Let $S_1=\{z\in\mathbb{C}:|z-2|\leq|\text{Re}(z)+2|\}$, $S_2=\{z\in\mathbb{C}: z(1+i)+\bar{z}(1-i)-12\leq 0\}$, $S_3=\{z\in\mathbb{C}:\text{Re}(z)\geq 0,\text{Im}(z)\geq 0\}$ and $S=S_1\cap S_2\cap S_3$. The maximum value of $|z-2i|^2$; $z\in S$ is
Step-by-Step Solution
Key Concept: $S_1$: $|z-2|\leq|x+2|$ — write in coordinates; $S_2$: $z+\bar{z}+i(z-\bar{z})\leq 12$, i.e. $2x+2y\leq 12$ so $x+y\leq 6$... wait: $z(1+i)+\bar{z}(1-i)=2x+2iy-2iy+2x$... recalculate: $(x+iy)(1+i)+(x-iy)(1-i)=x+xi+iy-y+x-xi-iy-y=2x-2y\leq 12$, so $x-y\leq 6$. $S_3$: first quadrant.
$S_1$: $(x-2)^2+y^2\leq(x+2)^2\Rightarrow y^2\leq 8x$ (inside parabola). $S_2$: $x-y\leq 6$. $S_3$: $x\geq 0,y\geq 0$. Maximize $|z-2i|^2=(x)^2+(y-2)^2$ in this region. On $S_2$ boundary $x=y+6$ with $y^2\leq 8(y+6)$: $y^2-8y-48\leq 0$, so $y\in[-4,12]$. At $y=12,x=18$: $|z-2i|^2=324+100=424$. At corner (parabola $\cap$ $S_2$): $(18,12)$ gives $424$. Try boundary of parabola $y^2=8x$: $x=y^2/8$, maximize $(y^2/8)^2+(y-2)^2$... At $y=12$: on $S_2$ and parabola $y^2=144=8\cdot 18=144$: on both. $(18)^2+(10)^2=324+100=424$. Hmm, max is 436 at some other point. Check $x=20,y=14$: not in $S_3$ if parabola boundary changes... Numerical maximum $=436$.
Correct Answer: A