If `z ne 0` lies on the circle `|z-1| = 1 and omega = 5//z`, then `omega` lies on

Options

A. a circle
B. an ellipse
C. a straight line
D. a parabola

Answer

To solve the problem, we need to analyze the given conditions and derive the relationship between \( z \) and \( \omega \).

### Step-by-Step Solution:

1. **Understanding the Circle Condition**:
The condition \( |z - 1| = 1 \) indicates that \( z \) lies on a circle centered at \( 1 \) in the complex plane with a radius of \( 1 \). This can be expressed as:
\[ z = 1 + e^{i\theta} \quad \text{for } \theta \in [0, 2\pi) \]

2. **Expressing \( \omega \)**:
We are given that \( \omega = \frac{5}{z} \). We will substitute \( z \) from the previous step into this expression.

3. **Substituting \( z \)**:
Substitute \( z = 1 + e^{i\theta} \) into \( \omega \):
\[ \omega = \frac{5}{1 + e^{i\theta}} \]

4. **Finding the Modulus of \( \omega \)**:
To find the modulus of \( \omega \), we can calculate:
\[ |\omega| = \left| \frac{5}{1 + e^{i\theta}} \right| = \frac{5}{|1 + e^{i\theta}|} \]

5. **Calculating \( |1 + e^{i\theta}| \)**:
We can express \( |1 + e^{i\theta}| \) using the formula for the modulus of a complex number:
\[ |1 + e^{i\theta}| = \sqrt{(1 + \cos\theta)^2 + \sin^2\theta} = \sqrt{1 + 2\cos\theta + 1} = \sqrt{2 + 2\cos\theta} = \sqrt{2(1 + \cos\theta)} = \sqrt{4\cos^2\left(\frac{\theta}{2}\right)} = 2|\cos\left(\frac{\theta}{2}\right)| \]

6. **Substituting Back**:
Now substituting this back into the expression for \( |\omega| \):
\[ |\omega| = \frac{5}{2|\cos\left(\frac{\theta}{2}\right)|} \]

7. **Finding the Locus of \( \omega \)**:
Since \( z \) lies on a circle, \( |\omega| \) will vary as \( \theta \) changes. We can express the relationship between \( \omega \) and its components:
\[ 5 - |\omega| = 5 - \frac{5}{2|\cos\left(\frac{\theta}{2}\right)|} \]
This leads to a relationship that describes a straight line in the complex plane.

8. **Conclusion**:
The locus of \( \omega \) as \( z \) varies over the circle \( |z - 1| = 1 \) is a straight line.

### Final Answer:
Thus, \( \omega \) lies on a straight line.

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