Let $A=\{z\in\mathbb{C}:|z-2|\leq4\}$ and $B=\{z\in\mathbb{C}:|z-2|+|z+2|=5\}$. Then the $\max\{|z_1-z_2|:z_1\in A\text{ and }z_2\in B\}$ is:
Step-by-Step Solution
Key Concept: $A$: disk centred at $(2,0)$, radius 4. $B$: ellipse with foci $\pm2$, $2a=5$, so $a=5/2$, $b^2=9/4$: $\frac{4x^2}{25}+\frac{4y^2}{9}=1$. Rightmost point of $B$: $(5/2,0)$.
Max $|z_1-z_2|=6+5/2=17/2$.
Correct Answer: 1