Let `Arg((z+1)/(z-1))=pi/4` then locus of z be a circle whose radius and centre respectively are ?

Options

A. `sqrt2 , (0,1)`
B. `sqrt2 , (0,-1)`
C. `sqrt2 , (0,0)`
D. `1 , (1,1)`

Answer

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

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

### Step 1: Express \( z \) in terms of \( x \) and \( y \)

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can rewrite the expression:

\[ \frac{z+1}{z-1} = \frac{(x+1) + iy}{(x-1) + iy}. \]

### Step 2: Multiply by the conjugate

To simplify the fraction, multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{((x+1) + iy)((x-1) - iy)}{((x-1) + iy)((x-1) - iy)}. \]

### Step 3: Simplify the denominator

The denominator simplifies to:

\[ (x-1)^2 + y^2. \]

### Step 4: Simplify the numerator

The numerator simplifies to:

\[ (x^2 - 1 + y^2) + i(2y). \]

So we have:

\[ \frac{(x^2 + y^2 - 1) + i(2y)}{(x-1)^2 + y^2}. \]

### Step 5: Find the argument

The argument of a complex number \( a + ib \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Therefore, we have:

\[ \text{Arg}\left(\frac{(x^2 + y^2 - 1) + i(2y)}{(x-1)^2 + y^2}\right) = \tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right). \]

Setting this equal to \( \frac{\pi}{4} \):

\[ \tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right) = \frac{\pi}{4}. \]

### Step 6: Solve the equation

From \( \tan\left(\frac{\pi}{4}\right) = 1 \), we get:

\[ \frac{2y}{x^2 + y^2 - 1} = 1. \]

Cross-multiplying gives:

\[ 2y = x^2 + y^2 - 1. \]

Rearranging this, we get:

\[ x^2 + y^2 - 2y - 1 = 0. \]

### Step 7: Complete the square

To rewrite this in standard form, we complete the square for the \( y \) terms:

\[ x^2 + (y^2 - 2y + 1) - 1 - 1 = 0 \implies x^2 + (y-1)^2 - 2 = 0. \]

Thus,

\[ x^2 + (y-1)^2 = 2. \]

### Step 8: Identify the center and radius

This is the equation of a circle with center \( (0, 1) \) and radius \( \sqrt{2} \).

### Final Answer

- **Center:** \( (0, 1) \)
- **Radius:** \( \sqrt{2} \)