If S = z ∈ C : | z - i | = | z + i | = | z - 1 | , then, n ( S ) is:

Given: $|\mathrm{z}-\mathrm{i}|=|\mathrm{z}+\mathrm{i}|=|\mathrm{z}-1|$

The given equation represents that $\mathrm{z}$ is a point such that it is equidistant from $\left(0,1\right), \left(0,-1\right) \& \left(1,0\right)$

We know that in a triangle only circumcentre is the point which is equidistant from the three vertices.

$ABC$ is a triangle. Hence its circum-centre will be the only point whose distance from $\mathrm{A},\mathrm{B},\mathrm{C}$ will be same.
So $\mathrm{n}\left(\mathrm{S}\right)=1$