Key Concepts and Formulas

• Geometric Interpretation of Modulus: $|z - a|$ represents the distance between complex numbers $z$ and $a$ in the complex plane.
• Perpendicular Bisector: The locus of points $z$ such that $|z - a| = |z - b|$ is the perpendicular bisector of the line segment joining the points representing $a$ and $b$ in the complex plane.
• Circumcenter: The circumcenter of a triangle is the point equidistant from all three vertices.

Step-by-Step Solution

We are given the set $S = \{z \in \mathbb{C} : |z-i| = |z+i| = |z-1|\}$ and we want to find $n(S)$, the number of elements in $S$. The condition $|z-i| = |z+i| = |z-1|$ means that the complex number $z$ is equidistant from the complex numbers $i$, $-i$, and $1$.

Step 1: Analyze the condition $|z-i| = |z+i|$

This condition implies that $z$ is equidistant from $i$ and $-i$. Geometrically, this means $z$ lies on the perpendicular bisector of the line segment joining $i$ and $-i$ in the complex plane. This line segment lies on the imaginary axis, and its midpoint is at $0$. Therefore, the perpendicular bisector is the real axis. Algebraically, we can express this as follows:

Let $z = x + iy$, where $x, y \in \mathbb{R}$. Substituting into the equation, we have: $|x + iy - i| = |x + iy + i|$ $|x + i(y-1)| = |x + i(y+1)|$ Using the definition of the modulus, we have: $\sqrt{x^2 + (y-1)^2} = \sqrt{x^2 + (y+1)^2}$ Squaring both sides, we get: $x^2 + (y-1)^2 = x^2 + (y+1)^2$ $x^2 + y^2 - 2y + 1 = x^2 + y^2 + 2y + 1$ Subtracting $x^2 + y^2 + 1$ from both sides gives: $-2y = 2y$ $4y = 0$ $y = 0$

This confirms that $z$ lies on the real axis, i.e., the imaginary part of $z$ is zero.

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

This condition implies that $z$ is equidistant from $-i$ and $1$. Geometrically, this means $z$ lies on the perpendicular bisector of the line segment joining $-i$ and $1$ in the complex plane. Algebraically, we can express this as follows:

Let $z = x + iy$, where $x, y \in \mathbb{R}$. Substituting into the equation, we have: $|x + iy + i| = |x + iy - 1|$ $|x + i(y+1)| = |(x-1) + iy|$ Using the definition of the modulus, we have: $\sqrt{x^2 + (y+1)^2} = \sqrt{(x-1)^2 + y^2}$ Squaring both sides, we get: $x^2 + (y+1)^2 = (x-1)^2 + y^2$ $x^2 + y^2 + 2y + 1 = x^2 - 2x + 1 + y^2$ Subtracting $x^2 + y^2 + 1$ from both sides gives: $2y = -2x$ $y = -x$

Step 3: Find the intersection of the two loci

The complex number $z$ must satisfy both $y = 0$ and $y = -x$. Substituting $y = 0$ into $y = -x$, we get: $0 = -x$ $x = 0$ Thus, the only solution is $x = 0$ and $y = 0$, which corresponds to the complex number $z = 0 + 0i = 0$.

Since $z=0$, $|z-i| = |-i| = 1$, $|z+i| = |i| = 1$, and $|z-1| = |-1| = 1$. Thus, $z=0$ satisfies the original equation.

Common Mistakes & Tips:

• Remember to square both sides after substituting $z=x+iy$ to eliminate the square roots in the modulus.
• Be careful with the signs when expanding the squared terms.
• Recognize the geometric interpretation of $|z-a|=|z-b|$ as the perpendicular bisector of the segment connecting $a$ and $b$.

Summary

By analyzing the given conditions geometrically and algebraically, we found that the only complex number $z$ that satisfies $|z-i| = |z+i| = |z-1|$ is $z=0$. Therefore, the set $S$ contains only one element, which is $0$. Thus, $n(S) = 1$.

Final Answer

The final answer is \boxed{1}, which corresponds to option (A).