Animated Solution for Mathematics - Complex Numbers: The number of elements in the set $\{z=a+ib\in \mathbb{C}:a,b\in \mathbb{Z}\text{ and }1<|z-3+2i|<4\}$ is \_\_\_\_\_.

Imagine standing on the complex plane. Usually, we think of it as a smooth, continuous sheet where every point is accessible.

But today, we are looking at something more rigid, more structured: the world of Gaussian integers. When we say $z=a+ib$ where $a,b\in \mathbb{Z}$, we are restricting our playground to a grid of lattice points. This is where the magic of number theory meets the elegance of geometry.

The condition $1<|z-(3-2i)|<4$ is not just an inequality; it is a geometric command. The expression $|z-(3-2i)|$ represents the distance between any point $z$ and the fixed center $(3,-2)$.

By saying the distance is strictly between $1$ and $4$, we are defining an annulus—a ring-shaped region. But remember, we aren't filling in the ring; we are hunting for the specific lattice points that live inside it.

Working with the center $(3,-2)$ is cumbersome. Let's simplify our lives.

By defining $x=a-3$ and $y=b+2$, we perform a coordinate shift. Because $a$ and $b$ are integers, $x$ and $y$ are guaranteed to be integers as well.

Our inequality transforms into the much friendlier:

We have effectively moved the center of our annulus to the origin $(0,0)$. Now, the problem becomes a systematic search for integer pairs $(x,y)$ such that their squared distance from the origin is between $1$ and $16$.

This is where we roll up our sleeves. We need to find all integer pairs $(x,y)$ such that $x^2+y^2=k$, where $1<k<16$.

Let's test the values:

For $x^2+y^2=2$: $(\pm 1,\pm 1)\Rightarrow 4$ points. For $x^2+y^2=4$: $(\pm 2,0),(0,\pm 2)\Rightarrow 4$ points. For $x^2+y^2=5$: $(\pm 1,\pm 2),(\pm 2,\pm 1)\Rightarrow 8$ points. For $x^2+y^2=8$: $(\pm 2,\pm 2)\Rightarrow 4$ points. For $x^2+y^2=9$: $(\pm 3,0),(0,\pm 3)\Rightarrow 4$ points. For $x^2+y^2=10$: $(\pm 1,\pm 3),(\pm 3,\pm 1)\Rightarrow 8$ points. * For $x^2+y^2=13$: $(\pm 2,\pm 3),(\pm 3,\pm 2)\Rightarrow 8$ points.

When we sum these up: $4+4+8+4+4+8+8$, we arrive at exactly $40$.

It is a beautiful result. We navigated the geometry, simplified the algebra, and systematically counted the discrete reality of the complex plane.

You have successfully mapped the lattice points within the annulus. The final answer is 40. Keep this systematic approach in your toolkit—it is the key to mastering JEE Advanced problems.