If z is a complex number satisfying |z - 3| l | vedclass

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(A) The condition $|z - 3| \leq 5$ represents a disk in the complex plane centered at $A = (3, 0)$ with radius $R = 5$.
We want to find the range of

Vedclass · Accepted Answer

$[0, 5 + 3\sqrt{2}]$

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(A) The condition $|z - 3| \leq 5$ represents a disk in the complex plane centered at $A = (3, 0)$ with radius $R = 5$.
We want to find the range of $|z - (-3i)|$,which represents the distance of points $z$ in the disk from the point $C = (0, -3)$.
The distance between the center of the disk $A(3, 0)$ and the point $C(0, -3)$ is $d = \sqrt{(3 - 0)^2 + (0 - (-3))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.
Since the point $C(0, -3)$ lies inside the disk (because the distance $d = 3\sqrt{2} \approx 4.24 < 5$),the minimum distance from $C$ to any point in the disk is $0$.
The maximum distance from $C$ to any point in the disk is $R + d = 5 + 3\sqrt{2}$.
Thus,the range of $|z + 3i|$ is $[0, 5 + 3\sqrt{2}]$.

If $z$ is a complex number satisfying $|z - 3| \leq 5$,then the range of $|z + 3i|$ is (where $i = \sqrt{-1}$).

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