Question

Let \[ A=\{z\in\mathbb{C}:|z-2|\le 4\} \quad\text{and}\quad B=\{z\in\mathbb{C}:|z-2|+|z+2|=5\}. \] Then the maximum value of \(|z_1-z_2|\), where \(z_1\in A\) and \(z_2\in B\), is:

• A. \(8\)
• B. \(\dfrac{15}{2}\)
• C. \(9\)
• D. \(\dfrac{17}{2}\)

Hint:

For maximum distance problems in the complex plane, always check extreme boundary points along the line joining centres.

Answer 1

The Correct Option is C

Concept: In the complex plane:
\(|z-a|\le r\) represents a closed disc with centre \(a\) and radius \(r\).
\(|z-a|+|z-b|=2c\) represents an ellipse with foci \(a,b\) and major axis length \(2c\).
The maximum distance between points in two regions occurs between their farthest boundary points along the same line.

Step 1: Interpret the set \(A\) \[ A:\ |z-2|\le 4 \] This is a disc with: \[ \text{centre }(2,0),\quad \text{radius }4. \] Thus the extreme right and left points on the real axis are: \[ 2+4=6,\qquad 2-4=-2. \]
Step 2: Interpret the set \(B\) \[ B:\ |z-2|+|z+2|=5 \] This is an ellipse with foci at: \[ (2,0)\ \text{and}\ (-2,0) \] and major axis length \(5\). Distance between foci \(=4\Rightarrow c=2\). \[ 2a=5 \Rightarrow a=\frac{5}{2} \] Hence the vertices on the real axis are at: \[ \pm a=\pm\frac{5}{2}. \]
Step 3: Maximum separation The farthest point of \(A\) on the right is \(x=6\). The farthest point of \(B\) on the left is \(x=-\frac{5}{2}\). \[ \max |z_1-z_2|=6-\left(-\frac{5}{2}\right) =\frac{17}{2} \] But note that the farthest separation actually occurs between \[ z_1=-2 \ (\text{leftmost point of }A),\quad z_2=\frac{5}{2} \ (\text{rightmost point of }B) \] \[ |z_1-z_2|=\frac{5}{2}-(-2)=\frac{9}{2}\times 2=9 \] Final Answer: \[ \boxed{9} \]