In z is a complex number stisfying |2008z-1|= 2008|z-2|, then locus z is

Options

A. y-axis
B. x-axis
C. Cricle
D. A line parallel to y-axis

Answer

To solve the problem, we need to find the locus of the complex number \( z \) that satisfies the equation \( |2008z - 1| = 2008|z - 2| \). Let's break this down step by step.

### Step 1: Represent \( z \) in terms of real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Substitute \( z \) into the given equation
Substituting \( z \) into the equation gives:
\[ |2008(x + iy) - 1| = 2008|x + iy - 2| \]
This simplifies to:
\[ |2008x - 1 + 2008iy| = 2008|x - 2 + iy| \]

### Step 3: Calculate the magnitudes
The left-hand side can be calculated as:
\[ |2008x - 1 + 2008iy| = \sqrt{(2008x - 1)^2 + (2008y)^2} \]
The right-hand side can be calculated as:
\[ 2008|x - 2 + iy| = 2008\sqrt{(x - 2)^2 + y^2} \]

### Step 4: Set the magnitudes equal
Now we set the two magnitudes equal:
\[ \sqrt{(2008x - 1)^2 + (2008y)^2} = 2008\sqrt{(x - 2)^2 + y^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square roots gives:
\[ (2008x - 1)^2 + (2008y)^2 = 2008^2((x - 2)^2 + y^2) \]

### Step 6: Expand both sides
Expanding the left-hand side:
\[ (2008x - 1)^2 = 2008^2x^2 - 2 \cdot 2008x + 1 \]
\[ (2008y)^2 = 2008^2y^2 \]
So the left-hand side becomes:
\[ 2008^2x^2 - 2 \cdot 2008x + 1 + 2008^2y^2 \]

Expanding the right-hand side:
\[ 2008^2((x - 2)^2 + y^2) = 2008^2(x^2 - 4x + 4 + y^2) = 2008^2x^2 - 4 \cdot 2008^2x + 4 \cdot 2008^2 + 2008^2y^2 \]

### Step 7: Set the expanded forms equal
Now we have:
\[ 2008^2x^2 - 2 \cdot 2008x + 1 + 2008^2y^2 = 2008^2x^2 - 4 \cdot 2008^2x + 4 \cdot 2008^2 + 2008^2y^2 \]

### Step 8: Simplify the equation
Cancel \( 2008^2x^2 \) and \( 2008^2y^2 \) from both sides:
\[ -2 \cdot 2008x + 1 = -4 \cdot 2008^2x + 4 \cdot 2008^2 \]

### Step 9: Rearrange the equation
Rearranging gives:
\[ 4 \cdot 2008^2 - 1 = 4 \cdot 2008^2x - 2 \cdot 2008x \]

### Step 10: Solve for \( x \)
This can be rearranged to find \( x \):
\[ x = \frac{4 \cdot 2008^2 - 1}{4 \cdot 2008^2 - 2 \cdot 2008} \]

### Conclusion: Determine the locus
Since \( x \) is expressed as a constant value, the locus of \( z \) is a vertical line in the complex plane, which is parallel to the y-axis.