If `z = x + iotay` and `amp((z-1)/(z+1)) = pi/3` , then locus of z is

Options

A. a straight line
B. a circle
C. a pair of lines
D. None of these

Answer

To solve the problem where \( z = x + i y \) and \( \text{amp}\left(\frac{z-1}{z+1}\right) = \frac{\pi}{3} \), we will follow these steps:

### Step 1: Express the given condition in terms of \( z \)
We start with the expression:
\[ \text{amp}\left(\frac{z-1}{z+1}\right) = \frac{\pi}{3} \]
This can be rewritten using the property of amplitude:
\[ \text{amp}(a/b) = \text{amp}(a) - \text{amp}(b) \]
Thus, we have:
\[ \text{amp}(z-1) - \text{amp}(z+1) = \frac{\pi}{3} \]

### Step 2: Substitute \( z = x + i y \)
Substituting \( z \) into the expression gives:
\[ \text{amp}((x - 1) + i y) - \text{amp}((x + 1) + i y) = \frac{\pi}{3} \]

### Step 3: Calculate the amplitudes
The amplitude of a complex number \( a + ib \) is given by:
\[ \text{amp}(a + ib) = \tan^{-1}\left(\frac{b}{a}\right) \]
Thus, we can express the amplitudes:
\[ \text{amp}(z-1) = \tan^{-1}\left(\frac{y}{x-1}\right) \]
\[ \text{amp}(z+1) = \tan^{-1}\left(\frac{y}{x+1}\right) \]

### Step 4: Set up the equation
Now substituting these into our equation gives:
\[ \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right) = \frac{\pi}{3} \]

### Step 5: Use the tangent subtraction formula
Using the formula for the difference of two arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
we set:
\[ a = \frac{y}{x-1}, \quad b = \frac{y}{x+1} \]
Thus:
\[ \tan^{-1}\left(\frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \frac{y}{x-1} \cdot \frac{y}{x+1}}\right) = \frac{\pi}{3} \]

### Step 6: Simplify the expression
Calculating \( a - b \):
\[ \frac{y}{x-1} - \frac{y}{x+1} = y\left(\frac{1}{x-1} - \frac{1}{x+1}\right) = y\left(\frac{(x+1) - (x-1)}{(x-1)(x+1)}\right) = \frac{2y}{(x-1)(x+1)} \]

Calculating \( 1 + ab \):
\[ 1 + \frac{y^2}{(x-1)(x+1)} = 1 + \frac{y^2}{x^2 - 1} \]

### Step 7: Set the equation equal to \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \)
Setting the left-hand side equal to \( \sqrt{3} \):
\[ \frac{\frac{2y}{(x-1)(x+1)}}{1 + \frac{y^2}{x^2 - 1}} = \sqrt{3} \]

### Step 8: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2y = \sqrt{3}\left(1 + \frac{y^2}{x^2 - 1}\right)(x^2 - 1) \]
This leads to:
\[ 2y = \sqrt{3}(x^2 - 1) + \frac{\sqrt{3}y^2}{x^2 - 1}(x^2 - 1) \]

### Step 9: Rearranging to find the locus
Rearranging terms leads to:
\[ x^2 + y^2 - 2\sqrt{3}y - 1 = 0 \]
This is the equation of a circle.

### Final Answer
The locus of \( z \) is a circle.