If `|omega|=2`, then the set of points `z=omega-1/omega` is contained in or equal to the set of points z satisfying

Options

A. `"Im"(z)=0`
B. `|"Im"(z)| le 1`
C. `|"Re"(z)| le 2`
D. `|z| le 3`

Explanation

To solve the problem, we need to find the set of points \( z = \omega - \frac{1}{\omega} \) given that \( |\omega| = 2 \).

### Step-by-Step Solution:

1. **Express \( \omega \)**:
Since \( |\omega| = 2 \), we can express \( \omega \) in polar form:
\[ \omega = 2(\cos \theta + i \sin \theta) = 2e^{i\theta} \]

2. **Calculate \( \frac{1}{\omega} \)**:
The reciprocal of \( \omega \) is given by:
\[ \frac{1}{\omega} = \frac{1}{2(\cos \theta + i \sin \theta)} = \frac{1}{2}(\cos(-\theta) + i \sin(-\theta)) = \frac{1}{2}(\cos \theta - i \sin \theta) \]

3. **Substitute into \( z \)**:
Now substitute \( \omega \) and \( \frac{1}{\omega} \) into the expression for \( z \):
\[ z = \omega - \frac{1}{\omega} = 2(\cos \theta + i \sin \theta) - \frac{1}{2}(\cos \theta - i \sin \theta) \]

4. **Combine the terms**:
Simplifying the expression for \( z \):
\[ z = 2\cos \theta + 2i\sin \theta - \frac{1}{2}\cos \theta + \frac{1}{2}i\sin \theta \]
\[ z = \left(2 - \frac{1}{2}\right)\cos \theta + \left(2 + \frac{1}{2}\right)i\sin \theta \]
\[ z = \frac{3}{2}\cos \theta + \frac{5}{2}i\sin \theta \]

5. **Find the modulus of \( z \)**:
The modulus of \( z \) is given by:
\[ |z| = \sqrt{\left(\frac{3}{2}\cos \theta\right)^2 + \left(\frac{5}{2}\sin \theta\right)^2} \]
\[ |z| = \sqrt{\frac{9}{4}\cos^2 \theta + \frac{25}{4}\sin^2 \theta} \]
\[ |z| = \frac{1}{2}\sqrt{9\cos^2 \theta + 25\sin^2 \theta} \]

6. **Analyze the maximum value of \( |z| \)**:
To find the maximum value of \( |z| \), we can express it as:
\[ |z| = \frac{1}{2}\sqrt{9\cos^2 \theta + 25\sin^2 \theta} \]
We can use the Cauchy-Schwarz inequality or analyze the expression directly. The maximum occurs when \( \sin^2 \theta \) and \( \cos^2 \theta \) take on values that maximize the expression.

The maximum value of \( |z| \) can be calculated as:
\[ |z| \leq \frac{1}{2} \sqrt{9 + 25} = \frac{1}{2} \sqrt{34} \]
Since \( \sqrt{34} \approx 5.83 \), we find:
\[ |z| \leq \frac{5.83}{2} \approx 2.915 \]

Thus, we conclude:
\[ |z| \leq 3 \]

### Conclusion:
The set of points \( z = \omega - \frac{1}{\omega} \) is contained in or equal to the set of points satisfying \( |z| \leq 3 \).