The locus of the points z satisfying the condition arg `((z-1)/(z+1))=pi/3` is, a

Options

A. parabola
B. circle
C. pair of straight lines
D. none of these

Answer

To solve the problem of finding the locus of points \( z \) satisfying the condition \( \arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Express \( z \) in terms of \( x \) and \( y \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Set up the equation**:
We have:
\[ \arg\left(\frac{z-1}{z+1}\right) = \arg\left(\frac{(x-1) + iy}{(x+1) + iy}\right) = \frac{\pi}{3} \]

3. **Calculate the argument**:
The argument of a complex number \( \frac{a + ib}{c + id} \) is given by:
\[ \arg\left(\frac{a + ib}{c + id}\right) = \tan^{-1}\left(\frac{b}{a}\right) - \tan^{-1}\left(\frac{d}{c}\right) \]
For our case:
\[ \arg\left(\frac{(x-1) + iy}{(x+1) + iy}\right) = \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right) = \frac{\pi}{3} \]

4. **Use the tangent subtraction formula**:
Using the formula for the tangent of the difference of angles:
\[ \tan\left(\tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right)\right) = \frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \frac{y}{x-1} \cdot \frac{y}{x+1}} \]
Setting this equal to \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \).

5. **Simplify the equation**:
This gives us:
\[ \frac{y\left(\frac{(x+1) - (x-1)}{(x-1)(x+1)}\right)}{1 + \frac{y^2}{(x-1)(x+1)}} = \sqrt{3} \]
Simplifying this leads to:
\[ \frac{2y}{(x^2 - 1)} = \sqrt{3} \left(1 + \frac{y^2}{(x^2 - 1)}\right) \]

6. **Rearranging the equation**:
Rearranging gives:
\[ 2y = \sqrt{3}(x^2 - 1) + \frac{\sqrt{3}y^2}{(x^2 - 1)} \]
Multiplying through by \( (x^2 - 1) \) to eliminate the fraction leads to:
\[ 2y(x^2 - 1) = \sqrt{3}(x^2 - 1)^2 + \sqrt{3}y^2 \]

7. **Final form**:
Rearranging and simplifying this will yield the equation of a circle. After completing the square, we will find the center and radius.

8. **Conclusion**:
The locus of the points \( z \) satisfying the condition is a circle.