Let `z=x+iy` where x,y are real variable `i=sqrt(-1).` If `|2z-1|=|z-2|`,then the point z describ es :

Options

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

Explanation

To solve the equation \( |2z - 1| = |z - 2| \) where \( z = x + iy \) and \( i = \sqrt{-1} \), we can follow these steps:

### Step 1: Substitute \( z \)
Let \( z = x + iy \). Then, we can express \( 2z - 1 \) and \( z - 2 \):
\[ 2z - 1 = 2(x + iy) - 1 = 2x + 2iy - 1 = (2x - 1) + 2iy \]
\[ z - 2 = (x + iy) - 2 = (x - 2) + iy \]

### Step 2: Write the modulus equations
Now, we can write the modulus of both sides:
\[ |2z - 1| = |(2x - 1) + 2iy| = \sqrt{(2x - 1)^2 + (2y)^2} \]
\[ |z - 2| = |(x - 2) + iy| = \sqrt{(x - 2)^2 + y^2} \]

### Step 3: Set the moduli equal
Setting the two modulus expressions equal gives us:
\[ \sqrt{(2x - 1)^2 + (2y)^2} = \sqrt{(x - 2)^2 + y^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (2x - 1)^2 + (2y)^2 = (x - 2)^2 + y^2 \]

### Step 5: Expand both sides
Now we expand both sides:
\[ (2x - 1)^2 = 4x^2 - 4x + 1 \]
\[ (2y)^2 = 4y^2 \]
So the left side becomes:
\[ 4x^2 - 4x + 1 + 4y^2 \]
For the right side:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ y^2 = y^2 \]
So the right side becomes:
\[ x^2 - 4x + 4 + y^2 \]

### Step 6: Combine and simplify
Now we can set the left side equal to the right side:
\[ 4x^2 - 4x + 1 + 4y^2 = x^2 - 4x + 4 + y^2 \]
Rearranging gives:
\[ 4x^2 - x^2 + 4y^2 - y^2 + 1 - 4 = 0 \]
This simplifies to:
\[ 3x^2 + 3y^2 - 3 = 0 \]
Dividing through by 3:
\[ x^2 + y^2 = 1 \]

### Step 7: Identify the geometric figure
The equation \( x^2 + y^2 = 1 \) represents a circle with center at the origin (0,0) and radius 1.

### Final Answer
The point \( z \) describes a circle with center at (0,0) and radius 1.

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