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 y-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. We can substitute this into the equation:

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

This simplifies to:

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

### Step 2: Multiply both sides by the denominator
We multiply both sides by the denominator \( (x + 1) + iy \):

\[ (x - 1) + iy = k((x + 1) + iy) \]

### Step 3: Expand the right side
Expanding the right side gives:

\[ (x - 1) + iy = k(x + 1) + k(iy) \]

This can be rewritten as:

\[ (x - 1) + iy = kx + k + i(ky) \]

### Step 4: Separate real and imaginary parts
Now, we can separate the real and imaginary parts:

Real part:
\[ x - 1 = kx + k \]

Imaginary part:
\[ y = ky \]

### Step 5: Solve for \( y \)
From the imaginary part, we can rearrange:

\[ y(1 - k) = 0 \]

This gives us two cases:
1. \( y = 0 \)
2. \( 1 - k = 0 \) (which is not possible since \( k \) is non-zero)

Thus, we have \( y = 0 \).

### Step 6: Substitute \( y = 0 \) into the real part equation
Substituting \( y = 0 \) into the real part equation:

\[ x - 1 = kx + k \]

Rearranging gives:

\[ x - kx = k + 1 \]

Factoring out \( x \):

\[ x(1 - k) = k + 1 \]

### Step 7: Solve for \( x \)
Thus,

\[ x = \frac{k + 1}{1 - k} \]

### Conclusion
The locus of the complex number \( z \) is a horizontal line along the real axis where \( y = 0 \) and \( x \) takes the value \( \frac{k + 1}{1 - k} \). Therefore, the locus is:

\[ z = \frac{k + 1}{1 - k} + 0i \]