The locus of complex number z for which `((z-1)/(z+1))=k` , where k is non-zero real, is

Options

A. a circle with center on y-axis
B. a circle with center on x-axis
C. a straight line parallel to x-axis
D. a straight line making `pi//3` angle with the x-axis.

Answer

To find the locus of the complex number \( z \) for which \( \frac{z-1}{z+1} = k \) (where \( k \) is a non-zero real number), we can follow these steps:

### 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 equation as:
\[ \frac{(x + iy) - 1}{(x + iy) + 1} = k \]
This simplifies to:
\[ \frac{(x - 1) + iy}{(x + 1) + iy} = k \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ (x - 1) + iy = k \left( (x + 1) + iy \right) \]
Expanding the right side:
\[ (x - 1) + iy = k(x + 1) + kiy \]

### Step 3: Separate real and imaginary parts
Now, we can separate the real and imaginary parts:
- Real part: \( x - 1 = k(x + 1) \)
- Imaginary part: \( y = k y \)

### Step 4: Solve the imaginary part
From the imaginary part \( y = k y \), we can rearrange it:
\[ y(1 - k) = 0 \]
Since \( k \) is non-zero, this implies:
\[ y = 0 \]

### Step 5: Substitute \( y = 0 \) into the real part equation
Substituting \( y = 0 \) into the real part equation:
\[ x - 1 = k(x + 1) \]
Expanding this gives:
\[ x - 1 = kx + k \]
Rearranging terms, we have:
\[ x - kx = k + 1 \]
Factoring out \( x \):
\[ x(1 - k) = k + 1 \]
Thus,
\[ x = \frac{k + 1}{1 - k} \]

### Step 6: Conclusion about the locus
Since \( y = 0 \) and \( x \) is a constant determined by \( k \), the locus of \( z \) is a straight line parallel to the x-axis at the height \( y = 0 \).

### Final Answer
The locus of the complex number \( z \) is a straight line parallel to the x-axis.

---