Question

If \( z = x + iy \) is a complex number such that \( |z - 1| = |z + 1| \), then the locus of \( z \) represents:

• A. A circle with center at origin
• B. The real axis
• C. The imaginary axis
• D. A line parallel to the x-axis

Hint:

\textbf{Tip:} When the modulus of distances from two points is equal, the locus is the perpendicular bisector of the segment joining them.

Answer 1

The Correct Option is C

Given the complex number \( z = x + iy \), we need to find the locus such that \( |z - 1| = |z + 1| \). This equation states that the distance from \( z \) to 1 is equal to the distance from \( z \) to -1. 

Let's expand these distances:

\(|z - 1| = \sqrt{(x - 1)^2 + y^2}\) and \(|z + 1| = \sqrt{(x + 1)^2 + y^2}\).

Set them equal:\(\sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2}\).

Square both sides to eliminate the square roots:\((x - 1)^2 + y^2 = (x + 1)^2 + y^2\).

Expand both sides:\(x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2\).

Cancel common terms (\(x^2\), \(y^2\), \(1\)) from both sides:\(-2x = 2x\).

Combine like terms:\(4x = 0\).

Solve for \(x\): \(x = 0\).

This implies that the complex number \(z = x + iy\) lies on the line where \(x = 0\), which is the imaginary axis.

Answer 2

We are given a complex number in the form \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.

The condition provided is: \[ |z - 1| = |z + 1| \]

🔍 Step 1: Use the Definition of Modulus of a Complex Number

The modulus (or absolute value) of a complex number \( z = a + bi \) is given by: \[ |z| = \sqrt{a^2 + b^2} \]

Applying this to the given condition, substitute \( z = x + iy \) into both sides: \[ |(x - 1) + iy| = |(x + 1) + iy| \] Which gives: \[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]

🧮 Step 2: Square Both Sides to Eliminate Square Roots

To simplify the equation, square both sides: \[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \] Expand both sides: \[ x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2 \] Cancel out common terms from both sides:

• \( x^2 \) appears on both sides and cancels out
• \( y^2 \) appears on both sides and cancels out
• \( +1 \) appears on both sides and cancels out

Now we are left with: \[ -2x = 2x \] Solving for \( x \): \[ -2x = 2x \Rightarrow 0 = 4x \Rightarrow x = 0 \]

 

📌 Step 3: Interpret the Result Geometrically

The result \( x = 0 \) implies that the real part of the complex number is zero. Hence, the complex number lies entirely on the imaginary axis.

✅ Final Answer: The Locus of \( z \) is the Imaginary Axis

Therefore, the locus of the complex number \( z \) satisfying the equation \( |z - 1| = |z + 1| \) is: the imaginary axis.