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

The correct answer is Option 1: n(S) = 1.

We are given the set $S=\{z\in C:|z-i|=|z+i|=|z-1\left|\right\}$ and we need to find how many complex numbers satisfy all three conditions simultaneously.

Let $z=x+\mathrm{iy}$, where $x,y\in ℝ$. We will interpret each modulus condition geometrically — the modulus $|z-a|$ represents the distance from the point $z$ in the complex plane to the point $a$.

Step 1: Apply the condition |z - i| = |z + i|

This says the distance from $z$ to the point $i$ (which is $(0,1)$) equals the distance from $z$ to the point $-i$ (which is $(0,-1)$).

Geometrically, all such points lie on the perpendicular bisector of the segment joining $(0,1)$ and $(0,-1)$, which is simply the real axis.

Algebraically:
$|z-i|=|z+i|$
$\sqrt{x^2+{(}^y}=\sqrt{x^2+{(}^y}$

Squaring both sides:
$x^2+y^2-2y+1=x^2+y^2+2y+1$
$-2y=2y$
$4y=0$
$y=0$

So from this condition, $z$ must lie on the real axis, i.e., $y=0$.

Step 2: Apply the condition |z + i| = |z - 1|

This says the distance from $z$ to $-i$ (i.e., $(0,-1)$) equals the distance from $z$ to $1$ (i.e., $(1,0)$).

Algebraically:
$\sqrt{x^2+{(}^y}=\sqrt{{(}^x+y^2}$

Squaring both sides:
$x^2+y^2+2y+1=x^2-2x+1+y^2$
$2y=-2x$
$y=-x$

So from this condition, $z$ must lie on the line $y=-x$.

Step 3: Solve the system simultaneously

From Step 1: $y=0$
From Step 2: $y=-x$

Substituting $y=0$ into $y=-x$:
$0=-x\Rightarrow x=0$

Therefore, the unique solution is $z=0+0i=0$.

Step 4: Verify the solution z = 0

$|0-i|=|-i|=1$
$|0+i|=\left|i\right|=1$
$|0-1|=|-1|=1$

All three distances are equal to 1. ✓ So $z=0$ is the only element of $S$.

Conclusion: The set $S=\left\{0\right\}$ contains exactly one element.

Therefore, $n\left(S\right)=1$.