If `z=x + yi and omega= (1-zi)/(z-i)` show that `|omega|=1 rArr z` is purely real

Answer

To solve the problem, we need to show that if \( |\omega| = 1 \), then \( z \) is purely real, where \( z = x + yi \) and \( \omega = \frac{1 - zi}{z - i} \).

### Step-by-Step Solution:

1. **Start with the definition of \( \omega \)**:
\[ \omega = \frac{1 - zi}{z - i} \]

2. **Substitute \( z = x + yi \)** into \( \omega \):
\[ \omega = \frac{1 - (x + yi)i}{(x + yi) - i} \]
Simplifying the numerator:
\[ 1 - (x + yi)i = 1 - xi - y(-1) = 1 + y - xi \]
Simplifying the denominator:
\[ (x + yi) - i = x + (y - 1)i \]
Thus,
\[ \omega = \frac{1 + y - xi}{x + (y - 1)i} \]

3. **Find the magnitude of \( \omega \)**:
The magnitude \( |\omega| \) is given by:
\[ |\omega| = \frac{|1 + y - xi|}{|x + (y - 1)i|} \]
We need to compute the magnitudes of the numerator and denominator.

4. **Calculate the magnitude of the numerator**:
\[ |1 + y - xi| = \sqrt{(1 + y)^2 + (-x)^2} = \sqrt{(1 + y)^2 + x^2} \]

5. **Calculate the magnitude of the denominator**:
\[ |x + (y - 1)i| = \sqrt{x^2 + (y - 1)^2} \]

6. **Set the magnitudes equal to 1**:
Since \( |\omega| = 1 \), we have:
\[ \frac{\sqrt{(1 + y)^2 + x^2}}{\sqrt{x^2 + (y - 1)^2}} = 1 \]
Squaring both sides gives:
\[ (1 + y)^2 + x^2 = x^2 + (y - 1)^2 \]

7. **Simplify the equation**:
Cancel \( x^2 \) from both sides:
\[ (1 + y)^2 = (y - 1)^2 \]
Expanding both sides:
\[ 1 + 2y + y^2 = y^2 - 2y + 1 \]
Cancel \( y^2 \) and \( 1 \):
\[ 2y = -2y \]
Thus,
\[ 4y = 0 \implies y = 0 \]

8. **Conclusion**:
Since \( y = 0 \), we have \( z = x + 0i = x \), which means \( z \) is purely real.