If |z - 3i| le 5,then the minimum value of |z | vedclass

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(A) The given inequality $|z - 3i| \le 5$ represents a disk in the complex plane with center $C = (0, 3)$ and radius $R = 5$.We want to find the minim

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$0$

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(A) The given inequality $|z - 3i| \le 5$ represents a disk in the complex plane with center $C = (0, 3)$ and radius $R = 5$.We want to find the minimum value of $|z - (-2)|$,which represents the distance between the point $z$ and the point $A = (-2, 0)$.The distance between the center $C(0, 3)$ and the point $A(-2, 0)$ is $d = \sqrt{(0 - (-2))^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.Since the point $A(-2, 0)$ lies inside the disk (because $d = \sqrt{13} However,if we interpret the question as the distance from the boundary or if the point were outside,we would use $d - R$. Since $A$ is inside,the minimum distance is $0$.

If $|z - 3i| \le 5$,then the minimum value of $|z + 2|$ is equal to

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