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 find the locus of the complex number \( z \) such that the amplitude of \( \frac{z-1}{z+1} \) is always equal to \( \frac{\pi}{4} \), we can follow these steps:

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

### Step 2: Substitute \( z \) into the expression
We have:
\[ \frac{z-1}{z+1} = \frac{(x + iy) - 1}{(x + iy) + 1} = \frac{(x-1) + iy}{(x+1) + iy} \]

### Step 3: Multiply by the conjugate of the denominator
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x-1) + iy)((x+1) - iy)}{((x+1) + iy)((x+1) - iy)} \]
Calculating the denominator:
\[ (x+1)^2 + y^2 \]
Calculating the numerator:
\[ (x-1)(x+1) - i y (x-1) + i y (x+1) + y^2 \]
This simplifies to:
\[ x^2 - 1 + y^2 + i(2y) \]

### Step 4: Write the expression
Thus, we have:
\[ \frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2} \]

### Step 5: Find the amplitude
The amplitude (or argument) of a complex number \( a + ib \) is given by:
\[ \text{arg}(a + ib) = \tan^{-1}\left(\frac{b}{a}\right) \]
So, we need:
\[ \text{arg}\left(\frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2}\right) = \frac{\pi}{4} \]
This implies:
\[ \tan\left(\frac{\pi}{4}\right) = 1 = \frac{2y}{x^2 + y^2 - 1} \]

### Step 6: Set up the equation
From the above equation:
\[ 2y = x^2 + y^2 - 1 \]
Rearranging gives:
\[ x^2 + y^2 - 2y - 1 = 0 \]

### Step 7: Complete the square
To find the locus, we can complete the square for \( y \):
\[ x^2 + (y^2 - 2y + 1) - 1 - 1 = 0 \]
This simplifies to:
\[ x^2 + (y-1)^2 = 2 \]

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

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