Let |z_1 - 8 - 2i| leq 1 and |z_2 - 2 + 6i| l | vedclass

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(B) The given inequalities represent two disks in the complex plane: $|z_1 - (8 + 2i)| \leq 1$ is a disk centered at $A(8, 2)$ with radius $r_1 = 1$.

Vedclass · Accepted Answer

$7$

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(B) The given inequalities represent two disks in the complex plane: $|z_1 - (8 + 2i)| \leq 1$ is a disk centered at $A(8, 2)$ with radius $r_1 = 1$. $|z_2 - (2 - 6i)| \leq 2$ is a disk centered at $B(2, -6)$ with radius $r_2 = 2$. The distance between the centers $A(8, 2)$ and $B(2, -6)$ is given by the distance formula: $d = \sqrt{(8 - 2)^2 + (2 - (-6))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The minimum distance between two points $z_1$ and $z_2$ in these disks is given by $d - r_1 - r_2$. $|z_1 - z_2|_{	ext{min}} = 10 - 1 - 2 = 7$.

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

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