Let $|z_1 - 8 - 2i| \leq 1$ and $|z_2 - 2 + 6i| \leq 2, z_1, z_2 \in \mathbf{C}$. Then the minimum value of $|z_1 - z_2|$ is:

Step 1: Identify the regions for $z_1$ and $z_2$

The given inequalities describe two circular regions in the complex plane. The first inequality, $|z_1 - (8 + 2i)| \leq 1$, means that $z_1$ lies within or on a circle centered at $C_1(8, 2)$ with a radius $r_1 = 1$. Similarly, the second inequality, $|z_2 - (2 - 6i)| \leq 2$, means that $z_2$ lies within or on a circle centered at $C_2(2, -6)$ with a radius $r_2 = 2$.

Step 2: Calculate the distance between the centers of the disks

To find the minimum distance between $z_1$ and $z_2$, we first need to calculate the distance between the centers of the two disks. The distance $d$ between $C_1(8, 2)$ and $C_2(2, -6)$ is found using the distance formula for complex numbers, which is equivalent to the Euclidean distance formula in the Cartesian plane.

Step 3: Determine the minimum distance between $z_1$ and $z_2$

The minimum distance between any two points, one from each disk, occurs along the line connecting the centers of the disks. This minimum distance is the distance between the centers minus the sum of their radii, provided the disks do not overlap. If the disks overlap, the minimum distance would be 0.

Step 4: Substitute values and calculate the minimum distance

Substituting the calculated distance between centers $d=10$, and the radii $r_1=1$ and $r_2=2$ into the formula, we find the minimum value of $|z_1 - z_2|$. Since $d > r_1 + r_2$ ($10 > 1+2=3$), the disks do not overlap, and the formula is valid.

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