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A complex number z satisfies |z - 3| < 5. Find the range of |z + 3i| (where i = sqrt(-1)).

1. [5 - 3*sqrt(2), 5 + 3*sqrt(2)]
2. [3*sqrt(2) - 5, 3*sqrt(2) + 5]
3. [0, 5 + 3*sqrt(2)]
4. [0, 5 - 3*sqrt(2)]

Correct answer: [0, 5 + 3*sqrt(2)]

Solution

|z - 3| < 5 means z lies strictly inside a circle centered at A = (3, 0) with radius 5. We need the range of |z + 3i| = |z - (-3i)|, which is the distance from z to point B = (0, -3). Distance |AB| = sqrt((3-0)² + (0-(-3))²) = sqrt(9+9) = 3*sqrt(2) approx 4.24. Since 3*sqrt(2) < 5, point B lies inside the disk. Therefore z can be arbitrarily close to B (making |z-B| approach 0) and can also be at most |AB| + 5 = 3*sqrt(2) + 5 away from B. So the range of |z+3i| is [0, 3*sqrt(2)+5). Since it says range (and the question uses inequality, the boundary is not achieved), but among the options the best match is [0, 5 + 3*sqrt(2)].

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