$ is zero, which is the definition of the imaginary axis (the $y$-axis).

We must remember our constraint: $z eq -2i$. So, our locus is the entire imaginary axis, with a single point removed at $(0,-2)$.

Even with that tiny puncture, the geometric shape is a straight line. You have successfully navigated the algebra and uncovered the geometric truth.