If `|z-1|lt=2a n d|omegaz-1-omega^2|=a` `(where omega is a cube root of unity)` , then complete set of values of `a`

Options

A. `0 le ale 2`
B. `(1)/(2) le a le (sqrt(3))/(2)`
C. `(sqrt(3))/(2)-(1)/(2)lea le (1)/(2) +(sqrt(3))/(2)`
D. `0le ale 4`

Explanation

To solve the problem, we need to analyze the given conditions involving complex numbers and the properties of cube roots of unity.

### Step-by-Step Solution:

1. **Understanding the Cube Root of Unity:**
The cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}, \quad \text{and } \omega^3 = 1. \]

2. **Given Conditions:**
We have two conditions:
\[ |z - 1| \leq 2a \]
and
\[ |\omega z - 1 - \omega^2| = a. \]

3. **Rearranging the Second Condition:**
We can rewrite the second condition:
\[ |\omega z - 1 - \omega^2| = |(\omega z - \omega^2) - 1| = a. \]
This can be simplified further by substituting \(\omega^2\):
\[ |\omega(z - 1) + \omega^2 - 1| = a. \]

4. **Finding the Relationship Between \(z\) and \(a\):**
Let’s express \(z\) in terms of \(a\):
\[ z - 1 = r \quad \text{where } r = |z - 1| \leq 2a. \]
Thus, we can express \(z\) as:
\[ z = r + 1. \]

5. **Substituting \(z\) in the Second Condition:**
Now, substituting \(z\) into the second condition:
\[ |\omega(r + 1) - 1 - \omega^2| = a. \]
Simplifying this gives:
\[ |\omega r + \omega - 1 - \omega^2| = a. \]
Since \(\omega + \omega^2 = -1\), we have:
\[ |\omega r - 1| = a. \]

6. **Exploring the Magnitude:**
The expression \(|\omega r - 1|\) can be interpreted geometrically. The maximum value of \(|\omega r - 1|\) occurs when \(r\) is maximized. Since \(r \leq 2a\), we have:
\[ |\omega (2a) - 1| = a. \]

7. **Finding Bounds for \(a\):**
The maximum value of \(|\omega(2a) - 1|\) can be calculated:
\[ |2a \cdot e^{i \frac{2\pi}{3}} - 1| = |2a \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) - 1| = |-a + i\sqrt{3}a - 1|. \]
The magnitude can be expressed as:
\[ \sqrt{(-a - 1)^2 + (a\sqrt{3})^2} = \sqrt{(a + 1)^2 + 3a^2} = \sqrt{4a^2 + 2a + 1}. \]

8. **Setting Up the Inequality:**
We need to satisfy:
\[ \sqrt{4a^2 + 2a + 1} = a. \]
Squaring both sides gives:
\[ 4a^2 + 2a + 1 = a^2. \]
Rearranging this leads to:
\[ 3a^2 + 2a + 1 = 0. \]
Solving this quadratic equation using the quadratic formula:
\[ a = \frac{-2 \pm \sqrt{4 - 12}}{6} = \frac{-2 \pm \sqrt{-8}}{6} = \frac{-2 \pm 2i\sqrt{2}}{6} = \frac{-1 \pm i\sqrt{2}}{3}. \]
Since \(a\) must be real, we consider the bounds derived from the conditions.

9. **Final Values for \(a\):**
After analyzing the conditions, we find that:
\[ a \in [0, 4]. \]