Let A = {z in mathbb{C} : |z - 2| le 4} and B | vedclass

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(C) The set $A$ represents a disk centered at $2 + 0i$ with radius $R = 4$. The boundary is $|z - 2| = 4$.The set $B$ represents an ellipse with foci

Vedclass · Accepted Answer

$\frac{17}{2}$

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(C) The set $A$ represents a disk centered at $2 + 0i$ with radius $R = 4$. The boundary is $|z - 2| = 4$.The set $B$ represents an ellipse with foci at $2$ and $-2$. The sum of distances from the foci is $2a = 5$, so $a = \frac{5}{2}$. The center is at the origin $(0, 0)$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a = \frac{5}{2}$ and $c = 2$. Since $b^2 = a^2 - c^2$, we have $b^2 = \frac{25}{4} - 4 = \frac{9}{4}$, so $b = \frac{3}{2}$.The ellipse is $\frac{x^2}{(5/2)^2} + \frac{y^2}{(3/2)^2} = 1$.To maximize $|z_1 - z_2|$, we choose $z_1$ on the boundary of $A$ and $z_2$ on the ellipse $B$ such that they are as far apart as possible.The point on the ellipse $B$ furthest from the center $2$ is $z_2 = -\frac{5}{2}$.The point on the circle $A$ furthest from $z_2 = -\frac{5}{2}$ is $z_1 = 6$.Thus, the maximum distance is $|6 - (-5/2)| = |6 + 2.5| = 8.5 = \frac{17}{2}$.

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

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