Animated Solution for Mathematics - Complex Numbers: Let $S=\{z\in \mathbb{C}:|z-3|\le 1\text{ and }z(4+3i)+\bar{z}(4-3i)\le 24\}$. If $\alpha +i\beta$ is the point in $S$ which is closest to $4i$, then $25(\alpha +\beta )$ is equal to \_\_\_\_\_.

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the geometry of the complex plane. Imagine you are standing on the Argand plane, a vast, two-dimensional playground where complex numbers live as coordinates.

Our goal is to find a specific point $\alpha +i\beta$ within a region $S$ that is closest to the imaginary point $4i$. Let us break this down systematically.

The first condition given is $|z-3|\le 1$. In the language of geometry, this is the definition of a solid disk.

The expression $|z-z_0|\le r$ represents all points $z$ whose distance from $z_0$ is at most $r$. Here, $z_0=3$, which is the point $(3,0)$ on the real axis, and the radius $r=1$.

So, we have a disk centered at $(3,0)$ with a radius of $1$. This is our foundational shape.

Now, look at the second condition: $z(4+3i)+\bar{z}(4-3i)\le 24$. It looks intimidating, but do not panic.

Let us substitute $z=x+iy$ and $\bar{z}=x-iy$. When we expand this, we get:

Expanding the brackets, we see the real parts combine and the imaginary parts cancel out beautifully:

This simplifies to $8x-6y\le 24$, or more elegantly:

This is a half-plane.

We are looking for the region $S$, which is the intersection of our disk and this half-plane. Notice something profound: if you plug the center of our circle, $(3,0)$, into the line equation $4x-3y=12$, you get $4(3)-3(0)=12$.

The line passes exactly through the center of the circle! This means the line cuts the disk into two perfect semicircles.

Our region $S$ is one of these semicircles.

We need the point in $S$ closest to $4i$, which is the coordinate $(0,4)$. Geometrically, the shortest distance from an external point to a circle always lies along the line connecting the external point to the center of the circle.

So, we draw a line from $(0,4)$ to $(3,0)$. The slope of this line is:

Using the point-slope form, the equation of this line is $y-0=-\frac{4}{3}(x-3)$, which simplifies to:

We need the intersection of this line $4x+3y=12$ and our circle $(x-3{)}^2+y^2=1$. Substituting $y=-\frac{4}{3}(x-3)$ into the circle equation, we get:

This leads to:

Solving this, we find $x-3=\pm \frac{3}{5}$, giving us two points: $P_1(2.4,0.8)$ and $P_2(3.6,-0.8)$.

We must check which point lies in our region $S$ ($4x-3y\le 12$). Testing $P_1$, we get $4(2.4)-3(0.8)=7.2\le 12$ (Valid). Testing $P_2$, we get $16.8>12$ (Invalid).

Thus, $(\alpha ,\beta )=(2.4,0.8)$. Finally:

You have conquered the geometry!