Animated Solution for Mathematics - Complex Numbers: If $z$ is any complex number satisfying $|z-3-2i|\le 2$, then the minimum value of $|2z-6+5i|$ is .........

The given condition is $|z-(3+2i)|\le 2$. In the complex plane, the expression $|z-z_0|$ represents the distance between a variable point $z$ and a fixed point $z_0$.

Therefore, this inequality describes a solid disk in the Argand plane. The center of this disk is at $C(3,2)$ and its radius is $R=2$.

We are tasked with finding the minimum value of $|2z-6+5i|$. To simplify this, we factor out the constant $2$:

This expression represents twice the distance between the variable point $z$ and the fixed point $A(3,-2.5)$. Our goal is to minimize the distance between $z$ (which lies within the disk) and the point $A$.

The point $A(3,-2.5)$ lies outside the disk centered at $C(3,2)$. The shortest distance from a point to a disk is found along the line segment connecting the point to the center of the disk.

Since both $A$ and $C$ share the same x-coordinate of $3$, the line segment $AC$ is a vertical line. The distance between $A$ and $C$ is calculated as:

The minimum distance from $A$ to any point $z$ on the disk is the distance to the center minus the radius of the disk:

Recall that our target expression is $2|z-(3-2.5i)|$. Having found the minimum value of the distance $|z-(3-2.5i)|$ to be $2.5$, we multiply by the factor of $2$:

The minimum value of the expression is 5.