If for the complex numbers z satisfying \(|z-2-2i|\le1\), the maximum value of \(|3iz+6|\) is attained at \(a+ib\), then \(a+b\) is equal to ______________.
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Correct Answer: 5
Solution and Explanation
Step 2: z lies in the disc with centre \(C(2,2)\) and radius 1. The point \(2i\) is \((0,2)\).
Step 3: The farthest point of the disc from \((0,2)\) lies on the line through \((0,2)\) and the centre \((2,2)\), beyond the centre: \((3,2)\), at distance 3.
Step 4: Maximum value is \(3\times3=9\) at \(z=3+2i\), so \(a=3,\ b=2\) and \(a+b=5\).
Answer: 5
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