Animated Solution for Mathematics - Complex Numbers: Let $S_1,S_2$ and $S_3$ be three sets defined as $S_1=\{z\in \mathbb{C}:|z-1|\le \sqrt{2}\}$, $S_2=\{z\in \mathbb{C}:Re((1-i)z)\ge 1\}$, $S_3=\{z\in \mathbb{C}:Im(z)\le 1\}$. Then the set $S_1\cap S_2\cap S_3$

Welcome, future engineer. Today, we aren't just solving an inequality; we are mapping a territory. When you look at complex numbers, don't just see $z=x+iy$. See a coordinate plane. See a canvas where equations become shapes.

This problem is a beautiful exercise in visualization. Let us walk through the constraints step-by-step to define the region in the complex plane.

The Disk of $S_1$

We start with the set $S_1=\{z\in \mathbb{C}:|z-1|\le \sqrt{2}\}$. This is the classic definition of a disk.

If you set $z=x+iy$, the modulus becomes $\sqrt{(x-1{)}^2+y^2}\le \sqrt{2}$. Squaring both sides, we obtain the inequality:

This represents a solid disk centered at $(1,0)$ with a radius of $\sqrt{2}$. It includes the boundary because of the $\le$ sign.

The Half-Plane of $S_2$

Next, we consider $S_2=\{z\in \mathbb{C}:Re((1-i)z)\ge 1\}$. To simplify this, let $z=x+iy$.

Expanding the product, we get:

The real part is $x+y$. Therefore, the condition simplifies to the linear inequality:

This represents a half-plane bounded by the line $x+y=1$, including the region above and to the right of this line.

The Constraint of $S_3$

Finally, we examine $S_3=\{z\in \mathbb{C}:Im(z)\le 1\}$. Since $Im(z)=y$, this is simply:

This defines a horizontal half-plane consisting of all points on or below the line $y=1$.

When you overlay these three constraints, you find a region bounded by the circle $(x-1{)}^2+y^2\le 2$, the line $x+y=1$, and the line $y=1$.

We are looking for the intersection of these sets. The resulting region lies inside the circle, above the line $x+y=1$, and below the line $y=1$.

Because this region has a non-zero area, it contains infinitely many points.

Remember, in JEE Advanced, visualization is your greatest weapon. By mapping these algebraic constraints to geometric shapes, we have successfully identified the nature of the solution set. Keep practicing, and keep visualizing.