If` |(z +4)/(2z -1)|=1,` where `z =x +iy.` Then the point (x,y) lies on a:

Options

A. circle with center (4,0)
B. circel with center `(-2,0)`
C. circle with center `(2,0)`
D. None of these

Explanation

To solve the problem, we start with the given equation:

\[ \left|\frac{z + 4}{2z - 1}\right| = 1 \]

where \( z = x + iy \).

### Step 1: Apply the Modulus Property
Using the property of modulus, we can rewrite the equation as:

\[ |z + 4| = |2z - 1| \]

### Step 2: Substitute \( z \)
Substituting \( z = x + iy \):

\[ | (x + 4) + iy | = | (2x - 1) + 2iy | \]

### Step 3: Calculate the Moduli
Now we calculate the moduli on both sides:

\[ \sqrt{(x + 4)^2 + y^2} = \sqrt{(2x - 1)^2 + (2y)^2} \]

### Step 4: Square Both Sides
Squaring both sides to eliminate the square roots gives:

\[ (x + 4)^2 + y^2 = (2x - 1)^2 + (2y)^2 \]

### Step 5: Expand Both Sides
Expanding both sides:

Left side:
\[ (x + 4)^2 + y^2 = x^2 + 8x + 16 + y^2 \]

Right side:
\[ (2x - 1)^2 + (2y)^2 = 4x^2 - 4x + 1 + 4y^2 \]

### Step 6: Set the Equation
Setting the two expanded sides equal to each other:

\[ x^2 + 8x + 16 + y^2 = 4x^2 - 4x + 1 + 4y^2 \]

### Step 7: Rearranging the Equation
Rearranging gives:

\[ x^2 + 8x + 16 + y^2 - 4x^2 + 4x - 1 - 4y^2 = 0 \]

Combine like terms:

\[ -3x^2 + 12x - 3y^2 + 15 = 0 \]

### Step 8: Divide by -3
Dividing the entire equation by -3:

\[ x^2 - 4x + y^2 - 5 = 0 \]

### Step 9: Completing the Square
Completing the square for \( x \):

\[ (x^2 - 4x + 4) + y^2 - 5 - 4 = 0 \]

This simplifies to:

\[ (x - 2)^2 + y^2 = 9 \]

### Conclusion
This represents a circle with center at \( (2, 0) \) and radius \( 3 \).

Thus, the point \( (x, y) \) lies on a circle.