If `| (z-2)/(z+2)|= ( pi )/( 6)`, then the locus of z is

Options

A. A straight line
B. A circle
C. A parabola
D. An ellipse

Answer

To solve the problem given by the equation \( | \frac{z-2}{z+2} | = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Understand the given equation
The equation \( | \frac{z-2}{z+2} | = \frac{\pi}{6} \) represents the modulus of a complex fraction. Here, \( z \) is a complex number, and we need to find the locus of \( z \) in the complex plane.

### Step 2: Rewrite the equation
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can rewrite the equation as:
\[ | \frac{(x + iy) - 2}{(x + iy) + 2} | = \frac{\pi}{6} \]
This simplifies to:
\[ | \frac{(x - 2) + iy}{(x + 2) + iy} | = \frac{\pi}{6} \]

### Step 3: Use the property of modulus
The modulus of a complex number \( \frac{a + bi}{c + di} \) is given by:
\[ | \frac{a + bi}{c + di} | = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \]
Applying this property, we have:
\[ \sqrt{(x - 2)^2 + y^2} = \frac{\pi}{6} \sqrt{(x + 2)^2 + y^2} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - 2)^2 + y^2 = \left( \frac{\pi}{6} \right)^2 \left( (x + 2)^2 + y^2 \right) \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ (x^2 - 4x + 4 + y^2) = \frac{\pi^2}{36} (x^2 + 4x + 4 + y^2) \]

### Step 6: Rearranging the equation
Rearranging the equation leads to:
\[ (1 - \frac{\pi^2}{36}) (x^2 + y^2) - 4(1 + \frac{\pi^2}{36}) x + 4 - \frac{4\pi^2}{36} = 0 \]

### Step 7: Identify the locus
This equation represents a conic section (specifically a circle or ellipse) in the complex plane, depending on the coefficients. To analyze the locus, we can complete the square or analyze the coefficients further.

### Final Step: Conclusion
The locus of \( z \) is a circle in the complex plane centered at a specific point with a certain radius, determined by the coefficients in the rearranged equation.