If $S=\{z\in\mathbb{C}: |z-i|=|z+i|=|z-1|\}$, then $n(S)$ is:
Step-by-Step Solution
Key Concept: $|z-i|=|z+i|$ means $z$ is equidistant from $(0,1)$ and $(0,-1)$, so $z$ lies on the real axis ($y=0$). $|z-i|=|z-1|$ means equidistant from $(0,1)$ and $(1,0)$, giving perpendicular bisector $x-y=0$. The circumcentre of triangle with vertices $A(1,0), B(0,1), C(0,-1)$ is the unique point satisfying all three equal-distance conditions.
$|z-i|=|z+i|\Rightarrow y=0$ (real axis). $|z-i|=|z-1|\Rightarrow$ perpendicular bisector of $A(1,0)$ and $B(0,1)$: $x=y$. Intersection: $x=y=0$. Check: $|0-i|=|0+i|=|0-1|=1$. ✓. So $S=\{0\}$, $n(S)=1$.
Correct Answer: 1