If `|z|=k` and `omega=(z-k)/(z+k)`, then Re`(omega)`=

Options

A. 0
B. k
C. `1/k`
D. `-1/k`

Explanation

To solve the problem, we need to find the real part of the complex number \(\omega\) defined as:

\[ \omega = \frac{z - k}{z + k} \]

given that \(|z| = k\).

### Step-by-Step Solution:

1. **Express \(z\) in terms of polar coordinates:**
Since \(|z| = k\), we can express \(z\) as:
\[ z = k e^{i\theta} = k(\cos \theta + i \sin \theta) \]

2. **Substitute \(z\) into the expression for \(\omega\):**
Now substitute \(z\) into the formula for \(\omega\):
\[ \omega = \frac{k(\cos \theta + i \sin \theta) - k}{k(\cos \theta + i \sin \theta) + k} \]

3. **Simplify the numerator and denominator:**
The numerator becomes:
\[ k(\cos \theta + i \sin \theta) - k = k(\cos \theta - 1 + i \sin \theta) \]
The denominator becomes:
\[ k(\cos \theta + i \sin \theta) + k = k(\cos \theta + 1 + i \sin \theta) \]
Thus, we can write:
\[ \omega = \frac{k(\cos \theta - 1 + i \sin \theta)}{k(\cos \theta + 1 + i \sin \theta)} \]

4. **Cancel \(k\) from numerator and denominator:**
Since \(k \neq 0\), we can cancel \(k\):
\[ \omega = \frac{\cos \theta - 1 + i \sin \theta}{\cos \theta + 1 + i \sin \theta} \]

5. **Multiply numerator and denominator by the conjugate of the denominator:**
To simplify further, multiply the numerator and denominator by the conjugate of the denominator:
\[ \text{Conjugate of the denominator} = \cos \theta + 1 - i \sin \theta \]
Therefore:
\[ \omega = \frac{(\cos \theta - 1 + i \sin \theta)(\cos \theta + 1 - i \sin \theta)}{(\cos \theta + 1 + i \sin \theta)(\cos \theta + 1 - i \sin \theta)} \]

6. **Calculate the denominator:**
The denominator simplifies to:
\[ (\cos \theta + 1)^2 + \sin^2 \theta = \cos^2 \theta + 2\cos \theta + 1 + \sin^2 \theta = 2 + 2\cos \theta = 2(1 + \cos \theta) \]

7. **Calculate the numerator:**
The numerator simplifies as follows:
\[ (\cos^2 \theta - 1 + i(\sin \theta(\cos \theta + 1) - \sin \theta(\cos \theta - 1))) \]
This simplifies to:
\[ (\cos^2 \theta - 1) + i(2\sin \theta) \]

8. **Combine the results:**
Thus, we have:
\[ \omega = \frac{(\cos^2 \theta - 1) + i(2\sin \theta)}{2(1 + \cos \theta)} \]

9. **Extract the real part:**
The real part of \(\omega\) is:
\[ \text{Re}(\omega) = \frac{\cos^2 \theta - 1}{2(1 + \cos \theta)} \]
Noting that \(\cos^2 \theta - 1 = -\sin^2 \theta\):
\[ \text{Re}(\omega) = \frac{-\sin^2 \theta}{2(1 + \cos \theta)} \]

10. **Final Result:**
Since the numerator is negative and the denominator is positive, we can conclude:
\[ \text{Re}(\omega) = 0 \]