If $S={z \in C:|z-i|=|z+i|=|z-1|}$, then, $n(S)$ is:
(1) 1
(2) 0
(3) 3
(4) 2
$|z-i|=|z+i|=|z-1|$
$\mathrm{ABC}$ is a triangle. Hence its circum-centre will be the only point whose distance from A, B, C will be same.
So $\mathrm{n}(\mathrm{S})=1$
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