Animated Solution for Mathematics - Complex Numbers: The equation $|z-i|=|z-1|$, $i=\sqrt{-1}$, represents:

Welcome, fellow traveler on the journey of JEE Advanced mathematics. Today, we are going to demystify a classic problem that often trips up students: the equation $|z-i|=|z-1|$.

At first glance, it looks like a simple algebraic expression, but beneath the surface lies a beautiful geometric truth. Imagine you are standing on a vast, flat plane—the Argand plane.

On this plane, we have two fixed landmarks: the point $i$ (which sits at coordinates $(0,1)$) and the point $1$ (which sits at $(1,0)$). We are looking for the path of a point $z$ that is always, at every single moment, exactly the same distance from these two landmarks.

Before we dive into the algebra, let's trust our geometric intuition. If you have two points, $A$ and $B$, and you want to find all points $P$ such that the distance $PA$ equals the distance $PB$, you are describing the perpendicular bisector of the segment $AB$.

It is the line that cuts the segment $AB$ exactly in half at a $90^{\circ }$ angle. In our case, the segment connects $(0,1)$ and $(1,0)$.

The midpoint of this segment is:

The slope of the segment itself is:

Therefore, the slope of our perpendicular bisector must be the negative reciprocal of $-1$, which is $1$. A line with slope $1$ passing through the midpoint $(\frac{1}{2},\frac{1}{2})$ is simply $y=x$. This is our geometric prediction: the locus is the line $y=x$.

Now, let's prove this with the rigor that JEE Advanced demands. We start by letting $z=x+iy$, where $x$ and $y$ are real numbers.

Our equation is $|z-i|=|z-1|$. Substituting our expression for $z$, we get:

We group the real and imaginary parts:

Now, we apply the definition of the modulus, which is the square root of the sum of the squares of the real and imaginary parts:

This is where the magic happens. We square both sides to remove the radicals:

Expanding the squares, we get:

Look closely at the terms on both sides. We have $x^2$ on both sides, $y^2$ on both sides, and $1$ on both sides. They all cancel out perfectly!

We are left with $-2y=-2x$. Dividing by $-2$, we arrive at $y=x$.

This confirms our geometric intuition perfectly. The locus is indeed a straight line passing through the origin with a slope of $1$. You have just navigated the bridge between geometry and algebra.