Animated Solution for Mathematics - Complex Numbers: The sum of the square of the modulus of the elements in the set $\{z=a+ib:a,b\in \mathbb{Z},z\in \mathbb{C},|z-1|\le 1,|z-5|\le |z-5i|\}$ is ________

Welcome, fellow explorers of the mathematical universe! Today, we are going to unravel a problem that beautifully bridges the gap between geometry and algebra.

We are dealing with a set $S$ of complex numbers $z=a+ib$, where $a$ and $b$ are integers. This integer constraint is our first major clue—it tells us that we are not dealing with a continuous region, but rather a discrete grid of points.

Imagine the complex plane as a graph paper; we are only interested in the intersections where the grid lines meet.

The Geometry of the Disk

The first condition given is $|z-1|\le 1$. In the language of complex numbers, $|z-z_0|$ represents the distance between $z$ and $z_0$.

So, $|z-1|\le 1$ tells us that the distance of $z$ from the point $1$ (which is $(1,0)$ on our grid) is at most $1$. Geometrically, this is a solid disk of radius $1$ centered at $(1,0)$.

If we write this in Cartesian coordinates, we get $(x-1{)}^2+y^2\le 1$. Since we are restricted to integer coordinates $(a,b)$, we only need to check points near the center $(1,0)$.

Let's scan the grid: - At the center, we have $(1,0)$. - Moving horizontally, we have $(0,0)$ and $(2,0)$. - Moving vertically, we have $(1,1)$ and $(1,-1)$.

These five points are our candidates. They are the only integer points that satisfy the first condition.

The Half-Plane Boundary

Now, let's look at the second condition: $|z-5|\le |z-5i|$. This looks intimidating, but let's use our geometric intuition.

The equation $|z-z_1|=|z-z_2|$ represents the perpendicular bisector of the segment joining $z_1$ and $z_2$. Here, $z_1=5$ (or $(5,0)$) and $z_2=5i$ (or $(0,5)$). The perpendicular bisector of the segment connecting $(5,0)$ and $(0,5)$ is the line $y=x$.

If you prefer the algebraic approach, substitute $z=x+iy$ into the inequality:

Squaring both sides to eliminate the square roots, we get:

Expanding these terms, we see a beautiful cancellation:

Canceling $x^2$, $y^2$, and $25$ from both sides leaves us with $-10x\le -10y$. Dividing by $-10$ and remembering to flip the inequality sign, we arrive at $x\ge y$. This confirms our geometric intuition: we are looking for the region on or below the line $y=x$.

The Final Verification and Summation

We have our five candidate points: $(0,0)$, $(1,0)$, $(2,0)$, $(1,1)$, and $(1,-1)$. Let's test each one against $x\ge y$:

- $(0,0):0\ge 0$ (True) - $(1,0):1\ge 0$ (True) - $(2,0):2\ge 0$ (True) - $(1,1):1\ge 1$ (True) - $(1,-1):1\ge -1$ (True)

All five points are valid! Now, the problem asks for the sum of the square of the modulus of these elements. The square of the modulus $|z{|}^2$ is simply $x^2+y^2$.

Let's calculate this for each point:

- For $(0,0):0^2+0^2=0$ - For $(1,0):1^2+0^2=1$ - For $(2,0):2^2+0^2=4$ - For $(1,1):1^2+1^2=2$ - For $(1,-1):1^2+(-1{)}^2=2$

Finally, we sum these values: $0+1+4+2+2=9$.

The sum of the squares of the moduli is 9. We have successfully navigated the geometry and the algebra to reach the solution.