Animated Solution for Mathematics - Complex Numbers: Let $S_1=\{z_1\in \mathbb{C}:|z_1-3|=\frac{1}{2}\}$ and $S_2=\{z_2\in \mathbb{C}:|z_2-|z_2+1||=|z_2+|z_2-1||\}$. Then, for $z_1\in S_1$ and $z_2\in S_2$, the least value of $|z_2-z_1|$ is:

Welcome, fellow problem solvers. Today, we are going to dissect a problem that looks like a nightmare of algebra but is actually a beautiful dance of geometry.

When you see complex numbers, your first instinct might be to write $z=x+iy$ and start grinding through equations. While that works, the JEE Advanced examiners love to hide geometric elegance behind algebraic masks. Let us peel back the layers.

The Circle $S_1$

We begin with $S_1=\{z_1\in \mathbb{C}:|z_1-3|=\frac{1}{2}\}$. This is the bread and butter of complex geometry.

The equation $|z-z_0|=r$ represents a circle centered at $z_0$ with radius $r$. Here, our center is $C(3,0)$ and our radius is $r=\frac{1}{2}$.

Imagine this circle sitting on the real axis, centered at $3$, extending from $2.5$ to $3.5$. It is a simple, well-behaved object.

Taming the Beast $S_2$

The set $S_2$ is defined by $|z_2-|z_2+1||=|z_2+|z_2-1||$. At first glance, this looks terrifying.

Take a deep breath. We have a powerful tool: squaring. Since both sides are moduli, they are non-negative, and squaring preserves the equality without introducing extraneous solutions.

We use the identity $|a\pm b{|}^2=|a{|}^2+|b{|}^2\pm 2\text{Re}(a\bar{b})$. Let $a=z_2$ and $b=|z_2+1|$. Since $b$ is a real number, its conjugate is itself. Expanding both sides, we get:

The $|z_2{|}^2$ terms cancel out beautifully, leaving us with:

The Geometric Revelation

Let $z_2=x+iy$, so $\text{Re}(z_2)=x$. The left side simplifies to $((x+1{)}^2+y^2)-((x-1{)}^2+y^2)=4x$.

Our equation becomes $4x=2x(|z_2+1|+|z_2-1|)$. Rearranging this, we get:

This gives us two distinct, elegant cases:

1. Case 1: $x=0$. This is the entire imaginary axis! Any point on the $y$-axis satisfies the original equation. 2. Case 2: $|z_2+1|+|z_2-1|=2$. This is the definition of an ellipse where the sum of distances to the foci $(-1,0)$ and $(1,0)$ equals the distance between the foci. This forces $z_2$ to lie on the line segment connecting $-1$ and $1$.

We now have our two sets: a circle $S_1$ and a set $S_2$ consisting of the $y$-axis and the segment $[-1,1]$. We want the minimum distance between $S_1$ and $S_2$.

The distance from the center $C(3,0)$ to the $y$-axis is $3$. The distance from $C(3,0)$ to the segment $[-1,1]$ is the distance to the point $(1,0)$, which is $2$.

Since $2<3$, the minimum distance from the center to the set $S_2$ is $2$. Finally, to find the minimum distance between the circle and the set, we subtract the radius of the circle from this distance:

And there you have it. What started as a daunting algebraic expression revealed itself to be a simple geometric problem. Trust the process, visualize the geometry, and the math will always guide you home.