$$\text{Let }A = \{ z \in \mathbb {C} : |z - 2| \le 4 \}\quad \text{and} \quad B = \{ z \in \mathbb{C} : |z - 2| + |z + 2| = 5 \}.\text{Then the maximum of }\left\{ |z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B \right\}text{ is:}$$

The sets are defined as follows:

Set A: $$\{ z \in \mathbb{C} : |z - 2| \le 4 \}$$, which is a closed disk centered at (2, 0) with radius 4.

Set B: $$\{ z \in \mathbb{C} : |z - 2| + |z + 2| = 5 \}$$, which is an ellipse with foci at (2, 0) and (-2, 0), and the sum of distances to the foci equal to 5.

The distance between the foci is |2 - (-2)| = 4. For an ellipse, the distance between foci is 2c, so 2c = 4, giving c = 2. The major axis length is 2a, so 2a = 5, giving a = 5/2. The relationship b² = a² - c² gives b² = (25/4) - 4 = (25/4) - (16/4) = 9/4, so b = 3/2. The center of the ellipse is the midpoint of the foci, which is (0, 0). Thus, the equation of the ellipse is $$\frac{x^2}{(5/2)^2} + \frac{y^2}{(3/2)^2} = 1$$, or $$\frac{x^2}{25/4} + \frac{y^2}{9/4} = 1$$.

To find the maximum of |z₁ - z₂| for z₁ ∈ A and z₂ ∈ B, note that both sets are compact and convex. The maximum distance between two convex sets occurs at boundary points. For a fixed z₂ ∈ B, the maximum distance to a point in A (a disk centered at C = (2, 0) with radius r = 4) is |z₂ - C| + r = |z₂ - 2| + 4.

Thus, the problem reduces to maximizing |z₂ - 2| + 4 for z₂ ∈ B.

On the ellipse B, |z - 2| + |z + 2| = 5. Let d₁ = |z - 2| and d₂ = |z + 2|, so d₁ + d₂ = 5. Maximizing d₁ is equivalent to minimizing d₂.

The minimum distance from a point on the ellipse to a focus occurs at the nearest vertex. For the focus at (2, 0), the vertices along the major axis are (-2.5, 0) and (2.5, 0). The distance from (2, 0) to (2.5, 0) is |2.5 - 2| = 0.5, and to (-2.5, 0) is |-2.5 - 2| = 4.5. Thus, |z - 2| ranges from 0.5 to 4.5 on B.

Therefore, the maximum of |z₂ - 2| is 4.5, achieved at z₂ = (-2.5, 0).

Then, max (|z₂ - 2| + 4) = 4.5 + 4 = 8.5 = 17/2.

This maximum is achieved when z₂ = (-2.5, 0) and z₁ is the point in A farthest from z₂. The direction from z₂ to the center C = (2, 0) is (4.5, 0), so the unit vector is (1, 0). Thus, z₁ = C + 4 × (1, 0) = (2, 0) + (4, 0) = (6, 0).

Verify:

• z₁ = (6, 0) ∈ A: |6 - 2| = 4 ≤ 4.
• z₂ = (-2.5, 0) ∈ B: | -2.5 - 2 | + | -2.5 + 2 | = | -4.5 | + | -0.5 | = 4.5 + 0.5 = 5.
• |z₁ - z₂| = |6 - (-2.5)| = 8.5 = 17/2.

No larger distance is possible, as |z₂ - 2| ≤ 4.5 for all z₂ ∈ B, so |z₂ - 2| + 4 ≤ 8.5, with equality only at z₂ = (-2.5, 0) and z₁ = (6, 0).

Thus, the maximum is 17/2.

The correct option is C.