If arg ` (z - 1)/( z + 1) = (pi)/( 4)` then the locus of z is

Options

A. straight line
B. circle
C. ellipse
D. hyperbola

Answer

To solve the problem, we need to find the locus of the complex number \( z \) such that

\[ \arg \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{4}. \]

### 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. **Substitute \( z \) into the argument expression**:
We have:
\[ \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}. \]

3. **Multiply numerator and denominator by the conjugate of the denominator**:
The conjugate of the denominator is \( (x + 1) - iy \). Thus,
\[ \frac{(x - 1) + iy}{(x + 1) + iy} \cdot \frac{(x + 1) - iy}{(x + 1) - iy} = \frac{((x - 1)(x + 1) + y^2) + i(y(x + 1) - y(x - 1))}{(x + 1)^2 + y^2}. \]

4. **Simplify the numerator**:
The numerator becomes:
\[ (x^2 - 1 + y^2) + i(2y). \]
Therefore,
\[ \frac{(x^2 - 1 + y^2) + i(2y)}{(x + 1)^2 + y^2}. \]

5. **Find the argument**:
The argument of a complex number \( a + ib \) is given by \( \tan^{-1} \left( \frac{b}{a} \right) \). Thus,
\[ \arg \left( \frac{z - 1}{z + 1} \right) = \tan^{-1} \left( \frac{2y}{x^2 - 1 + y^2} \right). \]

6. **Set the argument equal to \( \frac{\pi}{4} \)**:
Since \( \tan \left( \frac{\pi}{4} \right) = 1 \), we have:
\[ \frac{2y}{x^2 - 1 + y^2} = 1. \]

7. **Cross-multiply to eliminate the fraction**:
This gives:
\[ 2y = x^2 - 1 + y^2. \]

8. **Rearrange the equation**:
Rearranging gives:
\[ x^2 + y^2 - 2y - 1 = 0. \]

9. **Complete the square for \( y \)**:
Completing the square for \( y \):
\[ x^2 + (y - 1)^2 - 1 - 1 = 0 \implies x^2 + (y - 1)^2 = 2. \]

10. **Identify the locus**:
This is the equation of a circle with center at \( (0, 1) \) and radius \( \sqrt{2} \).

### Conclusion:
The locus of \( z \) is a circle centered at \( (0, 1) \) with a radius of \( \sqrt{2} \).

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