If `z = x + iy` and arg`((z-2)/(z+2))=pi/6,` then find the locus of `z.`

Answer

To find the locus of the complex number \( z = x + iy \) given that \( \arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{6} \), we can follow these steps:

### Step 1: Express the argument condition
Using the property of arguments, we know that:
\[ \arg\left(\frac{a}{b}\right) = \arg(a) - \arg(b) \]
Thus, we can rewrite the given condition:
\[ \arg(z - 2) - \arg(z + 2) = \frac{\pi}{6} \]

### Step 2: Substitute \( z \)
Substituting \( z = x + iy \):
\[ \arg((x - 2) + iy) - \arg((x + 2) + iy) = \frac{\pi}{6} \]

### Step 3: Use the formula for the argument
The argument of a complex number \( a + ib \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Therefore, we can express the arguments as:
\[ \tan^{-1}\left(\frac{y}{x - 2}\right) - \tan^{-1}\left(\frac{y}{x + 2}\right) = \frac{\pi}{6} \]

### Step 4: Use the tangent subtraction formula
Using the tangent subtraction formula:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A - B}{1 + AB}\right) \]
Let \( A = \frac{y}{x - 2} \) and \( B = \frac{y}{x + 2} \). Then we have:
\[ \tan^{-1}\left(\frac{\frac{y}{x - 2} - \frac{y}{x + 2}}{1 + \frac{y}{x - 2} \cdot \frac{y}{x + 2}}\right) = \frac{\pi}{6} \]

### Step 5: Simplify the expression
Calculating \( A - B \):
\[ \frac{y}{x - 2} - \frac{y}{x + 2} = y \left(\frac{(x + 2) - (x - 2)}{(x - 2)(x + 2)}\right) = y \left(\frac{4}{(x - 2)(x + 2)}\right) \]
Thus,
\[ A - B = \frac{4y}{(x - 2)(x + 2)} \]

Now calculating \( 1 + AB \):
\[ 1 + \frac{y^2}{(x - 2)(x + 2)} = 1 + \frac{y^2}{x^2 - 4} \]

### Step 6: Set up the equation
Now we have:
\[ \tan^{-1}\left(\frac{\frac{4y}{(x - 2)(x + 2)}}{1 + \frac{y^2}{x^2 - 4}}\right) = \frac{\pi}{6} \]
Taking the tangent of both sides gives:
\[ \frac{4y}{(x - 2)(x + 2) \left(1 + \frac{y^2}{x^2 - 4}\right)} = \frac{1}{\sqrt{3}} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4y \sqrt{3} = (x^2 - 4)(x - 2)(x + 2) + y^2 \]
This simplifies to:
\[ 4y \sqrt{3} = (x^2 - 4)(x^2 - 4) + y^2 \]

### Step 8: Rearranging to find the locus
Rearranging gives:
\[ x^2 + y^2 - 4\sqrt{3}y - 4 = 0 \]
This represents a circle.

### Conclusion
The locus of \( z \) is a circle.