Complex Numbers Part 2

Concept:Set $A$ is a closed disk (circle and its interior) centered at $(2,0)$ with radius $4$.Set $B$ is an ellipse with foci at $(2,0)$ and $(-2,0)$, and constant sum of distances $5$.Explanation:The circle $A$ extends along the real axis from $x=2-4=-2$ to $x=2+4=6$.For ellipse $B$, the sum of distances is $2a=5$, so semi-major axis $a=2.5$.Distance between foci is $2ae=4$, so eccentricity $e=\frac{4}{5}=0.8$.Semi-minor axis $b$ is found from $b^2=a^2(1-e^2)=6.25(0.36)=2.25$, so $b=1.5$.Ellipse center is at $(0,0)$, and it extends along the real axis from $x=-2.5$ to $x=2.5$.To maximize $|z_1-z_2|$ for $z_1$ in $A$ and $z_2$ in $B$, choose the farthest points along the real axis.The rightmost point of $A$ is $(6,0)$; the leftmost point of $B$ is $(-2.5,0)$.Distance $= |6 - (-2.5)| = 8.5 = \frac{17}{2}$.Answer:Option A: $\frac{17}{2}$