Let ∣z1−8−2i∣≤1 and ∣z2−2+6i∣≤2,z1,z2∈C. Then the minimum value of ∣z1−z2∣ is:
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Step-by-Step Solution
Step 1: Identify the regions for z1 and z2
The given inequalities describe two circular regions in the complex plane. The first inequality, ∣z1−(8+2i)∣≤1, means that z1 lies within or on a circle centered at C1(8,2) with a radius r1=1. Similarly, the second inequality, ∣z2−(2−6i)∣≤2, means that z2 lies within or on a circle centered at C2(2,−6) with a radius r2=2.
Step 2: Calculate the distance between the centers of the disks
To find the minimum distance between z1 and z2, we first need to calculate the distance between the centers of the two disks. The distance d between C1(8,2) and C2(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 z1 and z2
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 r1=1 and r2=2 into the formula, we find the minimum value of ∣z1−z2∣. Since d>r1+r2 (10>1+2=3), the disks do not overlap, and the formula is valid.