If ` omega = ((z - i)/( 1 + iz))^(n)` n integral, then ` omega` lies on the unit circle for

Options

A. only even n
B. only odd n
C. only positive n
D. all n

Answer

To solve the problem, we need to show that the expression for \( \omega \) lies on the unit circle for any integer \( n \). The unit circle is defined as the set of complex numbers \( z \) such that \( |z| = 1 \).

### Step-by-Step Solution:

1. **Define \( \omega \)**:
\[ \omega = \left( \frac{z - i}{1 + iz} \right)^n \]
where \( z \) is a complex number.

2. **Substitute \( z \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then,
\[ \omega = \left( \frac{(x + iy) - i}{1 + i(x + iy)} \right)^n = \left( \frac{x + (y - 1)i}{1 + ix - y} \right)^n \]

3. **Simplify the expression**:
We can rewrite \( \omega \) as follows:
\[ \omega = \left( \frac{x + (y - 1)i}{1 - y + ix} \right)^n \]

4. **Calculate the modulus**:
To show that \( \omega \) lies on the unit circle, we need to find \( |\omega| \):
\[ |\omega| = \left| \frac{x + (y - 1)i}{1 - y + ix} \right|^n \]
The modulus of a quotient is the quotient of the moduli:
\[ |\omega| = \frac{|x + (y - 1)i|}{|1 - y + ix|}^n \]

5. **Find the moduli**:
- For the numerator:
\[ |x + (y - 1)i| = \sqrt{x^2 + (y - 1)^2} \]
- For the denominator:
\[ |1 - y + ix| = \sqrt{(1 - y)^2 + x^2} \]

6. **Set up the equality**:
We need to show that:
\[ \frac{\sqrt{x^2 + (y - 1)^2}}{\sqrt{(1 - y)^2 + x^2}} = 1 \]

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

8. **Simplify**:
The \( x^2 \) terms cancel out:
\[ (y - 1)^2 = (1 - y)^2 \]
This is true for all values of \( y \).

9. **Conclusion**:
Since \( |\omega| = 1 \) for all \( n \), we conclude that \( \omega \) lies on the unit circle for any integer \( n \).

### Final Answer:
Thus, \( \omega \) lies on the unit circle for all integral values of \( n \).

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