z is such that `a r g ((z-3sqrt(3))/(z+3sqrt(3)))=pi/3` then locus z is

Options

A. `|z-3i|=6`
B. `|z-3i|=6, "Im"(z) gt 0`
C. `|z-3i|=6, "Im"(z) lt 0`
D. none of these

Answer

To find the locus of the complex number \( z \) such that

\[ \arg\left(\frac{z - 3\sqrt{3}}{z + 3\sqrt{3}}\right) = \frac{\pi}{3}, \]

we can follow these steps:

### Step 1: Express \( z \) in terms of \( x \) and \( y \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can rewrite the expression:

\[ \arg\left(\frac{(x + iy) - 3\sqrt{3}}{(x + iy) + 3\sqrt{3}}\right) = \frac{\pi}{3}. \]

### Step 2: Rewrite the argument
This can be rewritten as:

\[ \arg\left(\frac{(x - 3\sqrt{3}) + iy}{(x + 3\sqrt{3}) + iy}\right) = \frac{\pi}{3}. \]

### Step 3: Use the property of arguments
The argument of a complex number \( \frac{a + ib}{c + id} \) can be expressed as:

\[ \arg(a + ib) - \arg(c + id). \]

Thus, we have:

\[ \arg((x - 3\sqrt{3}) + iy) - \arg((x + 3\sqrt{3}) + iy) = \frac{\pi}{3}. \]

### Step 4: Apply the tangent function
Using the property of the tangent function, we can express the arguments as:

\[ \tan^{-1}\left(\frac{y}{x - 3\sqrt{3}}\right) - \tan^{-1}\left(\frac{y}{x + 3\sqrt{3}}\right) = \frac{\pi}{3}. \]

### Step 5: Use the tangent difference formula
Using the formula for the difference of two arctangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right), \]

we can set:

\[ a = \frac{y}{x - 3\sqrt{3}}, \quad b = \frac{y}{x + 3\sqrt{3}}. \]

Thus, we have:

\[ \tan^{-1}\left(\frac{\frac{y}{x - 3\sqrt{3}} - \frac{y}{x + 3\sqrt{3}}}{1 + \frac{y^2}{(x - 3\sqrt{3})(x + 3\sqrt{3})}}\right) = \frac{\pi}{3}. \]

### Step 6: Simplify the expression
This leads to:

\[ \frac{\frac{y}{x - 3\sqrt{3}} - \frac{y}{x + 3\sqrt{3}}}{1 + \frac{y^2}{(x - 3\sqrt{3})(x + 3\sqrt{3})}} = \sqrt{3}. \]

### Step 7: Cross-multiply and simplify
Cross-multiplying and simplifying gives us a relationship between \( x \) and \( y \):

\[ \frac{y\left((x + 3\sqrt{3}) - (x - 3\sqrt{3})\right)}{(x - 3\sqrt{3})(x + 3\sqrt{3}) + y^2} = \sqrt{3}. \]

This simplifies to:

\[ \frac{6y}{x^2 - 27 + y^2} = \sqrt{3}. \]

### Step 8: Rearranging the equation
Rearranging gives:

\[ 6y = \sqrt{3}(x^2 - 27 + y^2). \]

### Step 9: Final form of the locus
This leads to the equation of a circle:

\[ x^2 + (y - 3)^2 = 36. \]

### Conclusion
The locus of \( z \) is a circle with center at \( (0, 3) \) and radius \( 6 \), constrained to the upper half-plane since \( y > 0 \).