Animated Solution for Mathematics - Complex Numbers: Let $S=\{z:3\le |2z-3(1+i)|\le 7\}$ be a set of complex numbers. Then ${\min }_{z\in S}\left|z+\frac{1}{2}(5+3i)\right|$ is equal to :

Welcome, fellow explorer of the mathematical universe! Today, we are going to dismantle a complex number problem that, at first glance, might look like a dry algebraic exercise. We are not just manipulating symbols; we are navigating the Argand plane to turn an inequality into a beautiful, tangible geometric shape.

The inequality is given by $S=\{z:3\le |2z-3(1+i)|\le 7\}$. To interpret this, we must transform it into the standard form of a circle, $|z-z_0|=r$.

First, factor out the $2$ from the modulus:

Now, divide the entire inequality by $2$ to isolate the modulus:

The fog clears! We have a center $C$ at $(1.5,1.5)$. The inner radius is $r=1.5$, and the outer radius is $R=3.5$.

This region is an annulus—a perfect, shaded ring floating in the complex plane. Every complex number $z$ that satisfies our inequality lives somewhere within this ring.

The problem asks us to minimize $|z+\frac{1}{2}(5+3i)|$. In the complex plane, the modulus represents the distance between two points.

We rewrite the target expression as:

We now have a fixed point $P$ at $(-2.5,-1.5)$. We are looking for the minimum distance from this point $P$ to any point $z$ inside our annulus.

Imagine you are standing at point $P(-2.5,-1.5)$. You are looking at the ring centered at $C(1.5,1.5)$. The closest point on the ring must lie on the straight line connecting $P$ to $C$.

First, calculate the total distance $PC$ using the distance formula:

This simplifies as follows:

We have a total distance of $5$ units from $P$ to the center $C$. Since the outer radius of our ring is $R=3.5$, and $5>3.5$, our point $P$ is sitting outside the ring.

To find the minimum distance to the shaded region, we subtract the outer radius from the total distance to the center:

The final answer is $1.5$ (or $\frac{3}{2}$). Whenever you see a complex inequality, find the center and the radius, and let the geometry guide you to the solution.