Animated Solution for Mathematics - Complex Numbers: Let $\mathbb{C}$ be the set of all complex numbers. Let $S_1=\{z\in \mathbb{C}:|z-3-2i{|}^2=8\}$, $S_2=\{z\in \mathbb{C}:\text{Re}(z)\ge 5\}$ and $S_3=\{z\in \mathbb{C}:|z-\bar{z}|\ge 8\}$. Then the number of elements in $S_1\cap S_2\cap S_3$ is equal to

A Journey into the Intersection

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey into the heart of the complex plane. Many students fear complex numbers, seeing them as abstract, algebraic monsters.

But I want you to see them as they truly are: a beautiful, geometric playground. We have three sets, $S_1$, $S_2$, and $S_3$, and we are looking for their intersection. Think of this as finding the 'sweet spot' where three different worlds collide.

Let us decode them one by one.

Decoding $S_1$ - The Circle of Influence

The first set is defined as $S_1=\{z\in \mathbb{C}:|z-3-2i{|}^2=8\}$. When you see a modulus equation like this, do not panic.

Let $z=x+iy$. The expression becomes $|(x-3)+i(y-2){|}^2=8$. Squaring the modulus, we get:

Does this look familiar? It is the classic equation of a circle! Its center is at $(3,2)$ and its radius is $r=\sqrt{8}$, which is approximately $2.83$.

Imagine this circle sitting on your coordinate plane, centered at $(3,2)$. It is our first constraint.

Decoding $S_2$ - The Vertical Boundary

Next, we have $S_2=\{z\in \mathbb{C}:\text{Re}(z)\ge 5\}$. The real part of $z$ is simply $x$. So, this condition is just $x\ge 5$.

Graphically, this is a vertical line at $x=5$, and we are interested in everything to the right of it. This is a half-plane.

Now, combine this with our circle. The circle's center is at $x=3$, and its radius is $\approx 2.83$. The rightmost point of the circle is $3+2.83=5.83$.

Since $S_2$ requires $x\ge 5$, the intersection of $S_1$ and $S_2$ can only exist in the narrow strip where $5\le x\le 5.83$. We are already narrowing down our search!

Decoding $S_3$ - The Horizontal Strips

Now, let us tackle $S_3=\{z\in \mathbb{C}:|z-\bar{z}|\ge 8\}$. This looks intimidating, but let us break it down.

We know $z=x+iy$, so $\bar{z}=x-iy$. Thus, $z-\bar{z}=(x+iy)-(x-iy)=2iy$.

The condition becomes $|2iy|\ge 8$. Since the modulus of $i$ is $1$, this simplifies to $2|y|\ge 8$, or simply:

This means $y\ge 4$ or $y\le -4$. Graphically, these are two horizontal strips: one above the line $y=4$ and one below the line $y=-4$.

The Intersection - The Moment of Truth

We need points that satisfy all three conditions simultaneously. We know the intersection of $S_1$ and $S_2$ happens for $x\in [5,5.83]$.

Let us check the boundary $x=5$. Substituting $x=5$ into our circle equation $(x-3{)}^2+(y-2{)}^2=8$, we get:

This gives us two possible values for $y$: $y-2=2\Rightarrow y=4$, and $y-2=-2\Rightarrow y=0$. So, the circle intersects the line $x=5$ at the points $(5,4)$ and $(5,0)$.

Now, we must check if these points satisfy $S_3$ ($|y|\ge 4$). For the point $(5,4)$, $|4|\ge 4$ is true! This is a valid point. For the point $(5,0)$, $|0|\ge 4$ is false. So, $(5,0)$ is rejected.

What about $x>5$? If $x>5$, then $(x-3{)}^2>4$. From the circle equation $(x-3{)}^2+(y-2{)}^2=8$, this implies:

This means $|y-2|<2$, which implies $0<y<4$. In this range, $|y|$ is strictly less than 4, so the condition $|y|\ge 4$ can never be satisfied.

The Final Count

We have methodically eliminated all possibilities. The only point that survives the gauntlet of all three conditions is $(5,4)$.

Therefore, the number of elements in the intersection $S_1\cap S_2\cap S_3$ is exactly 1.

You see? It wasn't about guessing; it was about visualizing, simplifying, and testing the boundaries. You have mastered the geometry of the complex plane today. Keep this confidence, and carry it into your next problem.