If S = {z in mathbb{C} : |z - i| = |z + i| = | vedclass

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(A) Let $z = x + iy$. The given equations are $|z - i| = |z + i| = |z - 1|$.These represent the distances of a point $z$ from the points $A(1, 0)$,$B(

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$1$

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(A) Let $z = x + iy$. The given equations are $|z - i| = |z + i| = |z - 1|$.These represent the distances of a point $z$ from the points $A(1, 0)$,$B(0, 1)$,and $C(0, -1)$ in the complex plane.For $|z - i| = |z + i|$,the point $z$ must lie on the perpendicular bisector of the segment joining $i$ and $-i$,which is the real axis $(y = 0)$.For $|z - i| = |z - 1|$,the point $z$ must lie on the perpendicular bisector of the segment joining $i$ and $1$.Since $A, B, C$ form a non-degenerate triangle,there exists exactly one point (the circumcenter) that is equidistant from all three vertices.Thus,$n(S) = 1$.

If $S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}$,then $n(S)$ is:

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