Read the following writeup carefully:
If `z_1 = a+ib and z_2 =c + id` be two complex numbers such that `|z_1| = |z_2|=1 and "Re" (z_1 bar(z_2))=0`.
Now answer the following question
Let `W = a+ic`, then the locus of `|(W+1)/(W-1)|=1` is (where `W ne 1` )

Options

A. circle with centre `(0,1)` and radius 1
B. circle with centre `(0,-1)` and radius 1
C. x-axis
D. y-axis

Answer

To solve the problem, we need to analyze the given conditions and derive the locus of the complex number \( W = a + ic \) under the condition \( \left| \frac{W + 1}{W - 1} \right| = 1 \).

### Step-by-step Solution:

1. **Understanding the Condition**:
The condition \( \left| \frac{W + 1}{W - 1} \right| = 1 \) implies that the magnitudes of the numerator and denominator are equal:
\[ |W + 1| = |W - 1| \]

2. **Substituting \( W \)**:
Substitute \( W = a + ic \):
\[ |(a + ic) + 1| = |(a + ic) - 1| \]
This simplifies to:
\[ |(a + 1) + ic| = |(a - 1) + ic| \]

3. **Calculating the Magnitudes**:
The magnitudes can be expressed as:
\[ \sqrt{(a + 1)^2 + c^2} = \sqrt{(a - 1)^2 + c^2} \]

4. **Squaring Both Sides**:
To eliminate the square roots, we square both sides:
\[ (a + 1)^2 + c^2 = (a - 1)^2 + c^2 \]

5. **Simplifying the Equation**:
The \( c^2 \) terms cancel out:
\[ (a + 1)^2 = (a - 1)^2 \]
Expanding both sides gives:
\[ a^2 + 2a + 1 = a^2 - 2a + 1 \]

6. **Rearranging the Terms**:
Subtract \( a^2 + 1 \) from both sides:
\[ 2a + 1 = -2a + 1 \]
This simplifies to:
\[ 4a = 0 \]
Therefore, we find:
\[ a = 0 \]

7. **Finding the Locus**:
Since \( a = 0 \), we can substitute back into \( W \):
\[ W = 0 + ic = ic \]
This means that \( W \) is purely imaginary, which corresponds to the y-axis in the complex plane.

### Conclusion:
The locus of the complex number \( W \) is the y-axis.