A variable complex number z is such that the amplitude of `(z-1)/(z+1)` is always equal to `(pi)/(4)`. Illustrate the locus of z in the Argand plane

Answer

To solve the problem, we need to find the locus of the complex number \( z \) such that the amplitude (or argument) of \( \frac{z-1}{z+1} \) is always equal to \( \frac{\pi}{4} \).

### Step-by-Step Solution:

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

2. **Set up the equation for the argument**:
We know that:
\[ \text{arg}\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4} \]
This implies:
\[ \text{arg}(z-1) - \text{arg}(z+1) = \frac{\pi}{4} \]

3. **Substituting \( z \)**:
Substitute \( z = x + iy \) into the equation:
\[ \text{arg}((x - 1) + iy) - \text{arg}((x + 1) + iy) = \frac{\pi}{4} \]

4. **Use the definition of argument**:
The argument of a complex number \( a + ib \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Thus:
\[ \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right) = \frac{\pi}{4} \]

5. **Use the tangent subtraction formula**:
Using the formula for the difference of two arctangents:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1 + AB}\right) \]
Let \( A = \frac{y}{x-1} \) and \( B = \frac{y}{x+1} \):
\[ \frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \frac{y}{x-1} \cdot \frac{y}{x+1}} = 1 \]

6. **Simplify the left-hand side**:
The left-hand side becomes:
\[ \frac{y\left(\frac{(x+1) - (x-1)}{(x-1)(x+1)}\right)}{1 + \frac{y^2}{(x-1)(x+1)}} \]
Simplifying this gives:
\[ \frac{y \cdot \frac{2}{(x-1)(x+1)}}{1 + \frac{y^2}{(x-1)(x+1)}} \]

7. **Set the equation equal to 1**:
Set the simplified expression equal to 1:
\[ \frac{2y}{(x-1)(x+1) + y^2} = 1 \]

8. **Cross-multiply and rearrange**:
This leads to:
\[ 2y = (x^2 - 1) + y^2 \]
Rearranging gives:
\[ x^2 + y^2 - 2y - 1 = 0 \]

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

### Final Equation:
The locus of \( z \) is a circle with center at \( (0, 1) \) and radius \( \sqrt{2} \).