Question

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

• A. 1
• B. 0
• C. 3
• D. 4

Answer 1

The Correct Option is A

Let's solve the problem step-by-step to find the number of elements in the set \(S\). 

We are given the set:

\(S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}\)

Here, \(z\) is a complex number and \(\mathbb{C}\) denotes the set of all complex numbers.

First, consider the condition:

\(|z - i| = |z + i|\)

This condition represents a vertical line in the complex plane equidistant from the points \(i\) and \(-i\). The imaginary parts of \(i\) and \(-i\) are symmetric. Therefore, the line must be the real axis, i.e., the line where the imaginary part of \(z\) is 0. So \(z = x + 0i = x\), where \(x\) is real.

Next, consider the condition:

\(|z - i| = |z - 1|\)

This condition represents the set of points in the complex plane that are equidistant from the points \(i\) and \(1\). The geometric representation of this condition is the perpendicular bisector of the line segment joining \(i\) and \(1\).

The points \(i\) and \(1\) have coordinates \((0,1)\) and \((1,0)\) respectively. The midpoint is \(\left(\frac{0+1}{2}, \frac{1+0}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right)\).

The equation of the perpendicular bisector, which is a vertical line (since the line segment joining \(i\) and \(1\) is inclined at 45 degrees to the axes), is given by:

\(x = \frac{1}{2}\)

Combining these two conditions, the line representing \(|z - i| = |z + i|\) is the real axis, and \(|z - i| = |z - 1|\) is the vertical line \(x = \frac{1}{2}\). The intersection of these two lines is a single point.

The intersection yields \(z = \frac{1}{2}\).

Thus, the number of elements in the set \(S\), denoted by \(n(S)\), is:

Hence, the correct answer is:

1

Answer 2

Step 1. Interpret the Condition \( |z - i| = |z + i| = |z - 1| \): This condition implies that \( z \) is equidistant from the points \( (0, 1) \), \( (0, -1) \), and \( (1, 0) \).

Step 2. Geometric Interpretation: The points \( (0, 1) \), \( (0, -1) \), and \( (1, 0) \) form the vertices of an isosceles right triangle in the complex plane. The circumcenter of this triangle (the unique point equidistant from all three vertices) is the only point that satisfies the condition.

Step 3. Finding the Circumcenter: The circumcenter of a triangle with vertices \( (0, 1) \), \( (0, -1) \), and \( (1, 0) \) lies at the origin \( (0, 0) \).
Therefore, \( z = 0 \) is the only solution, so \( n(S) = 1 \).