If `z=x+iy` and `omega=(1-iz)/(z-i)`, then `|omega|=1` implies that in the complex plane

Options

A. z lies on imaginary axis
B. z lies on real axis
C. z lies on unit circle
D. none of these

Answer

To solve the problem, we need to analyze the given expression for \(\omega\) and the condition that \(|\omega| = 1\).

### Step-by-Step Solution:

1. **Substitute \( z \)**:
We have \( z = x + iy \). Therefore, we can substitute this into the expression for \(\omega\):
\[ \omega = \frac{1 - i(x + iy)}{(x + iy) - i} = \frac{1 - ix - y}{x + (y - 1)i} \]

2. **Simplify the numerator**:
The numerator simplifies to:
\[ 1 - ix - y = (1 - y) - ix \]

3. **Rewrite \(\omega\)**:
Now, substituting back, we have:
\[ \omega = \frac{(1 - y) - ix}{x + (y - 1)i} \]

4. **Multiply by the conjugate**:
To eliminate the imaginary part from the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \omega = \frac{[(1 - y) - ix][x - (y - 1)i]}{(x + (y - 1)i)(x - (y - 1)i)} \]

5. **Calculate the denominator**:
The denominator becomes:
\[ x^2 + (y - 1)^2 \]

6. **Calculate the numerator**:
Expanding the numerator:
\[ [(1 - y)x + (y - 1)ix + y - 1 - ix^2] = (1 - y)x + (y - 1)ix + y - 1 + x^2 \]

7. **Separate real and imaginary parts**:
Now, we can separate the real and imaginary parts:
\[ \text{Real part: } (1 - y)x + y - 1 + x^2 \]
\[ \text{Imaginary part: } (y - 1)x - x^2 \]

8. **Set the modulus condition**:
We know that \(|\omega| = 1\), which implies:
\[ \sqrt{\left(\frac{(1 - y)x + y - 1 + x^2}{x^2 + (y - 1)^2}\right)^2 + \left(\frac{(y - 1)x - x^2}{x^2 + (y - 1)^2}\right)^2} = 1 \]

9. **Square both sides**:
Squaring both sides gives:
\[ \left((1 - y)x + y - 1 + x^2\right)^2 + \left((y - 1)x - x^2\right)^2 = (x^2 + (y - 1)^2)^2 \]

10. **Simplify and analyze**:
After simplification, we find that the conditions lead us to a relationship between \(x\) and \(y\) that describes a circle or a line in the complex plane.

11. **Conclusion**:
The final result indicates that the locus of \(z\) is a circle or a line, specifically:
\[ z \text{ lies on the real axis or a point circle.} \]