Let `z= x+iy`, where x, y are real variable and `i = sqrt(-1)`. If `|2z -1| = |z-2|`, then the piont z describes

Options

A. A circle
B. An ellipse
C. A hyperbola
D. A parabola

Answer

To solve the problem, we need to analyze the given equation involving complex numbers. We have \( z = x + iy \), where \( x \) and \( y \) are real variables, and \( i = \sqrt{-1} \). The equation we need to work with is:

\[ |2z - 1| = |z - 2| \]

### Step-by-Step Solution:

1. **Substitute \( z \)**:
Substitute \( z = x + iy \) into the equation:
\[ |2(x + iy) - 1| = |(x + iy) - 2| \]
This simplifies to:
\[ |(2x - 1) + 2iy| = |(x - 2) + iy| \]

2. **Calculate the Moduli**:
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). So we calculate both sides:
- Left-hand side:
\[ |(2x - 1) + 2iy| = \sqrt{(2x - 1)^2 + (2y)^2} \]
- Right-hand side:
\[ |(x - 2) + iy| = \sqrt{(x - 2)^2 + y^2} \]

3. **Set the Moduli Equal**:
Set the two expressions equal to each other:
\[ \sqrt{(2x - 1)^2 + (2y)^2} = \sqrt{(x - 2)^2 + y^2} \]

4. **Square Both Sides**:
To eliminate the square roots, square both sides:
\[ (2x - 1)^2 + (2y)^2 = (x - 2)^2 + y^2 \]

5. **Expand Both Sides**:
Expanding both sides gives:
- Left-hand side:
\[ (2x - 1)^2 + 4y^2 = 4x^2 - 4x + 1 + 4y^2 \]
- Right-hand side:
\[ (x - 2)^2 + y^2 = x^2 - 4x + 4 + y^2 \]

6. **Combine and Simplify**:
Set the expanded forms equal:
\[ 4x^2 - 4x + 1 + 4y^2 = x^2 - 4x + 4 + y^2 \]
Rearranging gives:
\[ 4x^2 - x^2 + 4y^2 - y^2 + 4 - 1 = 0 \]
Simplifying further:
\[ 3x^2 + 3y^2 - 3 = 0 \]
Dividing through by 3:
\[ x^2 + y^2 = 1 \]

7. **Interpret the Result**:
The equation \( x^2 + y^2 = 1 \) represents a circle with a radius of 1 centered at the origin in the complex plane.

### Final Answer:
The point \( z \) describes a circle with radius 1 centered at the origin.