Find the number of complex numbers $z$ satisfying both $|z + \bar{z}| + |z - \bar{z}| = 2$ and $|z + i| + |z - i| = 2$.
Step-by-Step Solution
Key Concept: $|z+\bar{z}|+|z-\bar{z}|=2|x|+2|y|=2$ gives the square $|x|+|y|=1$; $|z+i|+|z-i|=2$ gives the segment $[-i,i]$ on imaginary axis.
$|z+\bar{z}|=2|x|$, $|z-\bar{z}|=2|y|$, so $2|x|+2|y|=2 \Rightarrow |x|+|y|=1$ (a square). $|z+i|+|z-i|=2 = $ distance between foci, so $z$ lies on the segment from $-i$ to $i$ on the imaginary axis ($x=0,y\in[-1,1]$). Intersection: $x=0$, $|y|=1$, giving $z=i$ and $z=-i$. Count: $2$.
Correct Answer: 2