Q. Let A = { z ∈ C : |z − 2| ≤ 4 } and B = { z ∈ C : |z − 2| + |z + 2| = 5 }. Then the max { |z₁ − z₂ : z₁ ∈ A and z₂ ∈ B } is :

Correct Answer: 17/2

Explanation

The set

A = { z : |z − 2| ≤ 4 }

represents a closed circle in the complex plane with centre at the point 2 on the real axis and radius 4.

Hence, the farthest point of A from the origin side is at

z = 2 + 4 = 6

and the farthest point in the opposite direction is at

z = 2 − 4 = −2

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Now consider

B = { z : |z − 2| + |z + 2| = 5 }

This represents an ellipse with foci at 2 and −2 on the real axis.

For an ellipse,

2a = 5 ⟹ a = 5/2

So the extreme points of B on the real axis are

z = ± 5/2

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To maximize |z₁ − z₂|, we take the farthest possible points in opposite directions.

That is,

z₁ = 6 (from A)
z₂ = −5/2 (from B)

Hence,

|z₁ − z₂| = |6 − (−5/2)| = 17/2

Therefore, the required maximum value is

17/2

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