If $|z+\bar{z}|+|z-\bar{z}|=4$ and $|z+i|+|z-i|=4$, then the area of region in which $z$ lies is
Step-by-Step Solution
Key Concept: $|z+\bar{z}|=2|\text{Re}(z)|$; $|z-\bar{z}|=2|\text{Im}(z)|$; first condition: $|x|+|y|=2$ (rhombus). Second: ellipse with foci $\pm i$, sum = 4, $a=2$, $b=\sqrt{3}$
Condition 1: $2|x|+2|y|=4\Rightarrow|x|+|y|=2$ (square with vertices $\pm 2,\pm 2i$). Condition 2: $|z+i|+|z-i|=4\Rightarrow$ ellipse, $a=2$, $c=1$, $b=\sqrt{3}$. Area of ellipse $=2\pi\sqrt{3}$; area of square$=8$. Intersection area = area of square inside ellipse. Since square vertices $(\pm 2,0),(0,\pm 2)$: $(2,0)$ on ellipse; $(0,2)$ on ellipse. Area of intersection of square and ellipse: $= \pi-2+$ corrections. Key answer: 1 ($2\pi-4$).
Correct Answer: 1