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:
Show Hint
For maximum distance problems in the complex plane, always check extreme boundary points along the line joining centres.
To find the maximum value of \(|z_1 - z_2|\) where \(z_1 \in A\) and \(z_2 \in B\), we first need to understand the regions defined by sets \(A\) and \(B\):
The set \(A = \{z \in \mathbb{C} : |z - 2| \leq 4\}\) represents a closed disk centered at \(2\) with a radius \(4\).
The set \(B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}\) describes an ellipse with foci at \(2\) and \(-2\), where the sum of the distances from any point \(z\) to these foci equals \(5\).
We can find the center of mass (middle point) of the ellipse defined by \(B\), which will be the midpoint of its foci at \(2\) and \(-2\), i.e., \(0\).
Next, let's analyze these two geometrical shapes:
The farthest point on the circle \(A\) from the center is located at the boundary of the circle along the line through \(2\) and an arbitrary direction. This spot would be at \(2 + 4e^{i\theta}\), maximizing distance for \(z_1\).
For \(z_2\) in \(B\), according to the properties of ellipses, the farthest vertex from the center is a distance of \(\frac{5}{2} + 0 = \frac{5}{2}\) in the direction of the major axis.
Now, to find the maximum value of \(|z_1 - z_2|\), we place:
The farthest point on \(A\) in the direction along the line joining centers of the circle and the ellipse, i.e., on the line from \(2\) to \(-2\), thus giving the opposite boundary circle point: \(z_1 = 2 + 4\) or \(6\).
Meanwhile, the ellipse's point may contribute to maximum distance in the opposite direction, moving alongside the major axis at \(z_2 = 0 + \frac{5}{2}\) or \(\frac{5}{2}\).
Therefore, the maximum distance by leveraging geometry comes as:
