If `z=x+ iy` such that the argument of `(z-1)/(z+1)` is always `pi/4`. Prove that `x^2 + y^2-2y=1`

Answer

To solve the problem, we start with the given condition that the argument of \((z - 1)/(z + 1)\) is always \(\pi/4\). Let's denote \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-step Solution:

1. **Substituting \(z\)**:
\[ \text{We have } z = x + iy. \]
Therefore,
\[ z - 1 = (x - 1) + iy \quad \text{and} \quad z + 1 = (x + 1) + iy. \]

2. **Forming the fraction**:
\[ \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}. \]

3. **Rationalizing the denominator**:
To simplify, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)}. \]
The denominator simplifies to:
\[ (x + 1)^2 + y^2. \]
The numerator expands to:
\[ (x - 1)(x + 1) - i y (x - 1) + i y (x + 1) + y^2 = x^2 - 1 + y^2 + i(2y). \]

4. **Combining results**:
Thus,
\[ \frac{z - 1}{z + 1} = \frac{x^2 + y^2 - 1 + 2iy}{(x + 1)^2 + y^2}. \]

5. **Finding the argument**:
The argument of a complex number \(a + bi\) is given by \(\tan^{-1}(\frac{b}{a})\). Therefore,
\[ \text{arg}\left(\frac{z - 1}{z + 1}\right) = \tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right). \]
Given that this argument is \(\frac{\pi}{4}\), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1. \]
Hence,
\[ \frac{2y}{x^2 + y^2 - 1} = 1. \]

6. **Cross-multiplying**:
\[ 2y = x^2 + y^2 - 1. \]

7. **Rearranging the equation**:
\[ x^2 + y^2 - 2y - 1 = 0 \implies x^2 + y^2 - 2y = 1. \]

Thus, we have proved that:
\[ x^2 + y^2 - 2y = 1. \]